# MP Board Class 12th Maths Important Questions Chapter 7A Integration

## MP Board Class 12th Maths Important Questions Chapter 7A Integration

### Integration Important Questions

Integration Objective Type Questions:

Question 1.
Choose the correct answer:

Question 1.
The value of $$\int { \frac { sec^{ 2 }x }{ 1+tanx } }$$ dx is:
(a) loge (l + tan x) + c
(b) tan x + c
(c) – cot x + c
(c) – cot x + c
(d) loge x + c
(a) loge (l + tan x) + c

Question 2.
The value of $$\int { \frac { x }{ 4+x^{ 4 } } }$$ dx is:
(a) $$\frac{1}{4}$$ x2 + c
(b) $$\frac{1}{4}$$ tan-1 $$\frac { x^{ 2 } }{ 2 }$$
(c) $$\frac{1}{2}$$ tan-1 $$\frac { x^{ 2 } }{ 2 }$$
(d) None of these
(a) $$\frac{1}{4}$$ x2 + c

Question 3.
If $$\int \frac{2^{\frac{1}{x}}}{x^{2}}$$ dx = k(2)1/x + c, then find the value of k is:
(a) $$\frac { -1 }{ log_{ e }2 }$$
(b) – loge 2
(c) -1
(d) $$\frac { 1 }{ 2 }$$
(a) $$\frac { -1 }{ log_{ e }2 }$$

Question 4.
The value $$\int \frac{e^{x}(1+x)}{\cos ^{2}\left(x e^{x}\right)}$$ dx is:
(a) 2 logecos(xex) + c
(b) sec(xex) + c
(c) tan(xex) + c
(d) tan(x + ex) + c
(c) tan(xex) + c

Question 5.
If $$\int { xsinxdx }$$ = -x cos x + α will be:
(a) sin x + c
(b) cos x + c
(c) c
(d) None of these
(a) sin x + c

Question 2.
Fill in the blanks :

1. $$\frac{1}{2}$$ tan-1 ( $$\frac{x}{a}$$ ) + c
2. log [x + $$\int { \frac { dx }{ \sqrt { x^{ 2 }-a^{ 2 } } } }$$ + c
3. log [x + $$\int { \frac { dx }{ \sqrt { a^{ 2 }-x^{ 2 } } } }$$ + c
4. $$\frac{x}{2}$$ $$\int { \frac { dx }{ \sqrt { a^{ 2 }-x^{ 2 } } } }$$ + $$\frac { a^{ 2 } }{ 2 }$$ sin-1 ( $$\frac{x}{a}$$ ) + c
5. log(sec x + tan x) + c
6. sin-1 x + $$\sqrt { 1-x }$$ + c
7. tan x + sec x.

Question 3.
Write True/False:

1. True
2. True
3. False
4. True
5. False
6. False

Integration Very Short Answer Type Questions

1. $$e^{ tan-1x }$$ + c
2. – cos x + c
3. sin2 x + c
4. $$\frac{1}{2a}$$ log $$\frac { a+x }{ a-x }$$
5. $$\frac{1}{2a}$$ log $$\frac { x-a }{ x+a }$$
6. $$\frac { e^{ x } }{ x }$$ + c
7. x log $$\frac { x }{ e }$$ + c
8. 2 tan-1$$\sqrt{x}$$
9. 1 + log x = t
10. a = – $$\frac { \pi }{ 4 }$$, b = 3
11. x = t2
12. – $$\frac{1}{a}$$ cos(ax + b)
13. $$\frac{1}{a}$$ log (ax + b)
14. tan x – x + c
15. sin-1 $$\frac{x}{a}$$
16. $$\frac { x^{ 3 } }{ 3 }$$ + c
17. $$\frac { x^{ 3 } }{ 2 }$$ + c
18. exf(x) + c
19. ex
20. ex log x + c.

Integration Short Answer Type Questions

Question 1.
Evaluate:
$$\int \frac{\cos 2 x+2 \sin ^{2} x}{\cos ^{2} x} dx$$? (CBSE 2018)
Solution:
Let I = $$\int \frac{\cos 2 x+2 \sin ^{2} x}{\cos ^{2} x} d x$$

= $$\int { sec^{ 2 }xdx }$$
= tan x + c.

Question 2.
Evaluate $$\int { \frac { 1-sinx }{ cos^{ 2 }x } }$$ dx? (NCERT)
Solution:
Let I = $$\int { \frac { 1-sinx }{ cos^{ 2 }x } }$$

Question 3.
Evaluate $$\int \frac{2-3 \sin x}{\cos ^{2} x} d x$$ dx? (NCERT)
Solution:
Let I = $$\int \frac{2-3 \sin x}{\cos ^{2} x} d x$$ dx

Question 4.
Evaluate $$\int \sin ^{-1}(\cos x) d x$$ dx? (NCERT)
Solution:
I = $$\int \sin ^{-1}(\cos x) d x$$ dx

Question 5.
Evaluate $$\int { \frac { dx }{ 1+cos2x } }$$?
Solution:
Let I =

Question 6.
(A) Evaluate $$\int { tan^{ -1 }xdx }$$?
Solution:
Let

(B) Evaluate $$\int { sin^{ -1 }xdx }$$ and $$\int { cos^{ -1 }xdx }$$?
(Do it by yourself)

Question 7.
Evaluate $$\int { sin^{ 2 }xdx }$$?
Solution:
Let

Question 8.
Evaluate $$\int { \frac { cosx }{ cos(x-\alpha ) } }$$ dx?
Solution:

Question 9.
(A) Evaluate $$\int { \frac { 1 }{ \sqrt { 1+cosx } } }$$ dx?
Solution:
Let

(B) Evaluate $$\int { \sqrt { 1+cos2xdx } }$$?
Solution:
Solve like Q.No 9 (B)

Question 10.
(A) Evaluate $$\int { \frac { e^{ x }(1+x) }{ cos^{ 2 }(xe^{ x }) } }$$ dx?
Solution:
Let

(B) Evaluate $$\int { \frac { e^{ tan-1x } }{ 1+x^{ 2 } } }$$ dx?
Solution:
Let

Question 11.
Evaluate $$\int { \frac { dx }{ 1-sinx } }$$?
Solution:
Let

Question 12.
Evaluate $$\int { \frac { logx }{ x } }$$ dx?
Solution:
Given:

Question 13.
(A) Evaluate $$\int { \frac { dx }{ 1-cosx } }$$?
Solution:
Let

(B) Evaluate $$\int { \frac { logx }{ x } }$$ dx?
Solution:
Let

Question 14.
Evaluate $$\int { \frac { dx }{ 1+sinx } }$$?
Solution:
Let

Question 15.
Evaluate $$\int { \frac { cos\sqrt { x } }{ \sqrt { x } } }$$ dx?
Solution:
Let

Question 16.
Evaluate $$\int { \frac { 1-cos2x }{ 1+cos2x } }$$ dx?
Solution:
Let

Question 17.
Integrate $$\frac { x^{ 4 } }{ x^{ 2 }+1 }$$ with respect to x?
Solution:
Let

Question 18.
Find the value of $$\int { \frac { sec^{ 2 }(logx) }{ x } }$$ dx?
Solution:
Let I = $$\int { \frac { sec^{ 2 }(logx) }{ x } }$$ dx
Put, log x = t
⇒ $$\frac { 1 }{ x }$$ dx = dt
∴ I = ∫sec2 t dt
⇒ I = tan x + c
⇒ tan(log x) + c

Question 19.
Find the value of $$\int { \frac { sin(logx) }{ x } }$$ dx?
Solution:
Let I = $$\int { \frac { sin(logx) }{ x } }$$ dx
Put, log x = t
⇒ $$\frac { 1 }{ x }$$ dx = dt
∴ I = ∫sin t dt = -cos t + c
⇒ I = – cos (log x) + c

Question 20.
Find the value of $$\int { \frac { cos(logx) }{ x } }$$ dx?
Solution:
Solve like Q.No.19

Question 21.
(A) Find the value of $$\int { tan^{ 2 }xdx }$$?
Solution:
Let I = $$\int { tan^{ 2 }xdx }$$ = ∫(sec2 x – 1) dx
= ∫sec2 x dx – ∫1dx = tan x – x.

(B) Find the value of $$\int { cot^{ 2 }xdx }$$?
Solution:
Let I = $$\int { cot^{ 2 }xdx }$$ = ∫(cosec2 x – 1) dx
= ∫cosec2 x dx – ∫1.dx = – cot x – x.

Question 22.
Find the value of $$\int { \frac { sinx }{ 1+cosx } }$$ dx?
Solution:
Let I = $$\int { \frac { sinx }{ 1+cosx } }$$ dx
= ∫$$\frac { 1 }{ t }$$ dt, (Put 1+cos x = dt ⇒ sin x dx = dt)
= log t
= log (1 + cos x).

Question 23.
Find the value $$\int { \frac { sin^{ -1 }x }{ \sqrt { 1-x^{ 2 } } } }$$ dx?
Solution:

Integration Long Answer Type Questions – I

Question 1.
Find the value of $$\int { \sqrt { \frac { a+x }{ a-x } } }$$ dx?
Solution:
Let I = $$\int { \sqrt { \frac { a+x }{ a-x } } }$$ dx
Again let x = a cos θ ⇒ dx = – a sin θ dθ

Question 2.
Evaluate $$\int\left[\frac{1}{(\log x)^{2}}-\frac{2}{(\log x)^{3}}\right]$$ dx?
Solution:
Let I = $$\int\left[\frac{1}{(\log x)^{2}}-\frac{2}{(\log x)^{3}}\right]$$
Again let log x = t ⇒ x = et ⇒ dx = etdt

Question 3.
Evaluate $$\int { sin^{ -1 } }$$ xdx?
Solution:
Let

Again let 1 – x2 t ⇒ – 2x dx = dt

Question 4.
Evaluate $$\int { cos^{ -1 }xdx }$$?
Solution:
Let

Question 5.
(A) Evaluate $$\int { \frac { x^{ 2 } }{ 1+x } }$$ dx?
Solution:
Let

(B) Evaluate $$\int { \frac { x }{ 1+x^{ 4 } } }$$ dx?
Solution:
Let

= $$\frac{1}{2}$$ tan-1 t,
= $$\frac{1}{2}$$ tan-1 x2

Question 6.
Evaluate $$\int { \frac { 1 }{ sinx-cosx } }$$ dx?
Solution:
Let I = $$\int { \frac { 1 }{ sinx-cosx } }$$ dx

Question 7.
Evaluate $$\int { \frac { dx }{ e^{ x }+1 } }$$?
Solution:
Let

Question 8.
Evaluate $$\int { sec^{ 3 }xdx }$$?
Solution:
Let

= sec x tan x – ∫sec x tan x tan x dx
= sec x tan x – ∫sec x tan2 x dx
= sec x tan x – ∫sec x (sec2 x – 1) dx
= sec x tan x – ∫sec3 xdx +∫sec x dx
⇒ I = sec x tan x – I + log(sec x + tan x)
⇒ 2I = sec x tan x + log (sec x + tan x)
⇒ I = $$\frac{1}{2}$$ [sec x tan x + log (sec x + tan x)].

Question 9.
Evaluate $$\int { \frac { dx }{ x^{ 2 }-a^{ 2 } } }$$?
Solution:
Let

We have to break the term with partial fraction,
img

Question 10.
Evaluate $$\int { \frac { 3x }{ (x-2)(x+1) } }$$ dx?
Solution:
Let $$\int { \frac { 3x }{ (x-2)(x+1) } }$$ = $$\frac { A }{ (x-2) }$$ + $$\frac { B }{ (x+1) }$$ ………………… (1)
⇒ $$\frac { 3x }{ (x-2)(x+1) }$$ = $$\frac { A(x+1)+B(x-2) }{ (x-2)(x+1) }$$
⇒ 3x = A(x + 1) + B (x – 2)
⇒ 3x = (A + B)x + (A – 2B) …………………… (2)
Comparing the coefficient of x from both sides,
3 = A + B
0 = A – 2B
3 = 3B ⇒ B = 1
and A = 2B = 2
image 40 40 40
= 2 log(x – 2) + log (x + 1) + c.

Question 11.
Evaluate $$\int { \frac { x^{ 2 }+1 }{ x^{ 4 }-x^{ 2 }+1 } }$$ dx?
Solution:

Question 12.
Evaluate $$\int { \frac { x^{ 2 }+1 }{ x^{ 4 }+x^{ 2 }+1 } }$$ dx?
Solution:
Solve like Q.No.11.

Question 13.
Evaluate $$\int { \frac { dx }{ \sqrt { x^{ 2 }+2x+3 } } }$$?
Solution:
Let

Question 14.
Integrate $$\frac { 1 }{ 1+sin^{ 2 }x }$$ with respect to x?
Solution:
Let

Question 15.
Evaluate $$\int { \frac { cos2x }{ (cosx+sinx)^{ 2 } } }$$ dx? (NCERT)
Solution:
Let

Taking cos x + sin x = t
$$\frac { d }{ dx }$$ (cosx + sinx) = $$\frac { dt }{ dx }$$
⇒ (-sinx + cos x) = $$\frac { dt }{ dx }$$
∴ I = ∫ $$\frac { dt }{ t }$$
⇒ I = logt + c
⇒ I = log(cos x + sinx) + c.

Question 16.
Evaluate $$\int { \frac { 1+tanx }{ x+logsecx } }$$dx?
Solution:
Let

Question 17.
Evaluate $$\int { \frac { cotx }{ log(sinx) } }$$ dx?
Solution:
Let I = $$\int { \frac { cotx }{ log(sinx) } }$$ dx
Again let log(sin x) = t
$$\frac{d}{dx}$$ log (sin x) = $$\frac{dt}{dx}$$
Taking sin x = u,

⇒ I = log t + c
⇒ I = log log(sin x) + c.

Question 18.
Evaluate $$\int { \frac { 2cosx-3sinx }{ 6cosx+4sinx } }$$ dx? (NCERT)
Solution:
Let I = $$\int { \frac { 2cosx-3sinx }{ 6cosx+4sinx } }$$ dx
Again let 6 cosx + 4 sin x = t,

Question 19.
Evaluate $$\int { e^{ 3logx } } (x^{ 4 }+1)^{ -1 }$$ dx? (NCERT)
Solution:
Let I = $$\int { e^{ 3logx } } (x^{ 4 }+1)^{ -1 }$$ dx

Put x4 + 1 = t,

Question 20.
Evaluate $$\int { \frac { dx }{ x-\sqrt { x } } }$$? (NCERT)
Solution:
Let I = $$\int { \frac { dx }{ x-\sqrt { x } } }$$

Put $$\sqrt{x}$$ – 1 = t

Question 21.
Evaluate $$\int { \frac { dx }{ 1+3sin^{ 2 }x } }$$?
Solution:
Let I = $$\int { \frac { dx }{ 1+3sin^{ 2 }x } }$$

Integration Long Answer Type Questions – II

Question 1.
Evaluate $$\int \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d x$$ dx?
Solution:
Let I = $$\int \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d x$$ dx
Again let x = tan θ ⇒ dx = sec2 θ dθ

Question 2.
Integrate $$\frac{e^{m \tan ^{-1} x}}{\left(1+x^{2}\right)^{\frac{3}{2}}}$$ with respect to x?
Solution:
Let tan-1 x = t ⇒ x = tan t
dx = sec2 t dt

= emtsin t – m[emt(-cos t) – $$\int { me^{ mt } }$$ (-cos t) dt]
= emtsin t + memtcos t – m2 $$\int { e^{ mt } }$$ cos t dt
= emt (sin t + m cos t) – m2 I

Question 3.
Evaluate $$\int { \frac { x^{ 2 }tan^{ -1 }x }{ 1+x^{ 2 } } }$$ dx?
Solution:
Let I = $$\int { \frac { x^{ 2 }tan^{ -1 }x }{ 1+x^{ 2 } } }$$ dx
Let x = tan θ ⇒ θ = tan-1 x
⇒ dx = sec2 θdθ

Question 4.
Evaluate $$\int { tan^{ -1 }\frac { 2x }{ 1+x^{ 2 } } }$$ dx?
Solution:
Let I = $$\int { tan^{ -1 }\frac { 2x }{ 1+x^{ 2 } } }$$ dx
Let x = tan θ ⇒ dx = sec2 θdθ

Question 5.
Evaluate $$\int { \frac { xtan^{ -1 }x }{ (1+x^{ 2 })^{ 3/2 } } }$$ dx?
Solution:
Let I = $$\int { \frac { xtan^{ -1 }x }{ (1+x^{ 2 })^{ 3/2 } } }$$ dx
Let x = tan θ ⇒ dx = sec2 θdθ

Question 6.
Evaluate $$\int { \frac { dx }{ 3+2cosx } }$$?
Solution:
Let I = $$\int { \frac { dx }{ 3+2cosx } }$$

Again let tan $$\frac { x }{ 2 }$$ = t,

Question 7.
Evaluate $$\int { \frac { dx }{ 4+5cosx } }$$?
Solution:
Solve like Q.No.6.

Question 8.
Evaluate $$\int { \frac { dx }{ 5-3cosx } }$$?
Solution:
Solve like Q.No.6.

Question 9.
Evaluate $$\int { \frac { 1 }{ 4+5sinx } }$$ dx?
Solution:

Let tan $$\frac { x }{ 2 }$$ = t ⇒ sec2 $$\frac { x }{ 2 }$$.$$\frac { 1 }{ 2 }$$.dx = dt

Question 10.
Evaluate $$\int { \frac { e^{ x }(1+sinx) }{ (1+cosx) } }$$ dx?
Solution:
Let I = $$\int { \frac { e^{ x }(1+sinx) }{ (1+cosx) } }$$ dx

Question 11.
Evaluate $$\int { \frac { xe^{ x } }{ (1+x)^{ 2 } } }$$ dx?
Solution:
Let I = $$\int { \frac { xe^{ x } }{ (1+x)^{ 2 } } }$$ dx = $$\int { \frac { (1+x-1)e^{ x } }{ (1+x)^{ 2 } } }$$

Question 12.
Evaluate $$\int { \frac { 2cosx }{ (1-sinx)(1+sin^{ 2 }x) } }$$ dx? (CBSE 2018)
Solution:
Let I = $$\int { \frac { 2cosx }{ (1-sinx)(1+sin^{ 2 }x) } }$$ dx
Put sin x = t, cos x dx = dt

⇒ 2 = A(1 + t2) + (1 – t) (Bt + 1)
⇒ 2 = A + At2 + Bt – Bt2 + 1 – t
⇒ 2 = (A – B)t2 + (B – 1) t + (A + 1)
Comparing the cofficient of like terms
∴ A – B = 0
B – 1 = 0
and A + 1 = 2
∴ B = 1, A = 1

Question 13.
Evaluate $$\int { \frac { dx }{ e^{ x }-1 } }$$? (NCERT)
Solution:
Let I = $$\int { \frac { dx }{ e^{ x }-1 } }$$ = $$\int { \frac { e^{ x }dx }{ e^{ x }(e^{ x }-1) } }$$
Again let ex = t, then exdx = dt

Question 14.
Evaluate $$\int { \frac { dx }{ x(x^{ n }+1) } }$$? (NCERT)
Solution:
Let I = $$\int { \frac { dx }{ x(x^{ n }+1) } }$$

Put xn = t,

Question 15.
Evaluate $$\int { (\sqrt { tanx } +\sqrt { cotx) } }$$ dx? (NCERT)
Solution:
Let

Again let, sin x – cos x = t
(cos x + sin x) dx = dt
(sin x – cos x)2 = t2
sin2x + cos2x – 2sinxcos x = t2
⇒ 1 – sin 2x = t2
⇒ sin 2x = 1 – t2