12th Grade > Mathematics
INDEFINITE INTEGRATION MCQs
Total Questions : 30
| Page 2 of 3 pages
Answer: Option B. -> 54(x−3x+1)1/5+c
:
B
I=∫dx(x−3)4/5(x+1)6/5Putt=x−3x+1dt=4(x+1)2dx.I=14∫dtt4/5=14(t−4/5+1−4/5+1)=54t1/5=54(x−3x+1)1/5
:
B
I=∫dx(x−3)4/5(x+1)6/5Putt=x−3x+1dt=4(x+1)2dx.I=14∫dtt4/5=14(t−4/5+1−4/5+1)=54t1/5=54(x−3x+1)1/5
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 C. -> tan x
:
C
∫1−7cos2xsin7xcos2xdx=∫(sec2xsin7x−7sin7x)dx=∫sec2xsin7xdx−∫7sin7dx=I1+I2Now,I1=∫sec2xsin7dx=tanxsin7x+7∫tanxcosxsin8x=tanxsin7x−I2∴I1+I2=tanxsin7x+C⇒f(x)=tanx
:
C
∫1−7cos2xsin7xcos2xdx=∫(sec2xsin7x−7sin7x)dx=∫sec2xsin7xdx−∫7sin7dx=I1+I2Now,I1=∫sec2xsin7dx=tanxsin7x+7∫tanxcosxsin8x=tanxsin7x−I2∴I1+I2=tanxsin7x+C⇒f(x)=tanx
Answer: Option A. -> ∑9n=1tannxn
:
A
We have In=∫tannxdx=∫tann−2x(sec2x−1)dx=tann−1xn−1−In−2i.e.In−2+In=tann−1xn−1(n≥2)
Thus, we have
I0+I1+2(I2+....+I8)+I9+I10=(I0+I2)+(I1+I3)+(I2+I4)+(I3+I5)+(I4+I6)+(I5+I7)+(I6+I8)+(I7+I9)+(I8+I10).=∑10n=2tann−1xn−1=∑9n=1tannxn
:
A
We have In=∫tannxdx=∫tann−2x(sec2x−1)dx=tann−1xn−1−In−2i.e.In−2+In=tann−1xn−1(n≥2)
Thus, we have
I0+I1+2(I2+....+I8)+I9+I10=(I0+I2)+(I1+I3)+(I2+I4)+(I3+I5)+(I4+I6)+(I5+I7)+(I6+I8)+(I7+I9)+(I8+I10).=∑10n=2tann−1xn−1=∑9n=1tannxn
Answer: Option C. -> (x2+1)In2+12(In2+1)+C
:
C
I=∫x2In(x2+1)dx let x2+1=t;xdx=dt2
Hence I=12∫2Intdt=12∫tIn2dt=12.tIn2+1In2+1+C=12.(x2+1)In2+1In2+1+C
:
C
I=∫x2In(x2+1)dx let x2+1=t;xdx=dt2
Hence I=12∫2Intdt=12∫tIn2dt=12.tIn2+1In2+1+C=12.(x2+1)In2+1In2+1+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 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 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. -> 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. -> 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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