12th Grade > Mathematics
INTEGRALS MCQs
Total Questions : 30
| Page 3 of 3 pages
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. -> (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 B. -> log|sec(x2+12)|+C
:
B
We have, ∫x√2sin(x2+1)−sin2(x2+1)2sin(x2+1)+sin2(x2+1)dx=∫x√2sin(x2+1)−2sin(x2+1)cos(x2+1)2sin(x2+1)+2sin(x2+1)cos(x2+1)dx=∫x√1−cos(x2+1)1+cos(x2+1)dx=∫xtan(x2+12)dx=∫tan(x2+12)d(x2+12)=log|sec(x2+12)|+C
:
B
We have, ∫x√2sin(x2+1)−sin2(x2+1)2sin(x2+1)+sin2(x2+1)dx=∫x√2sin(x2+1)−2sin(x2+1)cos(x2+1)2sin(x2+1)+2sin(x2+1)cos(x2+1)dx=∫x√1−cos(x2+1)1+cos(x2+1)dx=∫xtan(x2+12)dx=∫tan(x2+12)d(x2+12)=log|sec(x2+12)|+C
Answer: Option A. -> 12√2log|1+√2 sin 2x1−√2 sin 2x|−12(log|sec 2x+tan 2x|)+C
:
A
∫sin(4x−2x)dxsin(2x)cos(2x)cos(4x)=∫sin(4x)dxsin(2x)cos(4x)−∫sec2xdx=2∫cos2xdxcos4x−12(log|sec2x+tan2x|)
:
A
∫sin(4x−2x)dxsin(2x)cos(2x)cos(4x)=∫sin(4x)dxsin(2x)cos(4x)−∫sec2xdx=2∫cos2xdxcos4x−12(log|sec2x+tan2x|)
Answer: Option D. -> xex(1+x2)1/2+c
:
D
I=∫ex[x√1+x2+1(1+x2)3/2]dx
Let f(x)=x√1+x2⇒f′(x)=√1+x2−x2√1+x2(1+x2)
=1(1+x2)3/2
∴I=exf(x)+c
=exx√1+x2+c
:
D
I=∫ex[x√1+x2+1(1+x2)3/2]dx
Let f(x)=x√1+x2⇒f′(x)=√1+x2−x2√1+x2(1+x2)
=1(1+x2)3/2
∴I=exf(x)+c
=exx√1+x2+c
Answer: Option A. -> 1√2tan−1(1√2tan 2x)+C
:
A
∫dxsin4x+cos4x=∫dx(sin2x+cos2x)2−2sin2xcos2x=∫2dx2−sin22x=∫2sec22x2sec22x−tan22xdx=2∫sec22x2+tan22xdx=∫dt(√2)2+t2[puttingtan2x=t⇒2sec22xdx=dt]=1√2tan−1(t√2)+C=1√2tan−1(1√2tan2x)+C
:
A
∫dxsin4x+cos4x=∫dx(sin2x+cos2x)2−2sin2xcos2x=∫2dx2−sin22x=∫2sec22x2sec22x−tan22xdx=2∫sec22x2+tan22xdx=∫dt(√2)2+t2[puttingtan2x=t⇒2sec22xdx=dt]=1√2tan−1(t√2)+C=1√2tan−1(1√2tan2x)+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 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. -> ∑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. -> sinx−2(sinx)−1−6tan−1(sinx)+c
:
C
sinx=t;I=∫(1−t2)(2−t2)t2(1+t2)dt=∫(1+2t2−61+t2)dt
=sinx−2(sinx)−1−6tan−1(sinx)+c
:
C
sinx=t;I=∫(1−t2)(2−t2)t2(1+t2)dt=∫(1+2t2−61+t2)dt
=sinx−2(sinx)−1−6tan−1(sinx)+c
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