12th Grade > Mathematics
INDEFINITE INTEGRATION MCQs
Total Questions : 30
| Page 3 of 3 pages
Answer: Option B. -> (32,12,−1)
:
B
ddx(A|cosx+sinx−2|+Bx+C)=Acosx−sinxcosx+sinx−2+B=Acosx+Bcosx−Asinx+Bsinx−2Bcosx+sinx−2∴A+B=2,A+B=−1,−2B=λ∴A=32,B=12,λ=−1
:
B
ddx(A|cosx+sinx−2|+Bx+C)=Acosx−sinxcosx+sinx−2+B=Acosx+Bcosx−Asinx+Bsinx−2Bcosx+sinx−2∴A+B=2,A+B=−1,−2B=λ∴A=32,B=12,λ=−1
Answer: Option C. -> 2(1+x1/3)3/2+C
:
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=6tdt
=∫6t2dt
=2t3+c
=2(1+x3)3/2+c
:
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=6tdt
=∫6t2dt
=2t3+c
=2(1+x3)3/2+c
Answer: Option D. -> −√x2−1x2
:
D
∫x2−2x3√x2−1dx=∫dxx√x2−1−2∫dxx3√x2−1=sec−1x−2∫secθtanθsec3θtanθdθ=[putx=secθ⇒dx=secθtanθdθ]=sec−1x−2∫cos2θdθ=sec−1x−2∫(1+cos2θ)dθ=sec−1x−(θ+sin2θ2)+C=sec−1x−sec−1x−√x2−1x2+C=−√x2−1x2+C
:
D
∫x2−2x3√x2−1dx=∫dxx√x2−1−2∫dxx3√x2−1=sec−1x−2∫secθtanθsec3θtanθdθ=[putx=secθ⇒dx=secθtanθdθ]=sec−1x−2∫cos2θdθ=sec−1x−2∫(1+cos2θ)dθ=sec−1x−(θ+sin2θ2)+C=sec−1x−sec−1x−√x2−1x2+C=−√x2−1x2+C
Answer: Option A. -> sin−1[ax+bxc]+k
:
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
:
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
Answer: Option C. -> etan−1x.(sec−1(√1+x2))2+C
:
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
:
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
Answer: Option B. -> log|tan−1(secx+cosx)|+c
:
B
Put tan−1(secx+cosx)=f(x)⇒f′(x)=sin3xcos4x+3cos2x+1∴∫f′(x)f(x)dx=log|f(x)|+c∴Integral=log|tan−1(secx+cosx)|+c
:
B
Put tan−1(secx+cosx)=f(x)⇒f′(x)=sin3xcos4x+3cos2x+1∴∫f′(x)f(x)dx=log|f(x)|+c∴Integral=log|tan−1(secx+cosx)|+c
Answer: Option D. -> 0
:
D
fog(x)=√ex−1∴I=∫√ex−1dx=∫2t2t2+1dt{wheret=√ex−1}=2t−2tan−1t+C=2√ex−1−2tan−1(√ex−1)+C=2fog(x)−2tan−1(fog(x))+C∴A+B=2+(−2)=0
:
D
fog(x)=√ex−1∴I=∫√ex−1dx=∫2t2t2+1dt{wheret=√ex−1}=2t−2tan−1t+C=2√ex−1−2tan−1(√ex−1)+C=2fog(x)−2tan−1(fog(x))+C∴A+B=2+(−2)=0
Answer: Option B. -> 2f(x)g(x)1+(g(x))2
:
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)
:
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)
Answer: Option B. -> 2
:
B
I=∫cosx4dxsin3x(sin5x+cos5x)35=∫cosec2xcot4xdx(1+cot5x)3/51+cot5x=t.5cot4x(−cosec2x)dx=dt=−15∫dtt3/5=−t2/52=−12(1+cot5x)2/5+C.
:
B
I=∫cosx4dxsin3x(sin5x+cos5x)35=∫cosec2xcot4xdx(1+cot5x)3/51+cot5x=t.5cot4x(−cosec2x)dx=dt=−15∫dtt3/5=−t2/52=−12(1+cot5x)2/5+C.
Answer: Option A. -> x sin−1(x)+√1−x2+C
:
A
We have ϕ(x)=limn→∞x2n−1x2n+1=1,0<x<1∴∫sin−1x.ϕ(x)dx=∫sin−1xdx=[xsin−1x+√1−x2]+c
:
A
We have ϕ(x)=limn→∞x2n−1x2n+1=1,0<x<1∴∫sin−1x.ϕ(x)dx=∫sin−1xdx=[xsin−1x+√1−x2]+c
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