Chapter 7 Integrals | class 12th | Important Question for Maths

Class 12 Maths Chapter 7 Important Extra Questions Integrals

Integrals Important Extra Questions Very Short Answer Type

Question 1.
Find ∫3+3cosxx+sinxdx   (C.B.S.E. Sample Paper 2019-20)
Solution:
I = ∫3+3cosxx+sinxdx = 3 log lx + sin xl + c.
[∵ Num. = ddx denom.]

Question 2.
Find : ∫(cos² 2x – sin² 2x)dx.   (C.B.S.E. Sample Paper 2019-20)
Solution:
I = ∫cos 4x dx = sin4x4+ c.

Question 3.
Find : ∫ dx5−4x−2×2√   (C.B.S.E. Outside Delhi 2019)
Solution:

Question 4.
Evaluate ∫ x3−1×2 dx (N.C.E.R.T. C.B.S.E. 2010C)
Solution:

Question 5.
Find : ∫sin2x−cos2xsinxcosxdx (A.I.C.B.S.E. 2017)
Solution:

Question 6.
Write the value of ∫dxx2+16
Solution:

Question 7.
Evaluate: ∫ (x³ + 1)dx.   (C.B.S.E. Sample Paper 2019-20)
Solution:
I = ∫2−2x3dx+∫2−21⋅dx = I₁
⇒ 0 + [x]2−2 [∵ I₁ is an odd function] = 2 – (-2) = 4.
⇒ 2 – (-2) = 4.

Question 8.
Evaluate: ∫π/20 e^x (sin x -cosx)dx. (C.B.S.E. 2014)
Solution:
∫π/20 e^x (sin x -cosx)dx
∫π/20e^x (-cos x + sinx)dx
|“Form: ∫e^x (f(x) + f'(x) dx”
= [ex(−cosx)]π/20
= -e ^π/2cosπ2 + e⁰ cos 0
= -e ^π/2 (0) + (1) (1)
= -0 + 1 = 1

Question 9.
Evaluate: ∫204−x2−−−−−√dx. (A.I.C.B.S.E. 2014)
Solution:

= [0 + 2 sin^-1(1)] – [0 + 0]
= 2sin^-1(1)= 2(π/2) = π

Question 10.
Evaluate : If f(x) = ∫x0 t sin t dt, then write the value of f’ (x). (A.I. C.B.S.E. 2014)
Solution:
We have : f(x) = ∫x0 t sin t dt.
f'(x) = x sin x. ddx (x) – 0
[Property XII ; Leibnitz’s Rule]
= x sin x . (1)
= x sin x.

Question 11.
Prove that: ∫2a0 f(x)dx = ∫2a0 f(2a-x)dx. o o
Solution:
Put x = 2a – t so that dx = – dt.
When x = 0, t – 2a. When x = 2a, t – 0.
∫2a0 f(x)dx = ∫02a f(2a-t)(-dt)
=∫02a f{2a-t)dt = ∫2a0 f(2a-t)dt o
[Property II]
= ∫2a0 (2a – x) dx, [Property I]
which is true.

Integrals Important Extra Questions Short Answer Type

Question 1.
Evaluate :
∫cos2x+2sin2xcos2xdx (C.B.S.E)
Solution:

Question 2.
Find : ∫sec2xtan2x+4√dx
Solution:
I = ∫sec2xtan2x+4√dx
Put tan x = t so that sec² x dx = dt.
∴ I = ∫dtt2+22√
= log |t + t2+4−−−−−√| + C
= log |tan x + tan2+4−−−−−−−√| + C

Question 3.
Find : ∫1−sin2x−−−−−−−−√dx,π4<x<π2
Solution:

Question 4.
Find ∫sinx . log cos x dx (C.B.S.E 2019 C)
Solution:
∫sinx . log cos x dx
Put cox x = t
so that – sin x dx = dt
i.e., sin x dx = – dt.
∴ I = -∫log t.1dt
= -[ log t.t – ∫ 1/t. t dt ]
[Integrating by parts]
= – [t log t – t] + C = f(1 – log t) + C
= cos x (1 – log (cos x)) + C.

Question 5.
Find : ∫(x2+sin2x)sec2x1+x2dx (CBSE Sample Paper 2018-19)
Solution:

Question 6.
Evaluate ∫ex(x−3)(x−1)3dx (CBSE Sample Paper 2018-19)
Solution:
I = ∫ex(x−3)(x−1)3dx

Question 7.
Find ∫sin^-1 (2x)dx
Solution:

Question 8.
Evaluate : ∫π−π (1 – x²) sin x cos² x dx.
Solution:
Here, f(x)=( 1-x²) sin x cos² x.
f(x) = (1 – x²) sin (-x) cos² (-x)
= – (1 – x²) sin x cos² x
= -f(x)
⇒ f is an odd function.
Hence, I = 0.

Question 9.
Evaluate : ∫2−1|x|xdx dx.
Solution:

Question 10.
Find ∫3−5sinxcos2xdx   (C.B.S.E. 2018 C)
Solution:
∫3−5sinxcos2xdx
= 3∫sce² x dx – 5∫sec x tan x dx
= 3tan x – 5sec x + C

Question 11.
Find :
∫tan2xsec2x1−tan6xdx   (C.B.S.E. 2019 (Delhi))
Solution:
Let I = ∫tan2xsec2x1−tan6xdx
Put tan³ x = t
so that 3 tan² x sec² x dx = dt
i.e tan² x sec²x dx = dt3

Question 12.
Find : ∫ sin x .log cos x dx.   (CBSE 2019C)
Solution:
I = ∫ sin x .log cos x dx.
Put cos x = t
i.e. sinx dx = -dt
∴ I = – ∫log t.1 dt
= -[logt.t – ∫1/t . t. dt]
[Integrating by parts]
= – [t log t – t] + C
= t(1 – log t) + C
= cos x (1 – log (cos JC)) + C.

Question 13.
Evaluate : ∫π−π (1 – x²) sin x cos² x dx   (C.B.S.E. 2019 (Delhi))
Solution:
Here, f(x) = (1 – x²) sin x cos² x
∴ f(-x) – (1 – x²) sin (-x) cos² (-x)
= – (1 – x²) sin x cos² x
= -f(x)
⇒ f is an odd function.
Hence, I = 0.

Question 14.
Evaluate ∫2−1|x|xdx   (C.B.S.E. 2019 (Delhi))
Solution:
 

Question 15.
Find : ∫0−π/41+tanx1−tanxdx
Solution:

Integrals Important Extra Questions Long Answer Type 1

Question 1.
Evaluate : ∫sin6x+cos6xsin2xcos2xdx   (C.B.S.E. 2019 (Delhi))
Solution:

Question 2.
Integrate the function cos(x+a)sin(x+b) w.r.t. x. (C.B.S.E. 2019 (Delhi))
Solution:

Question 3.
Evaluate : ∫ x² tan^-1 x dx. (C.B.S.E. (F) 2012)
Solution:

Question 4.
Find : ∫[log (log x) + 1(logx)2 ] dx (N.C.E.R.T.; A.I.C.B.S.E. 2010 C)
Solution:
Let ∫[log (log x) + 1(logx)2 ] dx
= ∫ log(log x)dx + ∫1(logx)2 dx …… (1)
Let I = I₁ + I₂
Now I₁ = ∫ log (log x) dx
=∫ log (log x) 1 dx
= log (log x).x – ∫ 1logx⋅xx.dx
(Integrating by parts)
= xlog(logx) – ∫ 1logxdx ……….. (2)
Let I₁ = I₃ + I₄

Putting in (2),
I₁ = x log (x) – xlogx−∫1(logx)2 dx
Putting in (1),
I = x log (log x)

Question 5.
Integrate : ∫ e^x ( tan^-1 x + 11+x2 ) dx   (N.C.E.R.T.)
Solution:
∫ e^x ( tan^-1 x + 11+x2 ) dx
[From ∫ e^xf(x) + f'(x) ]dx”]
= ∫ e^x tan^-1 x dx +∫ e^x 11+x2 ) dx
= ∫ tan^-1 x. e^x dx +∫ 11+x2 ) e^x dx
= tan^-1 x. e^x – ∫ 11+x2 ) e^x dx
+∫ 11+x2 ) e^xdx
(integrating first integral by parts)
= e^x tan^-1x + c.