12th Grade > Mathematics
INTEGRALS MCQs
Total Questions : 30
| Page 2 of 3 pages
Answer: Option B. -> log|tan−1(secx+cosx)|+c
:
B
Put tan−1(secx+cosx)=f(x)f1(x)=sin3xcos4x+3cos2x+1∴∫f1(x)f(x)=log|f(x)|+c
:
B
Put tan−1(secx+cosx)=f(x)f1(x)=sin3xcos4x+3cos2x+1∴∫f1(x)f(x)=log|f(x)|+c
Answer: Option C. -> 12{f(x2) g′(x2)−g(x2) f′(x2)}+c
:
C
Put x2=t⇒I=12∫{f(t)g′′(t)−g(t)f′′(t)}dt=12{f(t)g′(t)−∫f′(t)g′(t)−g(t)f′(t)+∫g′(t)f′(t)dt}+c∴I=12{f(t)g′(t)−g(t)f′(t)}+c=12{f(x2)g′(x2)−g(x2)f′(x2)}+c
:
C
Put x2=t⇒I=12∫{f(t)g′′(t)−g(t)f′′(t)}dt=12{f(t)g′(t)−∫f′(t)g′(t)−g(t)f′(t)+∫g′(t)f′(t)dt}+c∴I=12{f(t)g′(t)−g(t)f′(t)}+c=12{f(x2)g′(x2)−g(x2)f′(x2)}+c
Answer: Option C. -> a=116
:
C
∫cos8x+1tan2x−cot2xdx=∫2cos24xsin22x−cos22x.sin2xcos2xdx=−∫sin4xcos4xdx=−12∫sin8xdx=116cos8x+C∴a=116
:
C
∫cos8x+1tan2x−cot2xdx=∫2cos24xsin22x−cos22x.sin2xcos2xdx=−∫sin4xcos4xdx=−12∫sin8xdx=116cos8x+C∴a=116
Answer: Option A. -> 3 sin x−(3x+4) cos x+c
:
A
In the interval (0,π), sin x is positive, therefore, (3x+4)|sinx|=(3x+4)sinx.
∴ The antiderivative of (3x+4)|sinx| is
=∫(3x+4)sinxdx=−(3x+4)cosx+∫3cosxdx=−(3x+4)cosx+3sinx+c
:
A
In the interval (0,π), sin x is positive, therefore, (3x+4)|sinx|=(3x+4)sinx.
∴ The antiderivative of (3x+4)|sinx| is
=∫(3x+4)sinxdx=−(3x+4)cosx+∫3cosxdx=−(3x+4)cosx+3sinx+c
Answer: Option D. -> tan−1(x)+13tan−1(x3)+c
:
D
I=∫x4+11+x6dx=∫(x4−x2+1)+x2(1+x6)dx=∫x4−x2+11+x6dx+∫x21+x6dx=∫11+x2dx+13∫3x21+x6dx=tan−1(x)+13tan−1x3+c
:
D
I=∫x4+11+x6dx=∫(x4−x2+1)+x2(1+x6)dx=∫x4−x2+11+x6dx+∫x21+x6dx=∫11+x2dx+13∫3x21+x6dx=tan−1(x)+13tan−1x3+c
Answer: Option A. -> 125
:
A
tanx=t⇒sec2xdx=dt∴f(x)=∫dxsin12xcos12x.cos4x=∫(1+tan2x)sec2x√tanxdx=∫(1+t2)√tdt=∫(t−1/2+t3/2)dt=2t1/2+25t5/2=2√tanx+25(tanx)5/2∴ϕ(π4)−ϕ(0)=2+25=125
:
A
tanx=t⇒sec2xdx=dt∴f(x)=∫dxsin12xcos12x.cos4x=∫(1+tan2x)sec2x√tanxdx=∫(1+t2)√tdt=∫(t−1/2+t3/2)dt=2t1/2+25t5/2=2√tanx+25(tanx)5/2∴ϕ(π4)−ϕ(0)=2+25=125
Answer: Option A. -> x+1x+1
:
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+1Putx+1+1x=t2(1−1x2)dx=2tdt=∫2tdt(t2+1)t=2tan−1t+c=2tan−1(√x+1x+1)+c
f(x) = x+1x+1
:
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+1Putx+1+1x=t2(1−1x2)dx=2tdt=∫2tdt(t2+1)t=2tan−1t+c=2tan−1(√x+1x+1)+c
f(x) = x+1x+1
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. -> extanx+c
:
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
:
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
Answer: Option A. -> logtan2x+√tanx+c
:
A
∫12sinxcosxdx+12∫√tanxsinxcosxdx=12∫sin2x+cos2xsinxcosxdx+12∫sec2x√tanxdx=log(tan2x)+√tanx+c
:
A
∫12sinxcosxdx+12∫√tanxsinxcosxdx=12∫sin2x+cos2xsinxcosxdx+12∫sec2x√tanxdx=log(tan2x)+√tanx+c
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