NCERT Class 12 Mathematics Chapter 7 Integrals

NCERT Class 12 Mathematics Chapter 7 Integrals Solutions, NCERT Solutions For Class 12 Maths, NCERT Class 12 Mathematics Chapter 7 Integrals Notes to each chapter is provided in the list so that you can easily browse throughout different chapter NCERT Class 12 Mathematics Chapter 7 Integrals Question Answer and select needs one.

NCERT Class 12 Mathematics Chapter 7 Integrals

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Also, you can read the CBSE book online in these sections Solutions by Expert Teachers as per (CBSE) Book guidelines. NCERT Class 12 Mathematics Chapter 7 Integrals Question Answer. These solutions are part of NCERT All Subject Solutions. Here we have given NCERT Class 12 Mathematics Chapter 7 Integrals Solutions for All Subject, You can practice these here.

Integrals

Chapter – 7

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

1. sin 2 x

Ans: 

2. cos 3 x

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3. e²ˣ

Ans: 

4. (ax + b)²

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5. sin 2x – 4 e³ˣ

Ans: 

Find the following integrals in Exercises 6 to 20:

6. ∫ (4 e³ˣ + 1)dx

Ans: 

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8. ∫(ax² + bx + c) dx

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9. ∫ (2x² + eˣ ) dx

Ans: 

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14. ∫ (1 – x)√x dx

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15. ∫ √x(3x² + 2x + 3) dx

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16. ∫ (2x – 3 cos x + eˣ) dx

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17. ∫ (2x² – 3sin x + 5√x) dx

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18. ∫ sec x (sec x + tan x) dx

Ans: ∫ sec x (sec x + tan x) dx = ∫ (sec² x + sec x tan x) dx 

= ∫ sec² x dx + ∫ sec x tan x dx 

= tan x + sec x + C

Ans: 

Ans: 

= ∫ 2sec² x dx – 3 ∫ tan x sec x dx

= 2 tan x – 3sec(x) + C

Choose the correct answer in Exercises 21 and 22

Ans:

Thus, the correct option is C.

Ans: 

Thus, the correct option is A.

Integrate the functions in Exercises 1 to 37:

Ans: Put 1 + x² = t 

Therefore, 2xdx = dt

= log|1 + x²| + C 

= log(1 + x²) + C

Ans: 

Ans: 

4. sin x sin (cos x).

Ans: Put cos x = t 

Therefore, – sin xd x = dt 

∫ sin x sin(cos x) dx = – ∫ sin tdt = – [- cos t] + C 

= cos t + C 

= cos(cos x) + C

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

Ans:

6. √ax + b.

Ans: Put ax + b = t 

Therefore,

7. x√x + 2.

Ans: Put, x + 2 = t 

∴ dx = dt 

⇒ ∫ x √x + 2 = ∫ (t – 2) √t dt

8. x√1 + 2x²

Ans: 

9. (4x + 2)√x² + x + 1

Ans: Put, x² + x + 1 = t 

∴ (2x + 1) dx = dt 

∫ (4x + 2) √x² + x + 1dx 

= ∫ 2√t dt 

= 2 ∫ √t dt

Ans: 

Ans: 

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16. e²ˣ ⁺ ³

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Dividing Nr and Dr by eˣ, we get

Ans: 

21. tan² (2x – 3)

Ans: tan² (2x – 3) = sec² (2x – 3) – 1 

Put, 2x – 3 = t

∴ 2 dx = dt 

⇒ ∫ tan² (2x – 3) dx = ∫ [sec² (2x – 3) – 1] dx

22. sec² (7 – 4x)

Ans: Put, 7 – 4x = t

∴ – 4  dx = dt

Ans: Put, sin¯¹ x = t

Ans: 

Ans: 

Ans:  Let √x = t

= 2sin t + C 

= 2sin √x + C

27. √sin 2x cos 2x

Ans: Put, sin 2x = t 

So, 2cos 2x dx = dt

Ans: Put, 1 + sin x = t

∴ cos xdx = dt

= 2√t + C 

= 2√1 + sin x + C

29. cos x log sin x

Ans: 

Ans: Put, 1 + cos x = t

∴ – sin xdx = dt

= – log|t| + C 

= – log|1+cos x| + C

Ans: Put, 1 + cos x = t

∴ – sin xdx = dt

Ans: 

Ans: 

Ans: 

Ans: 

Ans: 

Ans: Put, x⁴ = t 

∴ 4x³ dx = dt

Let tan⁻¹ t = u

Choose the correct answer in Exercises 38 and 39.

(A) 10ˣ – x¹⁰ + C

(B) 10ˣ + x¹⁰ + C

(C) (10ˣ – x¹⁰)⁻¹ + C

(D) log(10ˣ + x¹⁰) + C

Ans: Put, x¹⁰ + 10ˣ = t

= log t + C 

= log(10ˣ + x¹⁰) + C 

Thus, the correct option is D.

(A) tan x + cot x + C

(B) tan x – cot x + C

(C) tan x cot x + C

(D) tan x – cot 2x + C

Ans: Put, 

= ∫ sec² xdx + ∫ cos ec² dx 

= tan x- cot x + C

Thus, the correct option is B.

Find the integrals of the functions in Exercises 1 to 22:

1. sin² (2x + 5)

Ans: 

2. sin 3x  cos 4x

Ans: 

3. cos 2x cos 4x cos 6x

Ans: 

4. sin³ (2x + 1)

Ans: Put, I = ∫ sin³ (2x + 1) 

⇒ ∫ sin³ (2x + 1) dx = ∫ sin² (2x + 1) sin(2x + 1) dx 

= ∫ (1 – cos² (2x + 1))  sin(2x + 1) dx 

Let cos(2x + 1) = t 

⇒ – 2sin(2x + 1) dx = dt

5. sin³ x cos³ x

Ans: Let I = ∫ sin³ x cos³ xdx 

= ∫ cos³ x sin² x sin xdx 

= ∫ cos³ x  (1 – cos² x) sin xdx 

Let cosx = t 

⇒ – sin xdx = dt 

⇒ I = – ∫ t³ (1 – t²) dt

6. sin x sin 2x sin 3x

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7. sin 4x sin 8x

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10. sin⁴ x

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11. cos⁴ 2x

Ans: 

Ans: 

Ans: 

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15. tan³ 2x sec 2x

Ans: tan³ 2x sec 2x = tan² 2x tan 2x sec 2x 

= (sec² 2x – 1) tan 2x sec 2x 

= sec² 2x tan 2x sec 2x – tan 2x sec 2x 

∴ ∫ tan³ 2x sec 2 xdx = ∫ sec ² 2x  tan 2x sec 2x – ∫ tan 2x  sec 2x 

16. tan⁴ x

Ans: tan⁴ x 

= tan² x tan² x 

= (sec² x – 1)  tan² x 

= sec² x  tan² x – tan² x 

= sec² x  tan² x – (sec² x – 1) 

= sec² x tan² x – sec² x + 1 

∴ ∫ tan⁴ xdx = ∫ sec² x tan² xdx – ∫ sec² xdx + ∫ 1dx 

= ∫ sec² x  tan² xdx – tan x + x + C …(1) 

Consider sec² x tan² xdx 

Let tan x = t ⇒ sec² xdx = dt

Ans: 

Ans: 

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21. sin⁻¹ (cos x)

Ans: sin⁻¹ (cos x)

Let cos x = t

Then, sin x = √ 1 – t²

⇒ (- sin x) dx = dt

Let sin⁻¹ t = u

Substituting in equation (1), we get

Ans: 

Choose the correct answer in Exercises 23 and 24.

(A) tan x + cot x + C

(B) tan x + cosec x + C

(C) – tan x + cot x + C

(D) tan x + sec x + C

Ans: 

= ∫ (sec² x – cosec² x) dx

= tan x + cot x + C

Thus, the correct option is A.

(A) – cot (exˣ) + C

(B) tan (xeˣ) + C

(C) tan(eˣ) + C

(D) cot(eˣ) + C

Ans: 

Put, eˣ x = t 

⇒ (eˣ x + eˣ .1) dx = dt 

eˣ (x + 1) dx = dt

= ∫ sec² tdt 

= tan t + C

= tan(eˣ x)+C

Thus, the correct answer is B.

Integrate the functions in Exercises 1 to 23.

Ans: Put, x³ = t

∴ 3x² dx = dt

= tan⁻¹ t + C

= tan⁻¹ (x³) + C

Ans: 

Ans: Put, 2 – x = t

⇒ – dx = dt

Ans: Put, 5x = t

∴ 5dx = dt

Ans: Let √2x² = t

∴ 2 √2xdx = dt

Ans: Put, x³ = t

∴ 3x² dx = dt

Ans: 

Ans: 

Ans: Put, tan x = t

∴ sec² xdx = dt

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Ans: 7 – 6x – x² can be written as 7 – (x² + 6x + 9 – 9) 

Thus,

7 – (x² + 6x + 9 – 9) 

= 16 – (x² + 6x + 9)

= 16 – (x + 3)²

= (4)² – (x + 3)²

Ans: (x – 1)(x – 2) can be written as x² – 3x + 2

Thus, 

x² – 3x + 2

Ans: 

Thus,

Ans: (x – a)(x – b) = x² – (a + b) x + ab 

Thus,

x² – (a + b) x + ab

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⇒ 4x + 1 = A(4x + 1) + B 

⇒ 4x + 1 = 4Ax + A + B 

Equating the coefficients of x and constant term on both sides, we get 

4A = 4 ⇒ A =1 

A + B =1 ⇒ B = 0 

Let 2x² + x – 3 = t 

∴ (4x + 1) dx = dt

= 2√t + C

= 2√2x² + x – 3) + C

Ans: 

Ans: 

Ans: 

Ans: 

Ans: 

Ans: Let 

(x + 3) = A(2x – 2) + B 

Equating the coefficients of x and constant term on both sides, we get

Ans: 

Choose the correct answer in Exercises 24 and 25.

(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

Ans: 

Ans: 

Hence, the correct option is B.

Integrate the rational functions in Exercises 1 to 21.

Ans: 

Ans: Let 

1 = A(x – 3) + B(x + 3) 

Equating the coefficients of x and constants term, we get

A + B = 0

– 3A + 3B = 1

On solving, we get

Ans: 

Ans: Let

x = A(x – 2)(x – 3) + B(x – 1)(x – 3) + C(x – 1)(x – 2) …(1)

Equating the coefficients of x² , x and constant terms, we get

A + B + C = 0 

– 5A – 4B – 3C = 1 

6A + 4B + 2C = 0 

Solving these equations, we get

Ans: 

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

Therefore, on dividing (1 – x²)by x(1 – 2x), we get

Ans: 

Ans: 

Ans: 

3x + 5 = A(x – 1)(x + 1) + B(x + 1) + C (x – 1)²

3x + 5 = A(x – 1)(x + 1) + B(x + 1) + C(x² – 2x + 1) …(1) 

Equating the coefficients of x²,x and constant term, we get 

A + C = 0 

B – 2C = 3 

– A + B + C = 5 

On solving these equations, we get

Ans: 

⇒ (2x – 3) = A(x – 1)(2x – 3) + B(x + 1)(2x + 3) + C(x + 1)(x – 1) 

⇒ (2x – 3) = A(2x²+ x – 3) + B(2x² + 5x + 3) + C(x² – 1) 

⇒ (2x – 3) = (2A + 2B + C)  x² + (A + 5B) x + (- 3A + 3B – C) 

Equating the coefficients of x², x and constant term, we get  

2A + 2B + C = 0 

A + 5B = 2 

– 3A + 3B – C = – 3 

On solving, we get

Ans: 

5x = A(x + 2)(x – 2) + B(x + 1)(x – 2) + C(x + 1)(x + 2) …(1) 

Equating the coefficients of x²,x and constant term, we get 

A + B + C = 0 

– B + 3C = 5

– 4A – 2B + 2C = 0 

On solving, we get

Ans: On dividing (x³ + x + 1) by x² – 1 , we get

2x + 1 = A(x – 1) + B(x + 1) …(1)

Equating the coefficients of x and constant term, we get 

A + B = 2

– A + B = 1 

On solving, we get

Ans: Let 

2 = A(1 + x²) + (Bx + C)(1 – x)

2 = A + A x² + Bx – B x² + C – Cx 

Equating the coefficients of x², x and constant term, we get

A – B = 0 

B – C = 0 

A + C = 2 

On solving these equations, we get 

A = 1, B = 1 and C = 1

Ans: Let

⇒ 3x – 1 = A(x + 2) + B

Equating the coefficient of x and constant term, we get

A = 3 

2A + B = -1 ⇒ B = -7

Ans: 

l = A(x – 1)(1 + x²) + B(x + 1)(1 + x²) + (Cx + D)(x² – 1)

l = A(x³ + x – x² – 1) + B(x³ + x + x² + 1) + C x³ + D x² – Cx – D

I = (A + B + C) x³ + (- A + B + D)  x² + (A + B – C) x + (- A + B – D)

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

A + B + C = 0

– A + B + D = 0 

A + B – C = 0

– A + B – D = 1

On solving, we get

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

Ans: 

[Hint : Put sin x = t]

Ans: 

Ans: 

4x² + 10 = (Ax + B)(x² + 4) + (Cx + D)(x² + 3)

4x² + 10 = Ax³ + 4Ax + Bx² + 4B + Cx³ + 3Cx + Dx² + 3D 

4x² + 10 = (A + C) x³ + (B + D) x² + (4A + 3C) x + (4B + 3D) 

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

A + C = 0 

B + D = 4 

4A + 3C = 0

4B + 3D = 10 

On solving these equations, we get 

A = 0, B = – 2, C = 0 and D = 6 

Ans: 

Ans: 

Multiplying Nr and Dr by x³, we get

Ans: Put eˣ = t ⇒ eˣ dx = dt

1 = A(t – 1) + Bt …(1) 

Equating the coefficients of t and constant, we get

A = – 1 and B = 1

Ans: 

Ans: 

Integrate the functions in Exercises 1 to 22. 

1. x sin x.

Ans: Let I = ∫ x sin xdx 

Taking u = x and v = sin x and integrating by parts,

I = x ∫ sin xdx – ∫ {d/dx (x)) ∫ sin xdx} dx 

= x(- cos x) – ∫ 1. (- cos x) dx 

= – x cos x + sin x + C

2. x sin 3x

Ans: Let I = ∫ x sin 3 xdx

Taking u = x and v = sin 3x and integrating by parts,

3. x² eˣ

Ans: Let I =  ∫ x² eˣ dx

Taking u = x² and v = eˣ and integrating by parts, we get 

= x²eˣ – ∫ 2x.eˣ dx

=x²eˣ – 2 ∫ x.eˣ

Again using integration by parts, we get

= x²eˣ – 2xeˣ + 2eˣ + C

= eˣ(x² – 2x + 2) + C

4. x log x

Ans: Let I = ∫ x log xdx 

Taking u = log x and v = x and integrating by parts, we get

5. x log 2x 

Ans: Let I = ∫ x log 2xdx

Taking u =  log 2x and v = x and integrating by parts, we get

6. x² log x

Ans: Let I = ∫ x² log xdx

Taking u = log x and v = x² and integrating by parts, we get

7. x sin⁻¹ x

Ans: Let I = ∫ x sin⁻¹ xdx

Taking u = sin⁻¹ x and v = x and integrating by parts, we get

 8. x tan⁻¹ x

Ans: Let I = ∫ x tan⁻¹ xdx 

Taking tann⁻¹  x and v = x and integrating by parts, we get

9. x cos⁻¹ x 

Ans: Let I = ∫ x cos⁻¹ xdx 

Taking u = cos⁻¹ x and v = x and integrating by parts, we get

10. (sin⁻¹ x)²

Ans: Let I = ∫ (sin⁻¹ x)².1 dx 

Taking u = (sin⁻¹ x)²  and v = 1 and integrating by parts, we get

Ans: 

12. x sec² x

Ans: Let I = ∫ x sec² xdx

Taking u = x and v = sec² x and integrating by parts, we get

= x tan x –  ∫ 1.tan xdx

= x tan x + log |cosx| + C

13. tan⁻¹ x

Ans: Let I = ∫1.tan⁻¹ xdx

Taking u = tan⁻¹ x and v = 1 and integrating by parts, we get

14. x (log x)² 

Ans: Let I = ∫ x (log x)² dx

Taking u = (log x)² and v = x and integrating by parts, we get

Again, using integration by parts, we get

15. (x² + 1) log x

Ans: Let I = ∫(x² + 1) log xdx = ∫ x² log xdx + ∫ log xdx 

Let I = I₁ + I₂ ……….(1) 

Where, I₁ = ∫ x² log xdx and I₂ =  ∫ log xdx 

I₁ = ∫ x² log xdx

Taking u = log x and v = x² and integrating by parts, we get

I₂ =  ∫ log xdx

Taking u = log x and v = 1 and integrating by parts,

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

16. eˣ (sin x + cos x)

Ans: Let I = ∫eˣ (sin x + cos x) dx 

Let ∫ (x) = sin x

f’(x) = cos x 

I = ∫eˣ {f(x) + f'(x)} dx 

Since, ∫ eˣ {f(x) + f'(x)} dx 

= eˣ f (x) + c 

∴ I = eˣ sin x + C

Ans: 

Ans: 

It is known that, ∫ eˣ {f(x) + f ‘(x)} dx 

= eˣ f (x) + C 

From equation (1), we get

Ans: 

It is known that,

Ans: 

Ans: Let I = e²ˣ sin xdx………. (1) 

Taking u = sin x and v = e²ˣ  and integrating by parts, we get

Again, using integration by parts, we get

Ans: 

Using integration by parts, we get

Choose the correct answer in Exercises 23 and 24.

Ans: 

Thus, the correct option is A.

24. ∫ eˣ sec x (1+ tan x) dx equals

(A) eˣ cos x + C

(B) eˣ sec x + C

(C) eˣ sin x + C 

(D) eˣ tan x + C

Ans: ∫ eˣ sec x (1 + tan x) dx

Consider, I = ∫ eˣ sec x (1 + tan x)dx =  ∫ eˣ (sec x + sec x tan x)dx

 Let sec x = f (x) sec x tan x = f'(x)

It is known that, ∫ eˣ {f (x) + f'(x)} dx = eˣ f(x) + C 

∴ I = eˣ sec x + C 

Thus, the correct option is B.

Integrate the functions in Exercises 1 to 9

1. √ 4 – x²

Ans: 

2. √1 – 4x²

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3. √x² + 4x + 6

Ans: 

4. √x² + 4x + 1

Ans: 

5. √1 – 4x – x²

Ans: 

6. √x² + 4x – 5

Ans: 

7. √1 + 3x – x²

Ans: 

8. √x² + 3x

Ans:  

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Choose the correct answer in Exercises 10 to 11.

Ans: 

Ans: 

Evaluate the definite integrals in Exercises 1 to 20.

Ans: 

Ans: 

Ans: Let

Ans: 

Ans: 

Ans: Let

∫ eˣ dx = eˣ = F(x)

Using second fundamental theorem of calculus, we get 

I = F(5) – F(4)

= e⁵ – e⁴

= e⁴ (e – 1)

Ans: Let 

∫ tan xdx = – log|cos x| = F(x)

Using second fundamental theorem of calculus, we get

Ans: 

Ans: 

Using second fundamental theorem of calculus, we get

Ans: 

Ans: 

Ans: 

Ans: 

Ans: Let

Ans: 

Ans: 

= 2Ax + (4A + B) 

Equating the coefficients of x and constant term, we get 

A = 10 and B = – 25 

Let x² + 4x + 3 = t 

⇒ (2x + 4) dx = dt

Substituting the value I₁ in (1), we get

Ans: 

Ans: 

∫ cos xdx = sin x = F(x)

Using second fundamental theorem of calculus, we get 

I = F(π) – F(0)

= sin π – sin 0

= 0

Ans: 

Ans: Let

Choose the correct answer in Exercises 21 and 22.

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Evaluate the integrals in Exercises 1 to 8 using substitution.

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Ans: 

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Ans: 

Ans: 

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Choose the correct answer in Exercises 9 and 10.

9. The value of the integral

(A) 6

(B) 0

(C) 3

(D) 4

Ans: 

(A) cos x + x sin x

(B) x sin x

(C) x cos x

(D) sin x + x cos x

Ans: 

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

Ans: 

Ans: 

Ans: 

Ans: 

Ans: 

Ans: 

Ans: 

Ans:

Ans: 

Ans: 

Ans: 

Ans: 

Multiplying and Dividing by (1-sinx)

⇒ 2 I = π[(tan(π) – tan(0)) – (sec(π) – sec(0))] 

⇒ 2I = π[2]

⇒ I = π

Ans: 

Ans: 

Ans: 

Ans: 

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19. Show that f (x) dx, if f and g are defined as f(x) = f(a – x) and g(x) + g(a – x) = 4

Ans:

Choose the correct answer in Exercises 20 and 21.

(A) 0

(B) 2

(C) π

(D) 1

Ans: 

(A) 2

(B) ¾

(C) 0

(D) -2

Ans: 

Thus, the correct option is C.

Integrate the functions in Exercises 1 to 23.

Ans: 

⇒ 1 = A(1 – x²) + Bx(1 + x) + Cx(1 – x) 

⇒ 1 = A – Ax² + Bx + Bx² + Cx – Cx² 

Equating the coefficients of x², x and constant terms, we get 

– A + B – C = 0

B + C = 0 

A = 1 

On solving these equations, we get 

A = 1

Ans: 

Ans: 

Ans: 

Ans: 

Ans: Consider,

⇒ 5x = A(x² + 9) + (Bx + C)(x + 1)

⇒ 5x = Ax² + 9A + Bx² + Bx + Cx + C

Equating the coefficients of x²,x and constant term, we get 

A + B = 0

B + C = 5

9A + C = 0 

On solving these equations, we get 

A = – 1/2

B = 1/2 

C = 9/2 

From equation (1), we get

Ans: 

= t cos a + sin a log|sint| + C₁

= (x – a) cos a + sin a log|sin(x – a)| + C₁

= x cos a + sin a log|sin(x – a)| – a cos a + C₁

= sin a log|sin(x – a)| + x cos a + C

Ans: 

Ans:

Ans: 

Ans: 

Multiplying and dividing by sin(a – b), we get

Ans: 

Ans: 

Ans:

⇒ 1 = (Ax + B)(x² + 4) + (Cx + D)(x² + 1)

⇒ 1 = Ax³ + 4Ax + Bx² + 4B + Cx³ + Cx + Dx² + D

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

A + C = 0

B + D = 0

4A + C = 0

4B + D = 1

On solving these equations, we get

A = 0

B = 1/3

C = 0 

D = – 1/3 

From equation (1), we get

15. cos³ x eˡᵒᵍ ˢⁱⁿˣ

Ans: cos³ xeˡᵒᵍ ˢⁱⁿ ˣ = cos³ x × sin x 

Let cos x = t  ⇒ -sin xdx = dt 

⇒- ∫ cos³ xeˡᵒᵍ ˢⁱⁿ ˣ  dx = ∫ cos³ x sin xdx 

= – ∫ t³ dt

16. e³ ˡᵒᵍˣ (x⁴ + 1)⁻¹

Ans: 

Let x⁴ +1= t ⇒ 4x³ dx = dt

17. f’ (ax + b) [f (ax + b)]ⁿ

Ans: 

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Put, cos α + cot x sin α = t ⇒ – cos ec² x sin α dx = dt

Ans: 

= θ + √1-cos² θ. cos θ – 2√1 – cos² θ + C 

= cos⁻¹ √x + √1-x. √x – 2√1 – x + C 

= – 2√1 – x + cos⁻¹ √x + √x(1 – x) + C 

= – 2√1 – x + cos⁻¹ √x + √x – x² + C

Ans: 

= ∫ (sec² x + tan x) eˣ 

Let f(x) = tan x ⇒ f’ (x) = sec² x 

∴ I = ∫ (f(x) + f’ (x)) eˣ dx 

= eˣ f(x)+C 

= eˣ tan x +C

Ans: Let

⇒ x² + x + 1 = A(x + 1)(x + 2) + B(x + 2) + C(x² + 2x + 1) 

⇒ x² + x + 1 = A(x² + 3x + 2) + B(x + 2) + C(x² + 2x + 1) 

⇒ x² + x + 1 = (A + C) x² + (3A + B + 2C)  x + (2A + 2B + C) 

Equating the coefficients of x² ,x and constant term, we get 

A + C = 1 

3A + B + 2C = 1

2A + 2B + C = 1

On solving these equations, we get 

A = – 2 

B = 1 

C = 3 

From equation (1), we get

Ans:

Ans: 

Ans:

Ans: Let

Ans: Consider,

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Ans: 

⇒ 1 = Ax(x + 1) + B(x + 1) + C(x²)

⇒ 1 = Ax² + Ax + Bx + B + Cx²

Equating the coefficients of x², x and constant terms, we get

A + C = 0

A + B = 0

B = 1 

On solving these equations, we get

A = – 1

C = 1

B = 1

Ans: 

Ans: 

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(A) tan⁻¹ (eˣ) + C.

(B) tan⁻¹ (e⁻ˣ) + C.

(C) log(eˣ – e⁻ˣ) + C.

(D) log(eˣ + e⁻ˣ) + C.

Ans:

(B) log|sin x + cos x| + C

(C) log|sin x – cos x| + C

Ans: 

∴ I = ∫ dt/t

= log|t|+C

= log|cosx + sinx|+ C

Thus, the correct option is B.

Ans: 

Parimol

Hi! my Name is Parimal Roy. I have completed my Bachelor’s degree in Philosophy (B.A.) from Silapathar General College. Currently, I am working as an HR Manager at Dev Library. It is a website that provides study materials for students from Class 3 to 12, including SCERT and NCERT notes. It also offers resources for BA, B.Com, B.Sc, and Computer Science, along with postgraduate notes. Besides study materials, the website has novels, eBooks, health and finance articles, biographies, quotes, and more.