12th Grade > Mathematics
INDEFINITE INTEGRATION MCQs
Total Questions : 30
| Page 1 of 3 pages
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.
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 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 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
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. -> 13
:
A
Put<br> 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<br> 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 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 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. -> 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
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