Free Integrals 01 Practice Test - 12th Grade - Commerce 

Question 1

∫(2+sec x)sec x(1+2sec x)2dx

A. 12cosec x+cot x+C
B. 2cosec x+cot x+C
C. 12cosec x−cot x+C
D. 2cosec x−cot x+C
 

SOLUTION

Solution : A

∫(2cos x+1)(2+cos x)2dx=∫(2+cos x)cos x+sin2x(2+cos x)2dx=∫cos x2+cos xdx−∫−sin2x(2+cos x)2dx=sin x2+cos x+c

Question 2

Let g(x) be an antiderivative for f(x). Then  In (1+(g(x))2) is an antiderivate for

A. 2f(x)g(x)1+(f(x))2
B. 2f(x)g(x)1+(g(x))2
C. 2f(x)1+(f(x))2
D. None

SOLUTION

Solution : B

Given ∫f(x)dx=g(x)g′(x)=f(x)
Now ddx(In(1+g2(x))=2g(x)g′(x)1+g2(x)2f(x)g(x)1+g2(x)

Question 3

If ∫(x−1x+1)dx√x3+x2+x=2tan−1√f(x)+C, find f(x).

A. x+1x+1
B. x+1x+2
C. x−1x+1
D. x−1x−1

SOLUTION

Solution : A

I=∫(x−1)dx(x+1)x√x+1+1x=∫(x−1)(x+1)dx(x+1)2x√x+1+1x=∫(1−1x2)dx(x+1x+2)√x+1x+1Put x+1+1x=t2(1−1x2)dx=2t dt=∫2t dt(t2+1)t=2tan−1t+c=2tan−1(√x+1x+1)+c
f(x) = x+1x+1
 

Question 4

∫etan−1x(1+x2)[(sec−1√1+x2)2+cos−1(1−x21+x2)]dx(x>0)

A. etan−1x.tan−1x+C
 
B. etan−1x.(tan−1x)22+C
 
C. etan−1x.(sec−1(√1+x2))2+C
 
D. etan−1x.(cos−1(√1+x2))2+C

SOLUTION

Solution : C

note that sec−1√1+x2=tan−1x;cos−1(1−x21+x2)=2tan−1x for x > 0
I=∫etan−1x1+x2((tan−1x)2+2tan−1x)dx
Put tan−1x=t
=∫et(t2+2t)dt=et.t2=etan−1x((tan−1x)2)+C

Question 5

Let f(x)=2sin2x−1cosx+cosx(2sinx+1)1+sinx then ∫ex(f(x)+f′(x))dx equals
(where c is the constant of integeration)

A. extanx+c
B. excotx+c
C. excosec2x+c
D. None of these

SOLUTION

Solution : A

cosx(1+2sinx)1+sinx−cos2x−sin2xcosx=cos2x(1+2sinx)−(1+sinx)(cos2x−sin2x)cosx(1+sinx)
=sinxcos2x+sin2x(1+sinx)cosx(1+sinx)=sinx(1−sinx)+sin2xcosx=tanx

Question 6

∫(1+√tanx)(1+tan2x)2tanxdx equal to 

A. logtan2x+√tanx+c
B. logtan2x+12√tanx+c
C. log|tanx|+2√tanx+c
D. log|tanx|+√tanx+c

SOLUTION

Solution : A

∫12sinxcosxdx+12∫√tanxsinxcosxdx=12∫sin2x+cos2xsinxcosxdx+12∫sec2x√tanxdx=log(tan2x)+√tanx+c

Question 7

∫x2−2x3√x2−1dx is equal to

A. x2√x2−1
B. −x2√x2−1
C. √x2−1x2
D. −√x2−1x2

SOLUTION

Solution : D

∫x2−2x3√x2−1dx=∫dxx√x2−1−2∫dxx3√x2−1=sec−1x−2∫sec θ tan θsec3 θ tan θdθ=[put x=sec θ⇒dx=sec θ tan θ dθ]=sec−1x−2∫cos2 θ dθ=sec−1x−2∫(1+cos 2θ)dθ=sec−1x−(θ+sin 2 θ2)+C=sec−1x−sec−1x−√x2−1x2+C=−√x2−1x2+C

Question 8

∫√1+3√x3√x2dx is equal to

A. (1+x1/3)3/2+C
B. −(1+x1/3)3/2+C
C. 2(1+x1/3)3/2+C
D. None of these

SOLUTION

Solution : C

∫√1+3√x3√x2dx=∫x−23(1+x13)1/2dx
=∫x−2/3(1+x1/3)1/2dx
1+x1/3=t2
x−2/3dx=6 t dt
=∫6 t2dt
=2t3+c
=2(1+x3)3/2+c

Question 9

If Φ(x)=limn→∞xn−x−nxn+x−n,0<x<1,ϵN, then ∫(sin−1x)(Φ(x))dx is equal to

A. x sin−1(x)+√1−x2+C
B. −(x sin−1(x)+√1−x2)
C. x sin−1 x+√1−x2+C
D. None of these

SOLUTION

Solution : A

We have ϕ(x)=limn→∞x2n−1x2n+1=1,0<x<1∴∫sin−1x.ϕ(x) dx=∫sin−1x dx=[x sin−1 x+√1−x2]+c

Question 10

The value of ∫ax2−bx√c2x2−(ax2+b)2dx is equal to

A. sin−1[ax+bxc]+k
B. sin−1[ax2+bx2c]+k
C. cos−1[ax+bxc]+k
D. cos−1[ax2+bx2c]+k

SOLUTION

Solution : A

I=∫a−bx2√c2−(ax2+bx)2dx=∫(a−bx2)√c2−(ax+bx)2dx
Put ax+bx=t⇒(a−bx2)dx=dt∴I=∫dt√c2−t2=sin−1[ax+bxc]+k
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