Practice Start Test

Indefinite Integration

JEE Mathematics · 69 questions · Page 1 of 7 · Click an option or "Show Solution" to reveal answer

Q1
If ∫sin⁡xsin⁡(x−α)dx=Ax+Blog⁡sin⁡(x−α),+C,\int {{{\sin x} \over {\sin \left( {x - \alpha } \right)}}dx = Ax + B\log \sin \left( {x - \alpha } \right), + C,} then value of (A,B)(A, B) is
A (−cos⁡α,sin⁡α)\left( { - \cos \alpha ,\sin \alpha } \right)
B (cos⁡α,sin⁡α)\left( { \cos \alpha ,\sin \alpha } \right)
C (−sin⁡α,cos⁡α)\left( { - \sin \alpha ,\cos \alpha } \right)
D (sin⁡α,cos⁡α)\left( { \sin \alpha ,\cos \alpha } \right)
Correct Answer
Option B
Solution
∫sin⁡xsin⁡(x−α)dx\int {{{\sin x} \over {\sin \left( {x - \alpha } \right)}}} dx
=∫sin⁡(x−α+α)sin⁡(x−α)dx= \int {{{\sin \left( {x - \alpha + \alpha } \right)} \over {\sin \left( {x - \alpha } \right)}}} dx
=∫sin⁡(x−α)cos⁡α+cos⁡(x−α)sin⁡αsin⁡(x−α)= \int {{{\sin \left( {x - \alpha } \right)\cos \alpha + \cos \left( {x - \alpha } \right)\sin \alpha } \over {\sin \left( {x - \alpha } \right)}}}
=∫{cos⁡α+sin⁡α cot⁡(x−α)}dx= \int {\left\{ {\cos \alpha + \sin \alpha \,\cot \left. {\left( {x - \alpha } \right)} \right\}} \right.} dx
=(cos⁡α)x+(sin⁡α)log⁡ sin⁡(x−α)+C= \left( {\cos \alpha } \right)x + \left( {\sin \alpha } \right)\log \,\sin \left( {x - \alpha } \right) + C

∴\therefore

A=cos⁡α,A = \cos \alpha ,
B=sin⁡αB = \sin \alpha
Q2
The integral ∫2x3−1x4+xdx\int {{{2{x^3} - 1} \over {{x^4} + x}}} dx is equal to : (Here C is a constant of integration)
A log⁡e∣x3+1∣x2+C{\log _e}{{\left| {{x^3} + 1} \right|} \over {{x^2}}} + C
B 12log⁡e∣x3+1∣x2+C{1 \over 2}{\log _e}{{\left| {{x^3} + 1} \right|} \over {{x^2}}} + C
C log⁡e∣x3+1x∣+C{\log _e}\left| {{{{x^3} + 1} \over x}} \right| + C
D 12log⁡e(x3+1)2∣x3∣+C{1 \over 2}{\log _e}{{{{\left( {{x^3} + 1} \right)}^2}} \over {\left| {{x^3}} \right|}} + C
Correct Answer
Option C
Solution
∫2x3−1x4+xdx=∫2x−x−2x2+x−1dx=ln⁡(x2+x−1)+c\int {{{2{x^3} - 1} \over {{x^4} + x}}dx = \int {{{2x - {x^{ - 2}}} \over {{x^2} + {x^{ - 1}}}}dx = \ln ({x^2} + {x^{ - 1}}} } ) + c
⇒ln⁡(x3+1)−ln⁡x+c\Rightarrow \ln ({x^3} + 1) - \ln x + c
Q3
If ∫x5e−x2dx=g(x)e−x2+c\int {{x^5}} {e^{ - {x^2}}}dx = g\left( x \right){e^{ - {x^2}}} + c, where c is a constant of integration, then gg(–1) is equal to :
A 1
B - 1
C −52 - {5 \over 2}
D −12 - {1 \over 2}
Correct Answer
Option C
Solution

Let x2 = t

⇒12∫t2e−tdt\Rightarrow {1 \over 2}\int {{t^2}{e^{ - t}}dt}
⇒12[−t2e−t+∫2te−tdt]\Rightarrow {1 \over 2}\left[ { - {t^2}{e^{ - t}} + \int {2t{e^{ - t}}dt} } \right]
⇒−t2e−t2−te−t−e−t\Rightarrow {{ - {t^2}{e^{ - t}}} \over 2} - t{e^{ - t}} - {e^{ - t}}
⇒(−x42−x2−1)e−x2+c\Rightarrow \left( { - {{{x^4}} \over 2} - {x^2} - 1} \right){e^{ - {x^2}}} + c

Then

g(x)=−x42−x2−1g(x) = - {{{x^4}} \over 2} - {x^2} - 1
⇒g(−1)=−12−1−1\Rightarrow g( - 1) = - {1 \over 2} - 1 - 1
⇒−52\Rightarrow - {5 \over 2}
Q4
∫dxcos⁡x−sin⁡x\int {{{dx} \over {\cos x - \sin x}}} is equal to
A 12log⁡∣tan⁡(x2+3π8)∣+C{1 \over {\sqrt 2 }}\log \left| {\tan \left( {{x \over 2} + {{3\pi } \over 8}} \right)} \right| + C
B 12log⁡∣cot⁡(x2)∣+C{1 \over {\sqrt 2 }}\log \left| {\cot \left( {{x \over 2}} \right)} \right| + C
C 12log⁡∣tan⁡(x2−3π8)∣+C{1 \over {\sqrt 2 }}\log \left| {\tan \left( {{x \over 2} - {{3\pi } \over 8}} \right)} \right| + C
D  12log⁡∣tan⁡(x2−π8)∣+C\,{1 \over {\sqrt 2 }}\log \left| {\tan \left( {{x \over 2} - {\pi \over 8}} \right)} \right| + C
Correct Answer
Option A
Solution
∫dxcos⁡x−sin⁡x\int {{{dx} \over {\cos x - \sin x}}}
=∫dx2cos⁡(x+π4)= \int {{{dx} \over {\sqrt 2 \cos \left( {x + {\pi \over 4}} \right)}}}
=12∫sec⁡(x+π4)dx= {1 \over {\sqrt 2 }}\int {\sec \left( {x + {\pi \over 4}} \right)dx}
=12log⁡∣tan⁡(π4+x2+π8)∣+C= {1 \over {\sqrt 2 }}\log \left| {\tan \left( {{\pi \over 4} + {x \over 2} + {\pi \over 8}} \right)} \right| + C
[ \left[ \, \right.

As

∫sec⁡x dx=log⁡∣tan⁡(π4+x2)∣\int {\sec x\,dx} = \log \left| {\tan \left( {{\pi \over 4} + {x \over 2}} \right)} \right|
 ]\left. \, \right]
=12log⁡∣tan⁡(x2+3π8)∣+C= {1 \over {\sqrt 2 }}\log \left| {\tan \left( {{x \over 2} + {{3\pi } \over 8}} \right)} \right| + C
Q5
∫{(log⁡x−1)1+(log⁡x)2}2  dx\int {{{\left\{ {{{\left( {\log x - 1} \right)} \over {1 + {{\left( {\log x} \right)}^2}}}} \right\}}^2}\,\,dx} is equal to
A log⁡x(log⁡x)2+1+C{{\log x} \over {{{\left( {\log x} \right)}^2} + 1}} + C
B xx2+1+C{x \over {{x^2} + 1}} + C
C xex1+x2+C{{x{e^x}} \over {1 + {x^2}}} + C
D x(log⁡x)2+1+C{x \over {{{\left( {\log x} \right)}^2} + 1}} + C
Correct Answer
Option D
Solution
∫(log⁡x−1)2(1+(log⁡x)2)2dx\int {{{{{\left( {\log x - 1} \right)}^2}} \over {{{\left( {1 + {{\left( {\log x} \right)}^2}} \right)}^2}}}} dx
=∫1+(log⁡x)2−2log⁡x[1+(log⁡x)2]2= \int {{{1 + {{\left( {\log x} \right)}^2} - 2\log x} \over {{{\left[ {1 + {{\left( {\log x} \right)}^2}} \right]}^2}}}}
=∫[1(1+(log⁡x)2)−2log⁡x(1+(log⁡x)2)2]dx= \int {\left[ {{1 \over {\left( {1 + {{\left( {\log x} \right)}^2}} \right)}} - {{2\log x} \over {{{\left( {1 + {{\left( {\log x} \right)}^2}} \right)}^2}}}} \right]} dx
=∫[et1+t2−2t et(1+t2)2]dt= \int {\left[ {{{{e^t}} \over {1 + {t^2}}} - {{2t\,{e^t}} \over {{{\left( {1 + {t^2}} \right)}^2}}}} \right]} dt

put

log⁡x=t⇒dx=et dt\log x = t \Rightarrow dx = {e^t}\,dt
=∫et[11+t2−2t(1+t2)2]dt= \int {{e^t}} \left[ {{1 \over {1 + {t^2}}} - {{2t} \over {{{\left( {1 + {t^2}} \right)}^2}}}} \right]dt
[ \left[ \, \right.

which is of the form

∫ex(f(x)+f′(x)dx)]\left. {\int {{e^x}\left( {f\left( x \right) + f'\left( x \right)dx} \right)} } \right]
=et1+t2+c=x1+(log⁡x)2+c= {{{e^t}} \over {1 + {t^2}}} + c = {x \over {1 + {{\left( {\log x} \right)}^2}}} + c
Q6
∫dxcos⁡x+3sin⁡x\int {{{dx} \over {\cos x + \sqrt 3 \sin x}}} equals
A log⁡ tan⁡ (x2+π12)+C\log \,\tan \,\left( {{x \over 2} + {\pi \over {12}}} \right) + C
B log⁡ tan⁡ (x2−π12)+C\log \,\tan \,\left( {{x \over 2} - {\pi \over {12}}} \right) + C
C  12 log⁡ tan⁡ (x2+π12)+C\,{1 \over 2}\,\log \,\tan \,\left( {{x \over 2} + {\pi \over {12}}} \right) + C
D  12 log⁡ tan⁡ (x2−π12)+C\,{1 \over 2}\,\log \,\tan \,\left( {{x \over 2} - {\pi \over {12}}} \right) + C
Correct Answer
Option C
Solution
I=∫dxcos⁡x+3sin⁡xI = \int {{{dx} \over {\cos x + \sqrt 3 \sin x}}}
⇒I=∫dx2[12cos⁡x+32sin⁡x]\Rightarrow I = \int {{{dx} \over {2\left[ {{1 \over 2}\cos x + {{\sqrt 3 } \over 2}\sin x} \right]}}}
=12∫dx[sin⁡π6cos⁡x+cos⁡π6sin⁡x]= {1 \over 2}\int {{{dx} \over {\left[ {\sin {\pi \over 6}\cos x + \cos {\pi \over 6}\sin x} \right]}}}
=12.∫dxsin⁡(x+π6)= {1 \over 2}.\int {{{dx} \over {\sin \left( {x + {\pi \over 6}} \right)}}}
⇒I=12.∫cos⁡ec(x+π6)dx\Rightarrow I = {1 \over 2}.\int {\cos ec\left( {x + {\pi \over 6}} \right)dx}

But we know that

∫cos⁡ec x dx=log⁡∣(tan⁡x/2)∣+C\int {\cos ec\,x\,dx} = \log \left| {\left( {\tan x/2} \right)} \right| + C

∴\therefore

I=12.log⁡ tan⁡(x2+π12)+CI = {1 \over 2}.\log \,\tan \left( {{x \over 2} + {\pi \over {12}}} \right) + C
Q7
The value of 2∫sin⁡xdxsin⁡(x−π4)\sqrt 2 \int {{{\sin xdx} \over {\sin \left( {x - {\pi \over 4}} \right)}}} is
A  x+log⁡ ∣ cos⁡(x−π4) ∣+c\,x + \log \,\left| {\,\cos \left( {x - {\pi \over 4}} \right)\,} \right| + c
B  x−log⁡ ∣ sin⁡(x−π4) ∣+c\,x - \log \,\left| {\,\sin \left( {x - {\pi \over 4}} \right)\,} \right| + c
C  x+log⁡ ∣ sin⁡(x−π4) ∣+c\,x + \log \,\left| {\,\sin \left( {x - {\pi \over 4}} \right)\,} \right| + c
D  x−log⁡ ∣ cos⁡(x−π4) ∣+c\,x - \log \,\left| {\,\cos \left( {x - {\pi \over 4}} \right)\,} \right| + c
Correct Answer
Option C
Solution

Let

I=2∫sin⁡ xdxsin⁡(x−π4)I = \sqrt 2 \int {{{\sin \,xdx} \over {\sin \left( {x - {\pi \over 4}} \right)}}}

Put

x−π4=tx - {\pi \over 4} = t
⇒dx=dt\Rightarrow dx = dt
⇒I=2∫sin⁡(t+π4)sin⁡ tdt\Rightarrow I = \sqrt 2 \int {{{\sin \left( {t + {\pi \over 4}} \right)} \over {\sin \,t}}} dt
            =22∫(sin⁡t+cos⁡tsin⁡t)  dt\,\,\,\,\,\,\,\,\,\,\,\, = {{\sqrt 2 } \over {\sqrt 2 }}\int {\left( {{{\sin t + \cos t} \over {\sin t}}} \right)} \,\,dt
⇒I=∫(1+cot⁡ t)dt\Rightarrow I = \int {\left( {1 + \cot \,t} \right)} dt
            =t+log⁡∣sin⁡t∣+c1\,\,\,\,\,\,\,\,\,\,\,\, = t + \log \left| {\sin t} \right| + {c_1}
=x−π4+log⁡∣sin⁡(x−π4)∣+c1= x - {\pi \over 4} + \log \left| {\sin \left( {x - {\pi \over 4}} \right)} \right| + {c_1}
=x+log⁡∣sin⁡(x−π4)∣+c= x + \log \left| {\sin \left( {x - {\pi \over 4}} \right)} \right| + c
( \left( \, \right.

where

c=c1−π4{c = {c_1} - {\pi \over 4}}
 )\left. \, \right)
Q8
If the ∫5tan⁡xtan⁡x−2dx=x+a ln⁡ ∣sin⁡x−2cos⁡x∣+k,\int {{{5\tan x} \over {\tan x - 2}}dx = x + a\,\ln \,\left| {\sin x - 2\cos x} \right| + k,} then aa is equal to :
A −1-1
B −2-2
C 11
D 22
Correct Answer
Option D
Solution
∫5tan⁡xtan⁡x−2dx\int {{{5\tan x} \over {\tan x - 2}}} dx
=∫5sin⁡xcos⁡xsin⁡xcos⁡x−2 dx= \int {{{5{{\sin x} \over {\cos x}}} \over {{{\sin x} \over {\cos x}} - 2}}} \,dx
=∫(5sin⁡xcos⁡x×cos⁡xsin⁡x−2cos⁡x) dx= \int {\left( {{{5\sin x} \over {\cos x}} \times {{\cos x} \over {\sin x - 2\cos x}}} \right)} \,dx
=∫5 sin⁡ x dxsin⁡x−2 cos⁡x= \int {{{5\,\sin \,x\,dx} \over {\sin x - 2\,\cos x}}}
=∫(4sin⁡x+sin⁡x+2cos⁡x−2cos⁡xsin⁡x−2cos⁡x) dx= \int {\left( {{{4\sin x + \sin x + 2\cos x - 2\cos x} \over {\sin x - 2\cos x}}} \right)} \,dx
=∫(sin⁡x−2cos⁡x)+(4sin⁡x+2cos⁡x)sin⁡x−2cos⁡x dx= \int {{{\left( {\sin x - 2\cos x} \right) + \left( {4\sin x + 2\cos x} \right)} \over {\sin x - 2\cos x}}} \,dx
=∫(sin⁡x−2cos⁡x)+2(cos⁡x+2sin⁡x)(sin⁡x−2cos⁡x) dx= \int {{{\left( {\sin x - 2\cos x} \right) + 2\left( {\cos x + 2\sin x} \right)} \over {\left( {\sin x - 2\cos x} \right)}}} \,dx
=∫sin⁡x−2cos⁡xsin⁡x−2cos⁡xdx+2∫(cos⁡x+2sin⁡xsin⁡x−2cos⁡x) dx= \int {{{\sin x - 2\cos x} \over {\sin x - 2\cos x}}dx + 2\int {\left( {{{\cos x + 2\sin x} \over {\sin x - 2\cos x}}} \right)} } \,dx
=∫dx+2∫cos⁡x+2sin⁡xsin⁡x−2cos⁡ x dx= \int {dx + 2\int {{{\cos x + 2\sin x} \over {\sin x - 2\cos \,x}}} } \,dx
=I1+I2= {I_1} + {I_2}

where

I1=∫dx{I_1} = \int {dx}

and

I2=2∫cos⁡x+2sin⁡xsin⁡x−2cos⁡x dx{I_2} = 2\int {{{\cos x + 2\sin x} \over {\sin x - 2\cos x}}\,dx}

Put

sin⁡x−2cos⁡x=t\sin x - 2\cos x = t
⇒(cos⁡x+2sin⁡x)dx=dt\Rightarrow \left( {\cos x + 2\sin x} \right)dx = dt

∴\therefore

I2=2∫dtt=2ln⁡ t+C{I_2} = 2\int {{{dt} \over t}} = 2\ln \,t + C
=2 ln⁡(sin⁡ x−2cos⁡ x)+C= 2\,\ln \left( {\sin \,x - 2\cos \,x} \right) + C

Hence,

I1+I2{I_1} + {I_2}
=∫dx+2ln⁡(sin⁡x−2cos⁡x)+c= \int {dx + 2\ln \left( {\sin x - 2\cos x} \right) + c}
=x+2ln⁡∣(sin⁡x−2cos⁡x)∣+k= x + 2\ln \left| {\left( {\sin x - 2\cos x} \right)} \right| + k
⇒a=2\Rightarrow a = 2
Q9
If ∫f(x)dx=ψ(x),\int {f\left( x \right)dx = \psi \left( x \right),} then ∫x5f(x3)dx\int {{x^5}f\left( {{x^3}} \right)dx} is equal to
A 13[x3ψ(x3)−∫x2ψ(x3)dx]+C{1 \over 3}\left[ {{x^3}\psi \left( {{x^3}} \right) - \int {{x^2}\psi \left( {{x^3}} \right)dx} } \right] + C
B 13x3ψ(x3)−3∫x3ψ(x3)dx+C{1 \over 3}{x^3}\psi \left( {{x^3}} \right) - 3\int {{x^3}\psi \left( {{x^3}} \right)dx} + C
C 13x3ψ(x3)−∫x2ψ(x3)dx+C{1 \over 3}{x^3}\psi \left( {{x^3}} \right) - \int {{x^2}\psi \left( {{x^3}} \right)dx} + C
D 13[x3ψ(x3)−∫x3ψ(x3)dx]+C{1 \over 3}\left[ {{x^3}\psi \left( {{x^3}} \right) - \int {{x^3}\psi \left( {{x^3}} \right)dx} } \right] + C
Correct Answer
Option C
Solution

Let

∫f(x)dx=ψ(x)\int {f\left( x \right)dx = \psi \left( x \right)}

Let

I=∫x5f(x3)dxI = \int {{x^5}} f\left( {{x^3}} \right)dx

put

x3=t⇒3x2dx=dt{x^3} = t \Rightarrow 3{x^2}dx = dt
I=13∫3.x2.x3.f(x3).dxI = {1 \over 3}\int {3.{x^2}} .{x^3}.f\left( {{x^3}} \right).dx
=13∫tf(t)dt= {1 \over 3}\int {tf} \left( t \right)dt
=13[t∫f(t)dt−∫f(t)dt]= {1 \over 3}\left[ {t\int {f\left( t \right)dt - \int {f\left( t \right)dt} } } \right]
=13[tψ(t)−∫ψ(t)dt]= {1 \over 3}\left[ {t\psi \left( t \right) - \int {\psi \left( t \right)dt} } \right]
=13[x3ψ(x3)−3∫x2ψ(x3)dx]+C= {1 \over 3}\left[ {{x^3}\psi \left( {{x^3}} \right) - 3\int {{x^2}\psi \left( {{x^3}} \right)dx} } \right] + C
=13x3ψ(x3)−∫x2ψ(x3)dx+C= {1 \over 3}{x^3}\psi \left( {{x^3}} \right) - \int {{x^2}} \psi \left( {{x^3}} \right)dx + C
Q10
The integral ∫dxx2(x4+1)3/4\int {{{dx} \over {{x^2}{{\left( {{x^4} + 1} \right)}^{3/4}}}}} equals :
A −(x4+1)14+c - {\left( {{x^4} + 1} \right)^{{1 \over 4}}} + c
B −(x4+1x4)14+c - {\left( {{{{x^4} + 1} \over {{x^4}}}} \right)^{{1 \over 4}}} + c
C (x4+1x4)14+c {\left( {{{{x^4} + 1} \over {{x^4}}}} \right)^{{1 \over 4}}} + c
D (x4+1)14+c {\left( {{x^4} + 1} \right)^{{1 \over 4}}} + c
Correct Answer
Option B
Solution
1=∫dxx2(x4+1)3/41 = \int {{{dx} \over {{x^2}{{\left( {{x^4} + 1} \right)}^{3/4}}}}}
=∫dxx3(1+x−4)3/4= \int {{{dx} \over {{x^3}{{\left( {1 + {x^{ - 4}}} \right)}^{3/4}}}}}

Let

x−4=y{x^{ - 4}} = y
⇒−4x−3 dx=dy\Rightarrow - 4{x^{ - 3}}\,dx = dy
⇒dx=−14x3dy\Rightarrow dx = {{ - 1} \over 4}{x^3}dy

∴\therefore

I=−14∫x3dyx3(1+y)3/4I = {{ - 1} \over 4}\int {{{{x^3}dy} \over {{x^3}{{\left( {1 + y} \right)}^{3/4}}}}}
=−14∫dy(1+y)3/4= {{ - 1} \over 4}\int {{{dy} \over {{{\left( {1 + y} \right)}^{3/4}}}}}
=−14×4(1+y)1/4= {{ - 1} \over 4} \times 4{\left( {1 + y} \right)^{1/4}}
=−1(1+x−4)1/4+C= - 1{\left( {1 + {x^{ - 4}}} \right)^{1/4}} + C
=−(x4+1x4)1/4+C= - {\left( {{{{x^4} + 1} \over {{x^4}}}} \right)^{1/4}} + C
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