12th Grade > Mathematics
INTEGRALS MCQs
Total Questions : 30
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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 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 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 A. -> 12cosec x+cot x+C
:
A
∫(2cosx+1)(2+cosx)2dx=∫(2+cosx)cosx+sin2x(2+cosx)2dx=∫cosx2+cosxdx−∫−sin2x(2+cosx)2dx=sinx2+cosx+c
:
A
∫(2cosx+1)(2+cosx)2dx=∫(2+cosx)cosx+sin2x(2+cosx)2dx=∫cosx2+cosxdx−∫−sin2x(2+cosx)2dx=sinx2+cosx+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
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 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 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 A. -> 13
:
A
Put 1−x3=t2⇒−3x2dx=2tdt∴∫dxx√1−x3=∫x2x3√1−x3dx=23∫dtt2−1=13log∣∣t−1t+1∣∣+C=13log∣∣∣√1−x3−1√1−x3+1∣∣∣+C∴a=13
:
A
Put 1−x3=t2⇒−3x2dx=2tdt∴∫dxx√1−x3=∫x2x3√1−x3dx=23∫dtt2−1=13log∣∣t−1t+1∣∣+C=13log∣∣∣√1−x3−1√1−x3+1∣∣∣+C∴a=13
Answer: Option B. -> b=1,a=−13
:
B
I=∫dxx√1−x3Put1−x3=t23x2dx=−2tdtI=−23∫dt(1−t2)=−13∫(11−t+11+t)dt.=−13log1+t1−t+c=−13log1+√1−x31−√1−x3+c.
:
B
I=∫dxx√1−x3Put1−x3=t23x2dx=−2tdtI=−23∫dt(1−t2)=−13∫(11−t+11+t)dt.=−13log1+t1−t+c=−13log1+√1−x31−√1−x3+c.