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Exercise 7.1 Page: 234

Find an anti derivative (or integral) of the following functions by the method of inspection.

Question 1 sin 2x 

Solution :

The anti derivative of sin 2x is a function of x whose derivative is sin 2x.

It is known that,

NCERT Solutions class 12 Maths Integrals

herefore, the anti derivative of sin 2x is -1/2 cos 2x.

Question 2. cos 3x

Solution :

The anti derivative of cos 3x is a function of x whose derivative is cos 3x.

It is known that,

NCERT Solutions class 12 Maths Integrals

Therefore, the anti derivative of cos 3x is 1/3 sin 3x.

Question 3. e2x

Solution :

The anti derivative of e2x is the function of x whose derivative is e2x.

It is known that,

chapter 7-Integrals Exercise 7.1

Therefore, the anti derivative of  e2x is 1/2 e2x.

Question 4. (ax + b)2

Solution :

The anti derivative of (ax + b)2 is the function of x whose derivative is (ax + b)2.

It is known that,

NCERT Solutions class 12 Maths Integrals
Question 5. sin 2x – 4 e3x

Solution :

The anti derivative of  sin 2x – 4 e3x is the function of x whose derivative is sin 2x – 4 e3x

It is known that,

chapter 7-Integrals Exercise 7.1

Evaluate the following integrals in Exercises 6 to 11.
Question 6.  ∫(4e3x+ 1) dx

Solution
NCERT Solutions class 12 Maths Integrals
Question 7. NCERT Solutions class 12 Maths Integrals

Solution :

NCERT Solutions class 12 Maths Integrals

Question 8. 

Solution

Question 9. 

Solution :

Question 10. 

Solution :


Question 11. 

Solution :

Evaluate the following integrals in Exercises 12 to 16.

Question 12. 

Solution :

Question 13. 

Solution :

Question 14. ∫(1 – x)√x dx

Solution : ∫(1 – x)√x dx

Question 15. 

Solution :


Question 16. NCERT Solutions class 12 Maths Integrals

Solution
NCERT Solutions class 12 Maths Integrals

Evaluate the following integrals in Exercises 17 to 20.
Question 17. 

Solution


Question 18.

Solution


Question 19. 

Solution :


Question 20. 

Solution

Question 21. Choose the correct answer:
The anti derivative of equals.

Solution

Therefore, option (C) is correct.

Question 22. Choose the correct answer:

If   such that f(2) = 0 Then f is:

Solution :

It is given that,

Therefore, option (A) is correct.

Exercise 7.2 Page: 240

Integrate the functions in Exercise 1 to 8.

Question 1.

Solution :

Let 1 + x2 = t

∴2x dx = dt
NCERT Solutions class 12 Maths Integrals

Question 2. chapter 7-Integrals Exercise 7.2

Solution :

Let log |x| = t

∴1/xdx=dt
chapter 7-Integrals Exercise 7.2
NCERT Solutions class 12 Maths Integrals

Question 3. 

Solution :

Question 4. sin x ⋅ sin (cos x)

Solution :

sin x ⋅ sin (cos x)

Let cos x = t

∴−sin xdx=dt
NCERT Solutions class 12 Maths Integrals

Question 5. sin(ax + b) cos(ax + b)

Solution :

Questionc 6. √ax + b

Solution :

Let ax + b = t

⇒ adx = dt

Question 7. x√x + 2

Solution :

Let  (x + 2) = t

∴ dx = dt

Question8. x√1+2x2
Solution :

Let 1 + 2x2 = t

∴ 4xdx = dt

Integrate the functions in Exercise 9 to 17.

Question 9.

Solution :

Question 10.

Solution :
chapter 7-Integrals Exercise 7.2

Question 11.

Solution :

Question 12.

Solution :

Let x3 – 1 = t

∴ 3x2 dx = dt

Question 13.

Solution :

Let 2 + 3x3 = t

∴ 9x2 dx = dt

Question14. 
Solution :

Let log x = t

∴ 1/x dx = dt


Question 15. x/9 – 4x2

Solution :

Let 9 – 4x2 =  t

∴ −8x dx = dt

Question 16. 

Solution :

Let 2x + 3 = t

∴ 2dx = dt

Question17. 
Solution :

Let x2 = t

∴ 2xdx = dt

Integrate the functions in Exercise 18 to 26.

Question 18. 

Solution :

Question 19.

Solution :

Dividing numerator and denominator by ex, we obtain

Question20. 
Solution :

Question 21. tan2 (2x – 3)

Solution :

Question 22. sec2 (7 – 4x)

Solution :

Let 7 − 4x = t

∴ −4dx = dt

Question 23. 

Solution :

chapter 7-Integrals Exercise 7.2
Question 24. 

Solution :

Question 25. 

Solution :

Question 26. 

Solution :

Let √x = t

Integrate the functions in Exercise 27 to 37.

Question 27.

Solution :

Let sin 2x = t

Question 28.

Solution :

Let 1 + sin x = t

∴ cos x dx = dt

Question 29. cot x log sin x

Solution :

Let log sin x = t


Question 30. sin x/1 + cos x

Solution :

Let 1 + cos x = t

∴ −sin x dx = dt

Question 31.  sin x/(1 + cos x)2

Solution :

Let 1 + cos x = t

∴ −sin x dx = dt

Question32.1/1+cot x
Solution :

Question 33. 1/1 – tan x

Solution :

Question 34. 

Solution :

Question 35. 

Solution :

Let 1 + log x = t

∴1/xdx=dt

Question 36. 

Solution :

Question 37.

Solution :

Let x4 = t

∴ 4x3 dx = dt

chapter 7-Integrals Exercise 7.2

Choose the correct answer in Exercise 38 and 39.

Question 38.  equals

(A)10x –x10 +C
(B) 10x +x10 +C
(C) (10x –x10)-1 +C
(D) log(10x + x10) + C

Solution :

Therefore, option (D) is correct.
Question 39.equals

(A) tanx+cotx+C
(B)tanx–cotx+C
(C)tanxcotx+C
(D) tan x – cot 2x + C

Solution :

Therefore, option (B) is correct.

Exercise 7.3 Page: 243

Find the integrals of the following functions in Exercises 1 to 9.

Question 1. sin2(2x + 5)

Solution :

chapter 7-Integrals Exercise 7.3

Question 2. sin 3x cos4x

Solution :

 

 

 

 

 

 

 

Question 3. cos 2x cos 4x cos 6x

Solution :

Question 4. sin3 (2x + 1)

Question 5. sin3 x cos3 x

Question 6. sin x sin 2x sin 3x

 

 

 

Question 7. sin 4x sin 8x

Solution: Itisknownthat,
sin A . sin B = 12cosA-B-cosA+B

∴∫sin4x sin8x dx=∫12cos4x-8x-cos4x+8xdx

=12∫cos-4x-cos12xdx

=12∫cos4x-cos12xdx

=12sin4x4-sin12x12+C

Question 8. 

 

Question 9. 

Solution :

Find the integrals of the following functions in Exercises 10 to 18.

Question 10. sin4 x

Solution :

Question 11. cos4 2x

Solution :

Question 12. 

Solution :

Question13. 

 

 

 

 

 

 

 

 

 

 

 

Question 14. 

Solution :

                                                                                                          

 

 

 

 

 

Question 15. tan3 2x sec2x

Solution :

Question 16. tan4x

Solution :Question 17. 

Solution :

Question 18. 

Solution :

                                             

.
 

Question 19. 

Solution :

 Question20. 

 

 

 

 

Solution :20

 

 

Question 21. sin−1 (cos x)

Solution :

NCERT Solutions class 12 Maths IntegralsNCERT Solutions class 12 Maths Integrals

Question 22. 

Solution : Choose the correct answer in Exercise 23 and 24.

Question 23.is equal to:

A. tan x + cot x + C

B. tan x + cosec x + C

C. − tan x + cot x + C

D. tan x + sec x + C

Solution :

Therefore, option (A) is correct.

Question 24. is equal to:

A. − cot (exx) + C

B. tan (xex) + C

C. tan (ex) + C

D. cot (ex) + C

Solution :
Let I = 

Let exx = t

Therefore, option (B) is correct.chapter 7-Integrals Exercise 7.3

Exercise 7.4 Page: 251

Integrate the following functions in Exercises 1 to 9.

Question 1. chapter 7-Integrals Exercise 7.4

Solution :

Let x3 = t

∴ 3x2 dx = dt

chapter 7-Integrals Exercise 7.4

Question 2. 

Solution :

Let 2x = t

∴ 2dx = dt

Question 3. 

Solution :

Let 2 − x = t

⇒ −dx = dt

 

 

Question 4. 

Solution :

Let 5x = t

∴ 5dx = dt

Question 5. 

Solution :

  

 

 

Question 6. 

Solution :

Let x3 = t

∴ 3x2 dx = dt

Question 7. 

Solution :

Question 8. 

Solution :

Let x3 = t

∴ 3x2 dx = dt

Question 9. 

Solution :

Let tan x = t

∴ sec2x dx = dt

Integrate the following functions in Exercises 10 to 18.

Question 10. 

Solution :

 

Question 11. 

Solution :

 

 

Question 12 .

Solution :

 

 

 

 

 

 

 

 

 

 

 

Question 13. 

 

 

 

Solution :

Question 14. 

 

 

 

 

 

 

 

 

 

Solution :

Question 15. 

 

 

 

Solution :

Question 16. 

Solution :

Question 17.

Solution :

  

 

 Question18. 

Solution :

Integrate the following functions in Exercises 19 to 23.
Question 19. 

Solution :

Question 20. 

Solution :

Question 21. 

Solution :

Using equations (2) and (3) in (1), we obtain

Question 22. 

Solution :

Question 23. 

Solution :

Choose the correct answer in Exercise 24 and 25.

Question 24. equals

A. x tan−1 (x + 1) + C

B. tan− 1 (x + 1) + C

C. (x + 1) tan−1 x + C

D. tan−1 x + C

Solution :

Therefore, option (B) is correct.

Question 25. equals

NCERT Solutions class 12 Maths Integrals

Solution :

Therefore,option(B)iscorrect.

Exercise 7.5 Page: 258

Question 1. 

Solution :

 chapter 7-Integrals Exercise 7.5

Question 2. 

Solution :

 

Question 3. chapter 7-Integrals Exercise 7.5

Solution :

 chapter 7-Integrals Exercise 7.5

uestion 4. chapter 7-Integrals Exercise 7.5

Solution :

chapter 7-Integrals Exercise 7.5

Question 5.

Solution :

Question 6. chapter 7-Integrals Exercise 7.5

Solution :

It can be seen that the given integrand is not a proper fraction.

Therefore, on dividing (1 − x2) by x(1 − 2x), we obtain

 

Question7. 

Solution :

chapter 7-Integrals Exercise 7.5

Question8. 

Solution :

  

Question9.  

Solution :

 

Question10. 

Solution :

Question11. 

Solution :

Question12. 

Solution :

It can be seen that the given integrand is not a proper fraction.

Therefore, on dividing (x3 + x + 1) by x2 − 1, we obtain

Integrate the following function in Exercises 13 to 17.

Question13. 

Solution :

 

Question 14. 

Solution :

 

Question15. 

Solution :

Question16.   [Hint: multiply numerator and denominator by xn − 1 and put xn = t]

Solution :

Multiplying numerator and denominator by xn − 1, we obtain

Question17.  [Hint: Put sin x = t]

Solution :

Integrate the following function in Exercises 18 to 21.

Question18. 

Solution :

Equating the coefficients of x3, x2, x, and constant term, we obtain

A + C = 0

B + D = 4

4A + 3C = 0

4B + 3D = 10

On solving these equations, we obtain

A = 0, B = −2, C = 0, and D = 6

 

Question19. 

Solution :

Question20. 

Solution :

Multiplying numerator and denominator by x3, we obtain

 

Question21.  [Hint: Put ex = t]

Solution :

Choose the correct answer in each of the Exercise 22 and 23.

Question22. equals:

Solution :

Therefore, option (B) is correct.

Question23. equals:

Solution :

Therefore, option (A) is correct.

Exercise 7.5 Page: 263

Question 1. x sin x

Solution : Let I =  ∫ x sin x dx

Taking x as first function and sin x as second function and integrating by parts, we obtain

NCERT Solutions class 12 Maths Integrals

Question2. x sin3x
Solution :
Let I = ∫ x sin 3x dx

Taking x as first function and sin 3x as second function and integrating by parts, we obtain

Question3. x2 ex
Solution :

Let  I = ∫ x2 ex dx

Taking x2 as first function and ex as second function and integrating by parts, we obtain

Again integrating by parts, we obtain

Question 4. x logx

Solution : Let  I = ∫ x logx dx

Taking log x as first function and x as second function and integrating by parts, we obtain

NCERT Solutions class 12 Maths Integrals


Question 5. x log 2x

Solution : Let  I = ∫ x log 2x dx

Taking log 2x as first function and x as second function and integrating by parts, we obtain


Question 6. x2 log x

Solution : Let  I = ∫ x2 log x dx

Taking log x as first function and x2 as second function and integrating by parts, we obtain

Question7. xsin-1 x
Solution :
Let I = ∫ x sin-1 x

Taking sin-1 x as first function and x as second function and integrating by parts, we obtain

Question8. xtan-1 x
Solution :
Let I = ∫ x tan-1 x

Taking tan-1 x as first function and x as second function and integrating by parts, we obtain

NCERT Solutions class 12 Maths Integrals

Integrate the functions in Exercises 9 to 15.
Question 9. x cos-1 x
Solution :
Let I = ∫ x cos-1 x

Taking cos−1 x as first function and x as second function and integrating by parts, we obtain

chapter 7-Integrals Exercise 7.6
Question10. 
Solution :
Let I = ∫ .1 dx

Taking  as first function and 1 as second function and integrating by parts, we obtain

chapter 7-Integrals Exercise 7.6
Question11. 
Solution :
Let  

Taking cos−1 x as first function and  as second function and integrating by parts, we obtain


Question12. xsec2 x
Solution : Let I = ∫ x sec2 x dx

Taking x as first function and sec2x as second function and integrating by parts, we obtain


Question 13. tan-1 x

Solution :
Let I = ∫ tan-1 x dx

Taking tan-1 x as first function and 1 as second function and integrating by parts, we obtain

chapter 7-Integrals Exercise 7.6
Question14. x(logx)2
Solution :
Let I = ∫ x (log x)2 dx

Taking (log x)2 as first function and x as second function and integrating by parts, we obtain

Question 15. (x2 + 1) log x

Solution :

Integrate the functions in Exercises 16 to 22.
Question16. 
Solution :

Question17.
Solution :

Question18. 
Solution :

Question19. 
Solution :

Question20. 
Solution :

Question21.e2x sinx
Solution :
Let I = ∫ e2x sin x

Integrating by parts, we obtain

chapter 7-Integrals Exercise 7.6
Question22.
Solution :


Choose the correct answer in Exercise 23 and 24.
Question23. equalsto
chapter 7-Integrals Exercise 7.6
Solution :
LetI= 

Therefore, option (A) is correct.
Question24. ∫equals:
(A) excosx+C
(B) ex secx+C
(C) ex sinx+C
(D) ex tanx+C
Solution :

Therefore, option (B) is correct.

Exercise 7.6 Page: 263

Question 1. x sin x

Solution : Let I =  ∫ x sin x dx

Taking x as first function and sin x as second function and integrating by parts, we obtain

NCERT Solutions class 12 Maths Integrals

Question2. x sin3x
Solution :
Let I = ∫ x sin 3x dx

Taking x as first function and sin 3x as second function and integrating by parts, we obtain

Question3. x2 ex
Solution :

Let  I = ∫ x2 ex dx

Taking x2 as first function and ex as second function and integrating by parts, we obtain

Again integrating by parts, we obtain

Question 4. x logx

Solution : Let  I = ∫ x logx dx

Taking log x as first function and x as second function and integrating by parts, we obtain

NCERT Solutions class 12 Maths Integrals

Question 5. x log 2x

Solution : Let  I = ∫ x log 2x dx

Taking log 2x as first function and x as second function and integrating by parts, we obtain


Question 6. x2 log x

Solution : Let  I = ∫ x2 log x dx

Taking log x as first function and x2 as second function and integrating by parts, we obtain

Question7. xsin-1 x
Solution :
Let I = ∫ x sin-1 x

Taking sin-1 x as first function and x as second function and integrating by parts, we obtain

Question8. xtan-1 x
Solution :
Let I = ∫ x tan-1 x

Taking tan-1 x as first function and x as second function and integrating by parts, we obtain

NCERT Solutions class 12 Maths Integrals

Integrate the functions in Exercises 9 to 15.
Question 9. x cos-1 x
Solution :
Let I = ∫ x cos-1 x

Taking cos−1 x as first function and x as second function and integrating by parts, we obtain

chapter 7-Integrals Exercise 7.6
Question10. 
Solution :
Let I = ∫ .1 dx

Taking  as first function and 1 as second function and integrating by parts, we obtain

chapter 7-Integrals Exercise 7.6
Question11. 
Solution :
Let  

Taking cos−1 x as first function and  as second function and integrating by parts, we obtain


Question12. xsec2 x
Solution : Let I = ∫ x sec2 x dx

Taking x as first function and sec2x as second function and integrating by parts, we obtain


Question 13. tan-1 x

Solution :
Let I = ∫ tan-1 x dx

Taking tan-1 x as first function and 1 as second function and integrating by parts, we obtain

chapter 7-Integrals Exercise 7.6
Question14. x(logx)2
Solution :
Let I = ∫ x (log x)2 dx

Taking (log x)2 as first function and x as second function and integrating by parts, we obtain

Question 15. (x2 + 1) log x

Solution :

Integrate the functions in Exercises 16 to 22.
Question16. 
Solution :

Question17.
Solution :

Question18. 
Solution :

Question19. 
Solution :

Question20. 
Solution :

Question21.e2x sinx
Solution :
Let I = ∫ e2x sin x

Integrating by parts, we obtain

chapter 7-Integrals Exercise 7.6
Question22.
Solution :


Choose the correct answer in Exercise 23 and 24.
Question23. equalsto
chapter 7-Integrals Exercise 7.6
Solution :
LetI= 

Therefore,option,(A)iscorrect.
Question24. ∫equals:
(A) excosx+C
(B) ex secx+C
(C) ex sinx+C
(D) ex tanx+C
Solution :

Therefore, option (B) is correct.

Exercise 7.7 Page: 266

Question 1.  chapter 7-Integrals Exercise 7.7

Solution :
chapter 7-Integrals Exercise 7.7

Question 2.    

Solution :

Question 3. 

Solution :

Question 4.    

Solution :

Question 5. 

Solution :

Question 6. 

Solution :

Question 7. 

Solution :
chapter 7-Integrals Exercise 7.7

Question 8. NCERT Solutions class 12 Maths Integrals

Solution :
NCERT Solutions class 12 Maths Integrals

Question 9. NCERT Solutions class 12 Maths Integrals

Solution :

Choose the correct answer in Exercise 10 to 11.

Question10.  is equal to:

Solution :

Therefore, option (A) is correct.

Question 11. is equal to:

NCERT Solutions class 12 Maths Integrals

NCERT Solutions class 12 Maths Integrals

Therefore, option (D) is correct.

Exercise 7.8 Page: 270

Question 1.  chapter 7-Integrals Exercise 7.8    

Solution :
It is know that

Question 2. 

Solution :
Let I = 

It is know that

Question 3. 

Solution :
We know that

Question 4. 

Solution :


We know that

chapter 7-Integrals Exercise 7.8

 

From equations (2) and (3), we obtain

Question 5. 

Solution: LetI= 
We know that

Question 6. 

Solution :
We know that

Exercise 7.9 Page: 273

Evaluate the definite integrals in Exercises 1 to 11.
Question1. 
Solution :

By second fundamental theorem of calculus, we obtain


Question2. chapter 7-Integrals Exercise 7.8
Solution :
chapter 7-Integrals Exercise 7.8

By second fundamental theorem of calculus, we obtain

I = F(3) – F(2)
= log|3| – log|2| = log 3/2

Question3. 
Solution :

By second fundamental theorem of calculus, we obtain

NCERT Solutions class 12 Maths Integrals
Question4. 
Solution :

By second fundamental theorem of calculus, we obtain


Question5. 
Solution :

By second fundamental theorem of calculus, we obtain

Question6. 
Solution :

By second fundamental theorem of calculus, we obtain


Question7. NCERT Solutions class 12 Maths Integrals
Solution :

By second fundamental theorem of calculus, we obtain

Question8. 
Solution :

By second fundamental theorem of calculus, we obtain
NCERT Solutions class 12 Maths Integrals
Question9. 
Solution :

By second fundamental theorem of calculus, we obtain

Question10. 
Solution :

By second fundamental theorem of calculus, we obtain

Question11. NCERT Solutions class 12 Maths Integrals
Solution :
NCERT Solutions class 12 Maths Integrals

By second fundamental theorem of calculus, we obtain

Evaluate the definite integrals in Exercises 12 to 20.
Question12. NCERT Solutions class 12 Maths Integrals
Solution :
NCERT Solutions class 12 Maths Integrals

By second fundamental theorem of calculus, we obtain


Question13. 
Solution :

By second fundamental theorem of calculus, we obtain

Question14. 
Solution :

Question15. 
Solution :

By second fundamental theorem of calculus, we obtain


Question16. 
Solution :

Equating the coefficients of x and constant term, we obtain


Question17. 
Solution :

By second fundamental theorem of calculus, we obtain

Question18. 
Solution :

By second fundamental theorem of calculus, we obtain

I = F(π) – F(0)

= sin π – sin 0

= 0
Question19. 
Solution :

By second fundamental theorem of calculus, we obtain

Question20. 
Solution :

By second fundamental theorem of calculus, we obtain

Choose the correct answer in Exercises 21 and 22.
Question21. equals:
(A)π/3
(B)2π/3
(C)π/6
(D)π/12
Solution :

Therefore, option (D) is correct.
Question22. equals:
(A)π/6
(B)π/12
(C)π/24
(D)π/4
Solution :

By second fundamental theorem of calculus, we obtain

 

Therefore, option (C) is correct.

Exercise 7.10 Page: 280

Question:1 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int_0^\frac{\pi}{2}\cos^2 x dx

Answer:

We have I\ =\ \int_0^\frac{\pi}{2}\cos^2 x dx ............................................................. (i)

By using

\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get :-

I\ =\ \int_0^\frac{\pi}{2}\cos^2 x dx\ =\ \int_0^\frac{\pi}{2}\cos^2\ (\frac{\pi}{2}- x) dx

or

I\ =\ \int_0^\frac{\pi}{2}\sin^2 x dx ................................................................ (ii)

Adding both (i) and (ii), we get :-

\int_0^\frac{\pi}{2}\cos^2 x dx +\ \int_0^\frac{\pi}{2}\sin^2 x dx\ =\ 2I

or \int_0^\frac{\pi}{2}\ (cos^2 x\ +\ sin^2 x) dx\ =\ 2I

or \int_0^\frac{\pi}{2}1. dx\ =\ 2I

or 2I\ =\ \left [ x \right ] ^\frac{\Pi }{2}_0\ =\ \frac{\Pi }{2}

or I\ =\ \frac{\Pi }{4}

 

Question:2 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

. \int_0^\frac{\pi}{2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+ \sqrt{\cos x}}dx

Answer:

We have I\ =\ \int_0^\frac{\pi}{2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+ \sqrt{\cos x}}dx .......................................................................... (i)

By using ,

\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get,

I\ =\ \int_0^\frac{\pi}{2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+ \sqrt{\cos x}}dx\ =\ \int_0^\frac{\pi}{2}\frac{\sqrt{\sin (\frac{\pi}{2}-x)}}{\sqrt{\sin (\frac{\pi}{2}-x)}+ \sqrt{\cos (\frac{\pi}{2}-x)}}dx

 

or I\ =\ \int_0^\frac{\pi}{2}\frac{\sqrt{\cos x}}{\sqrt{\cos x}+ \sqrt{\sin x}}dx .......................................................(ii)

Adding (i) and (ii), we get,

2I\ =\ \int_0^\frac{\pi}{2}\frac{\sqrt{\sin x}\ +\ \sqrt{\cos x}}{\sqrt{\sin x}+ \sqrt{\cos x}}dx

or 2I\ =\ \int_0^\frac{\pi}{2}1.dx

 

or 2I\ =\ \left [ x \right ]^\frac{\Pi }{2}_0\ =\ \frac{\Pi }{2}

Thus I\ =\ \frac{\Pi }{4}

 

 

Question:3 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int^{\frac{\pi}{2}}_0\frac{\sin^{\frac{3}{2}}xdx}{\sin^\frac{3}{2}x + \cos^{\frac{3}{2}}x}

Answer:

We have I\ =\ \int^{\frac{\pi}{2}}_0\frac{\sin^{\frac{3}{2}}xdx}{\sin^\frac{3}{2}x + \cos^{\frac{3}{2}}x} ..................................................................(i)

By using :

\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get,

I\ =\ \int^{\frac{\pi}{2}}_0\frac{\sin^{\frac{3}{2}}(\frac{\pi}{2}-x)dx}{\sin^\frac{3}{2}(\frac{\pi}{2}-x) + \cos^{\frac{3}{2}}(\frac{\pi}{2}-x)}

 

or I\ =\ \int^{\frac{\pi}{2}}_0\frac{\cos^{\frac{3}{2}}xdx}{\sin^\frac{3}{2}x + \cos^{\frac{3}{2}}x} . ............................................................(ii)

Adding (i) and (ii), we get :

2I\ =\ \int^{\frac{\pi}{2}}_0\frac{\ (sin^{\frac{3}{2}}x+cos^{\frac{3}{2}}x)dx}{\sin^\frac{3}{2}x + \cos^{\frac{3}{2}}x}

or 2I\ = \int_{0}^{{\frac{\pi}{2}}}1.dx

or 2I\ = \left [ x \right ]^{\frac{\pi}{2}}_ 0\ =\ {\frac{\pi}{2}}

Thus I\ =\ {\frac{\pi}{4}}

 

 

Question:4 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

. \int_0^\frac{\pi}{2} \frac{\cos^5 xdx}{\sin^5x + \cos^5x}

Answer:

We have I\ =\ \int_0^\frac{\pi}{2} \frac{\cos^5 xdx}{\sin^5x + \cos^5x} ..................................................................(i)

By using :

\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get,

I\ =\ \int_0^\frac{\pi}{2} \frac{\cos^5 (\frac{\pi}{2}-x)dx}{\sin^5(\frac{\pi}{2}-x) + \cos^5(\frac{\pi}{2}-x)}

or I\ =\ \int_0^\frac{\pi}{2} \frac{\sin^5 xdx}{\sin^5x + \cos^5x} . ............................................................(ii)

Adding (i) and (ii), we get :

2I\ =\ \int_0^\frac{\pi}{2} \frac{\ ( sin^5x \ +\ cos^5 x)dx}{\sin^5x + \cos^5x}

or 2I\ = \int_{0}^{{\frac{\pi}{2}}}1.dx

or 2I\ = \left [ x \right ]^{\frac{\pi}{2}}_ 0\ =\ {\frac{\pi}{2}}

Thus I\ =\ {\frac{\pi}{4}}

 

Question:5 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int_{-5}^5|x+2|dx

Answer:

We have, I\ =\ \int_{-5}^5|x+2|dx

For opening the modulas we need to define the bracket :

If (x + 2) < 0 then x belongs to (-5, -2). And if (x + 2) > 0 then x belongs to (-2, 5).

So the integral becomes :-

I\ =\ \int_{-5}^{-2} -(x+2)dx\ +\ \int_{-2}^{5} (x+2)dx

or I\ =\ -\left [ \frac{x^2}{2}\ +\ 2x \right ]^{-2} _{-5}\ +\ \left [ \frac{x^2}{2}\ +\ 2x \right ]^{5} _{-2}

This gives I\ =\ 29

 

Question:6 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int_2^8|x-5|dx

Answer:

We have, I\ =\ \int_{2}^8|x-5|dx

For opening the modulas we need to define the bracket :

If (x - 5) < 0 then x belongs to (2, 5). And if (x - 5) > 0 then x belongs to (5, 8).

So the integral becomes:-

I\ =\ \int_{2}^{5} -(x-5)dx\ +\ \int_{5}^{8} (x-5)dx

or I\ =\ -\left [ \frac{x^2}{2}\ -\ 5x \right ]^{5} _{2}\ +\ \left [ \frac{x^2}{2}\ -\ 5x \right ]^{8} _{5}

This gives I\ =\ 9

 

Question:7 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int^1_0x(1-x)^ndx

Answer:

We have I\ =\ \int^1_0x(1-x)^ndx

U sing the property : -

\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get : -

I\ =\ \int^1_0x(1-x)^ndx\ =\ \int^1_0(1-x)(1-(1-x))^ndx

or I\ =\ \int^1_0(1-x)x^n\ dx

or I\ =\ \int^1_0(x^n\ -\ x^{n+1}) \ dx

or =\ \left [ \frac{x^{n+1}}{n+1}\ -\ \frac{x^{n+2}}{n+2} \right ]^1_0

or =\ \left [ \frac{1}{n+1}\ -\ \frac{1}{n+2} \right ]

or I\ =\ \frac{1}{(n+1)(n+2)}

 

Question:8 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int_0^\frac{\pi}{4}\log(1+\tan x)dx

Answer:

We have I\ =\ \int_0^\frac{\pi}{4}\log(1+\tan x)dx

By using the identity

\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get,

I\ =\ \int_0^\frac{\pi}{4}\log(1+\tan x)dx\ =\ \int_0^\frac{\pi}{4}\log(1+\tan (\frac{\pi}{4}-x))dx

or I\ =\ \int_0^\frac{\pi}{4}\log(1+\frac{1-\tan x}{1+\tan x})dx

or I\ =\ \int_0^\frac{\pi}{4}\log(\frac{2}{1+\tan x})dx

or I\ =\ \int_0^\frac{\pi}{4}\log{2}dx\ -\ \int_0^\frac{\pi}{4}\log(1+ \tan x)dx

or I\ =\ \int_0^\frac{\pi}{4}\log{2}dx\ -\ I

or 2I\ =\ \left [ x\log2 \right ]^{\frac{\Pi }{4}}_0

or I\ =\ \frac{\Pi }{8}\log2

 

 

Question:9 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int_0^2x\sqrt{2-x}dx

Answer:

We have I\ =\ \int_0^2x\sqrt{2-x}dx

By using the identity

\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get :

I\ =\ \int_0^2x\sqrt{2-x}dx\ =\ \int_0^2(2-x)\sqrt{2-(2-x)}dx

or I\ =\ \int_0^2(2-x)\sqrt{x}dx

or I\ =\ \int_0^2(2\sqrt{x}\ -\ x^\frac{3}{2} dx

or =\ \left [ \frac{4}{3}x^\frac{3}{2}\ -\ \frac{2}{5}x^\frac{5}{2} \right ]^2_0

or =\ \frac{4}{3}(2)^\frac{3}{2}\ -\ \frac{2}{5}(2)^\frac{5}{2}

or I\ =\ \frac{16\sqrt{2}}{15}

 

Question:10 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int_0^\frac{\pi}{2} (2\log\sin x- \log\sin 2x)dx

Answer:

We have I\ =\ \int_0^\frac{\pi}{2} (2\log\sin x- \log\sin 2x)dx

or I\ =\ \int_0^\frac{\pi}{2} (2\log\sin x- \log(2\sin x\cos x))dx

or I\ =\ \int_0^\frac{\pi}{2} (\log\sin x- \log\cos x\ -\ \log2)dx ..............................................................(i)

By using the identity :

\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get :

I\ =\ \int_0^\frac{\pi}{2} (\log\sin (\frac{\pi}{2}-x)- \log\cos (\frac{\pi}{2}-x)\ -\ \log2)dx

or I\ =\ \int_0^\frac{\pi}{2} (\log\cos x- \log\sin x\ -\ \log2)dx ....................................................................(ii)

 

Adding (i) and (ii) we get :-

2I\ =\ \int_0^\frac{\pi}{2} (- \log 2 -\ \log 2)dx

or I\ =\ -\log 2\left [ \frac{\Pi }{2} \right ]

or I\ =\ \frac{\Pi }{2}\log\frac{1}{2}

 

Question:11 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int_\frac{-\pi}{2}^\frac{\pi}{2}\sin^2 x dx

Answer:

We have I\ =\ \int_\frac{-\pi}{2}^\frac{\pi}{2}\sin^2 x dx

We know that sin 2 x is an even function. i.e., sin 2 (-x) = (-sinx) 2 = sin 2 x.

Also,

I\ =\ \int_{-a}^af(x) dx\ =\ 2\int_{0}^af(x) dx

So,

I\ =\ 2\int_0^\frac{\pi}{2}\sin^2 x dx\ =\ 2\int_0^\frac{\pi}{2}\frac{(1-\cos2x)}{2} dx

or =\ \left [ x\ -\ \frac{\sin2x}{2} \right ]^{\frac{\Pi }{2}}_0

or I\ =\ \frac{\Pi }{2}

 

Question:12 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int_0^\pi\frac{xdx}{1+\sin x}

Answer:

We have I\ =\ \int_0^\pi\frac{xdx}{1+\sin x} ..........................................................................(i)

By using the identity :-

\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get,

I\ =\ \int_0^\pi\frac{xdx}{1+\sin x}\ =\ \int_0^\pi\frac{(\Pi -x)dx}{1+\sin (\Pi -x)}

or I\ =\ \int_0^\pi\frac{(\Pi -x)dx}{1+\sin x} ............................................................................(ii)

 

Adding both (i) and (ii) we get,

2I\ =\ \int_0^\pi\frac{\Pi}{1+\sin x} dx

or 2I\ =\ \Pi \int_0^\pi\frac{1-\sin x}{(1+\sin x)(1-\sin x)} dx\ =\ \Pi \int_0^\pi\frac{1-\sin x}{\cos^2 x} dx

or 2I\ =\ \Pi \int_0^\pi (\sec^2\ -\ \tan x \sec x) x dx

or I\ =\ \Pi

 

 

Question:13 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int_\frac{-\pi}{2}^\frac{\pi}{2}\sin^7xdx

Answer:

We have I\ =\ \int_\frac{-\pi}{2}^\frac{\pi}{2}\sin^7xdx

We know that \sin^7x is an odd function.

So the following property holds here:-

\int_{-a}^{a}f(x)dx\ =\ 0

Hence

I\ =\ \int_\frac{-\pi}{2}^\frac{\pi}{2}\sin^7xdx\ =\ 0

 

 

Question:14 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int_0^{2\pi}\cos^5xdx

Answer:

We have I\ =\ \int_0^{2\pi}\cos^5xdx

I t is known that :-

\int_0^{2a}f(x)dx\ =\ 2\int_0^{a}f(x)dx If f (2a - x) = f(x)

=\ 0 If f (2a - x) = - f(x)

Now, using the above property

\cos^5(\Pi - x)\ =\ - \cos^5x

Therefore, I\ =\ 0

 

 

Question:15 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int^\frac{\pi}{2} _0\frac{\sin x - \cos x }{1+\sin x\cos x}dx

Answer:

We have I\ =\ \int^\frac{\pi}{2} _0\frac{\sin x - \cos x }{1+\sin x\cos x}dx ................................................................(i)

 

By using the property :-

\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get ,

I\ =\ \int^\frac{\pi}{2} _0\frac{\sin (\frac{\pi}{2}-x) - \cos (\frac{\pi}{2}-x) }{1+\sin (\frac{\pi}{2}-x)\cos (\frac{\pi}{2}-x)}dx

or I\ =\ \int^\frac{\pi}{2} _0\frac{\cos x - \sin x }{1+\sin x\cos x}dx ......................................................................(ii)

 

Adding both (i) and (ii), we get

 

2I\ =\ \int^\frac{\pi}{2} _0\frac{0 }{1+\sin x\cos x}dx

Thus I = 0

 

Question:16 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int_0^\pi\log(1 +\cos x)dx

 

Answer:

We have I\ =\ \int_0^\pi\log(1 +\tan x)dx .....................................................................................(i)

By using the property:-

\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get,

 

or

I\ =\ \int_0^\pi\log(1 +\cos (\Pi -x))dx

I\ =\ \int_0^\pi\log(1 -\cos x)dx ....................................................................(ii)

 

Adding both (i) and (ii) we get,

2I\ =\ \int_0^\pi\log(1 +\cos x)dx\ +\ \int_0^\pi\log(1 -\cos x)dx

or 2I\ =\ \int_0^\pi\log(1 -\cos^2 x)dx\ =\ \int_0^\pi\log \sin^2 xdx

or 2I\ =\ 2\int_0^\pi\log \sin xdx

or I\ =\ \int_0^\pi\log \sin xdx ........................................................................(iii)

or I\ =\ 2\int_0^ \frac{\pi}{2} \log \sin xdx ........................................................................(iv)

or I\ =\ 2\int_0^ \frac{\pi}{2} \log \cos xdx .....................................................................(v)

Adding (iv) and (v) we get,

I\ =\ -\pi \log2

 

 

Question:17 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int_0^a \frac{\sqrt x}{\sqrt x + \sqrt{a-x}}dx

Answer:

We have I\ =\ \int_0^a \frac{\sqrt x}{\sqrt x + \sqrt{a-x}}dx ................................................................................(i)

By using, we get

 

\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get,

I\ =\ \int_0^a \frac{\sqrt x}{\sqrt x + \sqrt{a-x}}dx\ =\ \int_0^a \frac{\sqrt {(a-x)}}{\sqrt {(a-x)} + \sqrt{x}}dx .................................................................(ii)

 

Adding (i) and (ii) we get :

2I\ =\ \int_0^a \frac{\sqrt x\ +\ \sqrt{a-x}}{\sqrt x + \sqrt{a-x}}dx

or 2I\ =\ \left [ x \right ]^a_0 = a

or I\ =\ \frac{a}{2}

 

Question:18 By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

\int_0^4 |x-1|dx

Answer:

We have, I\ =\ \int_{0}^4|x-1|dx

For opening the modulas we need to define the bracket :

If (x - 1) < 0 then x belongs to (0, 1). And if (x - 1) > 0 then x belongs to (1, 4).

So the integral becomes:-

I\ =\ \int_{0}^{1} -(x-1)dx\ +\ \int_{1}^{4} (x-1)dx

or I\ =\ \left [ x\ -\ \frac{x^2}{2}\ \right ]^{1} _{0}\ +\ \left [ \frac{x^2}{2}\ -\ x \right ]^{4} _{1}

This gives I\ =\ 5

 

Question:19 Show that \int_0^a f(x)g(x)dx = 2\int_0^af(x)dx if f and g are defined as f(x) = f(a-x) and g(x) + g(a-x) = 4

 

Answer:

Let I\ =\ \int_0^a f(x)g(x)dx ........................................................(i)

This can also be written as :

I\ =\ \int_0^a f(a-x)g(a-x)dx

or I\ =\ \int_0^a f(x)g(a-x)dx ................................................................(ii)

Adding (i) and (ii), we get,

2I\ =\ \int_0^a f(x)g(a-x)dx +\ \int_0^a f(x)g(x)dx

2I\ =\ \int_0^a f(x)4dx

or I\ =\ 2\int_0^a f(x)dx

 

 

Question:20 Choose the correct answer in Exercises 20 and 21.

The value of is \int_\frac{-\pi}{2}^\frac{\pi}{2}(x^3 + x\cos x + \tan^5 x + 1)dx is

(A) 0

(B) 2

(C) \pi

(D) 1

Answer:

We have

I\ =\ \int_\frac{-\pi}{2}^\frac{\pi}{2}(x^3 + x\cos x + \tan^5 x + 1)dx

This can be written as :

I\ =\ \int_\frac{-\pi}{2}^\frac{\pi}{2}x^3dx +\ \int_\frac{-\pi}{2}^\frac{\pi}{2} x\cos x +\ \int_\frac{-\pi}{2}^\frac{\pi}{2} \tan^5 x +\ \int_\frac{-\pi}{2}^\frac{\pi}{2} 1dx

Also if a function is even function then \int_{-a}^{a}f(x)\ dx\ =\ 2\int_{0}^{a}f(x)\ dx

And if the function is an odd function then : \int_{-a}^{a}f(x)\ dx\ =\ 0

Using the above property I become:-

I\ =\ 0+0+0+ 2\int_{0}^{\frac{\Pi }{2}}1.dx

or I\ =\ 2\left [ x \right ]^{\frac{\Pi }{2}}_0

or I\ =\ \Pi

Question:21 Choose the correct answer in Exercises 20 and 21.

The value of \int_0^\frac{\pi}{2}\log\left(\frac{4+3\sin x}{4+3\cos x} \right )dx is

Answer:

We have

I\ =\ \int_0^\frac{\pi}{2}\log\left(\frac{4+3\sin x}{4+3\cos x} \right )dx .................................................................................(i)

 

By using :

\ \int_0^a\ f(x) dx\ =\ \ \int_0^a\ f(a-x) dx

We get,

I\ =\ \int_0^\frac{\pi}{2}\log\left(\frac{4+3\sin x}{4+3\cos x} \right )dx\ =\ \int_0^\frac{\pi}{2}\log\left(\frac{4+3\sin (\frac{\pi}{2}-x)}{4+3\cos (\frac{\pi}{2}-x)} \right )dx

or I\ =\ \int_0^\frac{\pi}{2}\log\left(\frac{4+3\cos x}{4+3\sin x} \right )dx .............................................................................(ii)

Adding (i) and (ii), we get:

2I\ =\ \int_0^\frac{\pi}{2}\log\left(\frac{4+3\sin x}{4+3\cos x} \right )dx\ +\ \int_0^\frac{\pi}{2}\log\left(\frac{4+3\cos x}{4+3\sin x} \right )dx

or 2I\ =\ \int_0^\frac{\pi}{2}\log1.dx

Thus I\ =\ 0

Chapter 7 Miscellaneous Exercise

NCERT Solutions for Class 12 Maths Chapter 7 Miscellaneous Exercise
Question 1:
ncert solutions class 12 maths Miscellaneous Questions Q 1
Solution:
ncert solutions class 12 maths Miscellaneous Questions Q 1 - i
ncert solutions class 12 maths Miscellaneous Questions Q 1 - ii

Question 2:
ncert solutions class 12 maths Miscellaneous Questions Q 2
Solution:
ncert solutions class 12 maths Miscellaneous Questions Q 2 - i

Question 3:
miscellaneous exercise on chapter 7 class 12 Q 3
Solution:
ncert solutions class 12 maths Miscellaneous Questions Q 3 - i

Question 4:
ncert solutions class 12 maths Miscellaneous Questions Q 4
Solution:
miscellaneous exercise on chapter 7 class 12 Q 4 - i
ncert solutions class 12 maths Miscellaneous Questions Q 4 - ii

Question 5:
ncert solutions class 12 maths Miscellaneous Questions Q 5
Solution:
ncert solutions class 12 maths Miscellaneous Questions Q 5 - i

Question 6:
ncert solutions class 12 maths Miscellaneous Questions Q 6
Solution:
ncert solutions class 12 maths Miscellaneous Questions Q 6 - i

Question 7:
ncert solutions class 12 maths Miscellaneous Question Q 7
Solution:
ncert solutions class 12 maths Miscellaneous Questions Q 7 - i

Question 8:
ncert solutions class 12 maths Miscellaneous Questions Q 8
Solution:
ncert solutions class 12 maths Miscellaneous Questions Q 8 - i

Question 9:
ncert solutions for class 12 maths chapter 7 miscellaneous exercise Q 9
Solution:
ncert solutions for class 12 maths miscellaneous Q 9 - i

Question 10:
ncert solutions class 12 maths Miscellaneous Questions Q 10
Solution:
ncert solutions class 12 maths Miscellaneous Questions Q 10 - i
Question 11:
ncert solutions class 12 maths Miscellaneous Questions Q 11
Solution:
ncert solutions class 12 maths Miscellaneous Questions Q 11 - i

Question 12:
ncert solutions class 12 maths Miscellaneous Questions Q 12
Solution:
ncert miscellaneous solutions class 12 Q 12 - i

Question 13:
ncert solutions class 12 maths Miscellaneous Questions Q 13
Solution:
ncert solutions class 12 maths Miscellaneous Questions Q 13 - i

Question 14:
ncert solutions class 12 maths Miscellaneous Questions Q 14
Solution:
ncert solutions for class 12 maths chapter 7 miscellaneous Q 14
ncert solutions class 12 maths Miscellaneous Questions Q 14 - i

Question 15:
ncert solutions class 12 maths Miscellaneous Questions Q 15
Solution:
ncert solutions class 12 maths Miscellaneous Questions Q 15 - i

Question 16:
ncert integrals miscellaneous solutions Q 16
Solution:
ncert solutions class 12 maths Miscellaneous Questions Q 16 - i

Question 17:
ncert solutions class 12 maths Miscellaneous Questions Q 17
Solution:
ncert solutions class 12 maths Miscellaneous Questions Q 17 - i

Question 18:
ncert solutions class 12 maths Miscellaneous Questions Q 18
Solution:
ncert solutions class 12 maths Miscellaneous Questions Q 18 - i

Question 20:
ncert solutions class 12 maths Miscellaneous Questions Q 20
Solution:
ncert solutions class 12 maths Miscellaneous Questions Q 20 -i
ncert solutions class 12 maths Miscellaneous Questions Q 20 - ii

Question 21:
ncert solutions class 12 maths Miscellaneous Questions Q 21
Solution:
ncert solutions class 12 maths Miscellaneous Questions Q 21 - i

Question 22:
ncert solutions class 12 maths Miscellaneous Questions Q 22
Solution:
ncert solutions class 12 maths Miscellaneous Questions Q 22 - i

Question 23:
ncert solutions class 12 maths Miscellaneous Questions Q 23
Solution:
ncert solutions class 12 maths Miscellaneous Questions Q 23 - i

Question 24:
ncert solutions class 12 maths Miscellaneous Questions Q 24
Solution:
ncert solutions class 12 maths Miscellaneous Questions Q 24 - i
ncert miscellaneous solutions class 12 Q 24 - ii

Question 25:
ncert solutions class 12 maths Miscellaneous Questions Q 25
Solution:
ncert solutions class 12 maths Miscellaneous Questions Q 25 - i

Question 26:
ncert solutions class 12 maths Miscellaneous Questions Q 26
Solution:
ncert solutions class 12 maths Miscellaneous Questions Q 26 - i

Question 27:
maths class 12 ncert solutions miscellaneous exercise 27
Solution:
maths class 12 ncert solutions miscellaneous exercise 27 - i
maths class 12 ncert solutions miscellaneous exercise 27 - ii

Question 28:
maths class 12 ncert solutions miscellaneous exercise 28
Solution:
maths class 12 ncert solutions miscellaneous exercise 28 - i
maths class 12 ncert solutions miscellaneous exercise 28 - ii

Question 29:
maths class 12 ncert solutions miscellaneous exercise 29
Solution:
maths class 12 ncert solutions miscellaneous exercise 29 - i

Question 30:
maths class 12 ncert solutions miscellaneous exercise 30
maths class 12 ncert solutions miscellaneous exercise 30 - i

Question 31:
maths class 12 ncert solutions miscellaneous exercise 31
Solution:
maths class 12 ncert solutions miscellaneous exercise 31 - i

Question 32:
maths class 12 ncert solutions miscellaneous exercise 32
Solution:
maths class 12 ncert solutions miscellaneous exercise 32- i
maths class 12 ncert solutions miscellaneous exercise 32- ii

Question 33:
Integration Class 12 NCERT Solutions Miscellaneous Exercise Q 33
Solution:

Integration Class 12 NCERT Solutions Miscellaneous Exercise Q 33 - ii

Question 34:
Integration Class 12 NCERT Solutions Miscellaneous Exercise Q 34
Solution:
Integration Class 12 NCERT Solutions Miscellaneous Exercise Q 34 - i
Integration Class 12 NCERT Solutions Miscellaneous Exercise Q 34 - ii

Question 35:
Integration Class 12 NCERT Solutions Miscellaneous Exercise Q 35
Solution:
Integration Class 12 NCERT Solutions Miscellaneous Exercise Q 35 -i

Question 36:
Integration Class 12 NCERT Solutions Miscellaneous Exercise Q 36
Solution:
Integration Class 12 NCERT Solutions Miscellaneous Exercise Q 36 -i

Question 37:
Integration Class 12 NCERT Solutions Miscellaneous Exercise Q 37
Solution:
Integration Class 12 NCERT Solutions Miscellaneous Exercise Q 37 - i

Question 38:
Integration Class 12 NCERT Solutions Miscellaneous Exercise Q 38
Solution:
Integration Class 12 NCERT Solutions Miscellaneous Exercise Q 38 -i

Question 39:
Integration Class 12 NCERT Solutions Miscellaneous Exercise Q 39
Solution:
Integration Class 12 NCERT Solutions Miscellaneous Exercise Q 39-i

Question 40:
Integration Class 12 NCERT Solutions Miscellaneous Exercise Q 40
Solution:
Integration Class 12 NCERT Solutions Miscellaneous Exercise Q 40 -i
Integration Class 12 NCERT Solutions Miscellaneous Exercise Q 40 -ii

Question 41:
Integration Class 12 NCERT Solutions Miscellaneous Exercise Q 41
Solution:
Integration Class 12 NCERT Solutions Miscellaneous Exercise Q 41 -i
Integration Class 12 NCERT Solutions Miscellaneous Exercise Q 41 -ii

Question 42:
Integration Class 12 NCERT Solutions Miscellaneous Exercise Q 42
Solution:
Integration Class 12 NCERT Solutions Miscellaneous Exercise Q 42 - i
Integration Class 12 NCERT Solutions Miscellaneous Exercise Q 42 - ii

Question 43:
Integration Class 12 NCERT Solutions Miscellaneous Exercise Q 43
Solution:
Integration Class 12 NCERT Solutions Miscellaneous Exercise Q 43-i
Integration Class 12 NCERT Solutions Miscellaneous Exercise Q 43-ii

Question 44:
Integration Class 12 NCERT Solutions Miscellaneous Exercise Q 44
Solution:
Integration Class 12 NCERT Solutions Miscellaneous Exercise Q 44 -i
Integration Class 12 NCERT Solutions Miscellaneous Exercise Q 44 -ii