JEE Main Indefinite Integrals Previous Year Questions With Solutions

JEE Previous year questions on Indefinite integrals gives students the opportunity to learn the right method of solving questions related to important concepts like indefinite integral, integration using partial fractions and integration by parts. The solutions provided here will serve as a great reference tool for students who are preparing for JEE exams. About 2-4 questions are asked from this topic in JEE Examination. Students are advised to download the set of past year important questions which will help them to go through and revise solutions quickly before the exam.

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JEE Main Indefinite Integration Past Year Questions With Solutions

Question 1: Evaluate x1(x+1)3ex dx\int_{{}}^{{}}{\frac{x-1}{{{(x+1)}^{3}}}{{e}^{x}}\ dx}

Solution:

x1(x+1)3  exdx\int \frac{x-1}{(x+1)^3} \; e^x dx =ex((x+1)(x+1)32(x+1)3)dx=\int_{{}}^{{}}{{{e}^{x}}\left( \frac{(x+1)}{{{(x+1)}^{3}}}-\frac{2}{{{(x+1)}^{3}}} \right)\,dx} =ex(1(x+1)22(x+1)3)dx=\int_{{}}^{{}}{{{e}^{x}}\left( \frac{1}{{{(x+1)}^{2}}}-\frac{2}{{{(x+1)}^{3}}} \right)\,dx} =ex(x+1)2+c=\frac{{{e}^{x}}}{{{(x+1)}^{2}}}+c

Using formula, ddx(1(x+1)2)=2(x+1)3\frac{d}{dx}\left( \frac{1}{{{(x+1)}^{2}}} \right) =- \frac{2}{(x+1)^3}

Question 2: If ex(1+sinx)dx1+cosx=exf(x)+c,\int_{{}}^{{}}{\frac{{{e}^{x}}(1+\sin x)dx}{1+\cos x}={{e}^{x}}f(x)+c}, then find f(x)f(x).

Solution:

I=ex(1+sinx1+cosx)dx=ex[1+2sin(x/2)cos(x/2)2cos2(x/2)]dxI=ex[12sec2(x/2)+tan(x/2)]dx=ex.tan(x/2)+cI=\int_{{}}^{{}}{{{e}^{x}}\left( \frac{1+\sin x}{1+\cos x} \right)\,dx}=\int_{{}}^{{}}{{{e}^{x}}\left[ \frac{1+2\sin (x/2)\,\cos (x/2)}{2{{\cos }^{2}}(x/2)} \right]dx}\\ I=\int_{{}}^{{}}{{{e}^{x}}\left[ \frac{1}{2}{{\sec }^{2}}(x/2)+\tan (x/2) \right]\,dx}={{e}^{x}}.\tan (x/2)+c

Because, ex[f(x)+f(x)]dx=ex.f(x)+c}\int_{{}}^{{}}{{{e}^{x}}[f(x)+{f}'(x)\,]dx={{e}^{x}}.\,f(x)+c\}}

Question 3: Evaluate xsin2x dx\int_{{}}^{{}}{x{{\sin }^{2}}x\ dx}

Solution:

xsin2xdx=x.(1cos2x)2dx=12[xdxx.cos2xdx]=x24x4sin2x18cos2x+c.\int_{{}}^{{}}{x{{\sin }^{2}}x\,dx}=\int_{{}}^{{}}{x\,.\,\frac{(1-\cos 2x)}{2}\,dx}\\ =\frac{1}{2}\left[ \int_{{}}^{{}}{x\,dx}-\int_{{}}^{{}}{x\,.\,\cos 2x\,dx} \right]\\ =\frac{{{x}^{2}}}{4}-\frac{x}{4}\sin 2x-\frac{1}{8}\cos 2x+c.

Question 4: Evaluate x2sin2x dx\int_{{}}^{{}}{{{x}^{2}}\sin 2x}\ dx

Solution:

I=x2sin2xdx=x2cos2x2+2xcos2x2dx+c=x2cos2x2+xsin2x2+cos2x4+c.I=\int_{{}}^{{}}{{{x}^{2}}\sin 2x\,dx}=\frac{-{{x}^{2}}\cos 2x}{2}+\int_{{}}^{{}}{\frac{2x\cos 2x}{2}\,dx}+c \\ =-\frac{{{x}^{2}}\cos 2x}{2}+\frac{x\sin 2x}{2}+\frac{\cos 2x}{4}+c.

Question 5: Evaluate exsinx dx\int_{{}}^{{}}{{{e}^{x}}\sin x\ dx}

Solution:

I=exsinxdx=exsinxexcosxdx+c=exsinxexcosxexsinxdx+c2I=ex(sinxcosx)+cI=12ex(sinxcosx)+c.I=\int_{{}}^{{}}{{{e}^{x}}\sin x\,dx}={{e}^{x}}\sin x-\int_{{}}^{{}}{{{e}^{x}}\cos x\,dx+c}\\ ={{e}^{x}}\sin x-{{e}^{x}}\cos x-\int_{{}}^{{}}{{{e}^{x}}\sin x\,dx+c}\\ 2I={{e}^{x}}(\sin x-\cos x)+c\\ I=\frac{1}{2}{{e}^{x}}(\sin x-\cos x)+c.

Question 6: Evaluate cosx dx\int_{{}}^{{}}{\cos \sqrt{x}\ dx}

Solution:

Put x=t12xdx=dtdx=2tdt,\sqrt{x}=t\Rightarrow \frac{1}{2\sqrt{x}}\,dx=dt\Rightarrow dx=2t\,dt,

It reduces to

2t.costdt=2[t.sintsintdt]=2tsint+2cost=2[xsinx+cosx]+c.\int_{{}}^{{}}{2t\,.\cos t\,dt}=2\left[ t\,.\,\sin t-\int_{{}}^{{}}{\sin t\,dt} \right]\\ =2t\sin t+2\cos t\\ =2[\sqrt{x}\sin \sqrt{x}+\cos \sqrt{x}]+c.

Question 7: Simplify xsin1x1x2 dx\int_{{}}^{{}}{\frac{x{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}\ }dx

Solution:

Putting sin1x=t11x2dx=dt,{{\sin }^{-1}}x=t\Rightarrow \frac{1}{\sqrt{1-{{x}^{2}}}}\,dx=dt,

We get

xsin1x1x2dx=tsintdt=tcost+sint+c=sin1xcos(sin1x)+sin(sin1x)+c=xsin1x1x2+c.\int_{{}}^{{}}{\frac{x{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}\,dx=}\int_{{}}^{{}}{t\sin t\,dt=-t\cos t+\sin t+c}\\ =-{{\sin }^{-1}}x\cos ({{\sin }^{-1}}x)+\sin ({{\sin }^{-1}}x)+c\\ =x-{{\sin }^{-1}}x\sqrt{1-{{x}^{2}}}+c.

Question 8: Evaluate dxsinx+cosx\int_{{}}^{{}}{\frac{dx}{\sin x+\cos x}}

Solution:

dxsinx+cosx=12dxsinxcosπ4+cosxsinπ4=12cosec (x+π4)dx=12logtan(π8+x2)+c.\int_{{}}^{{}}{\frac{dx}{\sin x+\cos x}}=\frac{1}{\sqrt{2}}\int_{{}}^{{}}{\frac{dx}{\sin x\cos \frac{\pi }{4}+\cos x\sin \frac{\pi }{4}}} \\ =\frac{1}{\sqrt{2}}\int_{{}}^{{}}{\text{cosec }\left( x+\frac{\pi }{4} \right)\,dx=\frac{1}{\sqrt{2}}\log \tan \left( \frac{\pi }{8}+\frac{x}{2} \right)}+c.

Question 9: Evaluate x2+a2dx\int{\sqrt{{{x}^{2}}+{{a}^{2}}}\,\,dx}

Solution:

I=x2+a2dx=x2+a2.1dx=x2+a21dx[ddx(x2+a2)1dx]dx=xx2+a2[2x2x2+a2x]dx=xx2+a2[x2+a2a2x2+a2]dx=xx2+a2[x2+a2a2x2+a2]dx=xx2+a2x2+a2dx+a2dxx2+a2=x2+a2I+a2log[x+x2+a2]+C2I=xx2+a2+a2log[x+x2+a2]+CI=x2x2+a2+a22log[x+x2+a2]+C.I=\int{\sqrt{{{x}^{2}}+{{a}^{2}}}\,dx}\\ =\int{\sqrt{{{x}^{2}}+{{a}^{2}}}\,.1\,dx}\\ =\sqrt{{{x}^{2}}+{{a}^{2}}}\int{1dx-\int{\left[ \frac{d}{dx}\left( \sqrt{{{x}^{2}}+{{a}^{2}}} \right)\int{1\,dx} \right]\,dx}}\\ =x\sqrt{{{x}^{2}}+{{a}^{2}}}-\int{\left[ \frac{2x}{2\sqrt{{{x}^{2}}+{{a}^{2}}}}x \right]}\,dx\\ =x\sqrt{{{x}^{2}}+{{a}^{2}}}-\int{\left[ \frac{{{x}^{2}}+{{a}^{2}}-{{a}^{2}}}{\sqrt{{{x}^{2}}+{{a}^{2}}}} \right]}dx\\ =x\sqrt{{{x}^{2}}+{{a}^{2}}}-\int{\left[ \sqrt{{{x}^{2}}+{{a}^{2}}}-\frac{{{a}^{2}}}{\sqrt{{{x}^{2}}+{{a}^{2}}}} \right]\,}dx \\ =x\sqrt{{{x}^{2}}+{{a}^{2}}}-\int{\sqrt{{{x}^{2}}+{{a}^{2}}}dx+{{a}^{2}}\int{\frac{dx}{\sqrt{{{x}^{2}}+{{a}^{2}}}}}} \\ =\sqrt{{{x}^{2}}+{{a}^{2}}}-I+{{a}^{2}}\log \left[ x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right]+C\\ 2I=x\sqrt{{{x}^{2}}+{{a}^{2}}+}{{a}^{2}}\log \left[ x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right]+C\\ I=\frac{x}{2}\sqrt{{{x}^{2}}+{{a}^{2}}}+\frac{{{a}^{2}}}{2}\log \left[ x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right]+C.

Question 10: Evaluate (x+1)2dxx(x2+1)\int{\frac{{{(x+1)}^{2}}\,\,dx}{x({{x}^{2}}+1)}}

Solution:

(x+1)2dxx(x2+1)\int{\frac{{{(x+1)}^{2}}\,\,dx}{x({{x}^{2}}+1)}} =x2+1+2xx(x2+1)dx=\int{\frac{{{x}^{2}}+1+2x}{x({{x}^{2}}+1)}dx} =x2+1x(x2+1)dx+2xx(x2+1)dx=\int{\frac{{{x}^{2}}+1}{x({{x}^{2}}+1)}dx+2\int{\frac{x}{x({{x}^{2}}+1)}dx}}

 

=dxx+2dxx2+1=logex+2tan1x+c=\int{\frac{dx}{x}+2\int{\frac{dx}{{{x}^{2}}+1}}}={{\log }_{e}}x+2{{\tan }^{-1}}x+c

Question 11: Find the value of A and B if sinxsin(xα)    dx=Ax+B  log  sin(xα)+C\int \frac{sinx}{sin(x – \alpha)} \; \; dx = Ax + B \; log \; sin(x – \alpha) + C

Choose the right option:

(a) (sin α, cos α)                    (b) (cos α, sin α)

(c) (- sinα, cosα)                   (d) (-cosα, sinα)

Solution: Correct option is (b)

Explanation:

Let x – α = m

=> sin(α+m)sin  m    dm=sinαcot  m  dm+cosαdm\int \frac{sin (\alpha + m)}{sin \; m} \; \; dm = sin \alpha \int cot \; m \; dm + cos \alpha \int dm

= cos α(x – α) + sin α ln|sin m| + c

Therefore, on comparing we have

A = cos α and B = sin α

Question 12: Solve ((log  x1)1+(log  x)2)2\int (\frac{(log \; x – 1)}{1 + (log \; x)^2})^2 dx

Solution:

((log  x1)1+(log  x)2)2\int (\frac{(log \; x – 1)}{1 + (log \; x)^2})^2 dx

= (log  x1)2(1+(log  x)2)2\int \frac{(log \; x – 1)^2}{(1 + (log \; x)^2)^2} dx

= [11+(logx)22logx1+(logx)2]\int[\frac{1}{1+(logx)^2} – \frac{2logx}{1+(logx)^2}]

Put log x = t so dx = e^t dt

=> [et1+t22tet(1+t2)2]  dt\int[\frac{e^t}{1+t^2} – \frac{2te^t}{(1+t^2)^2}] \; dt

=> et[11+t22t(1+t2)2]  dt\int e^t[\frac{1}{1+t^2} – \frac{2t}{(1+t^2)^2}] \; dt

= et1+t2+c\frac{e^t}{1+t^2} + c

= x1+(log  x)2+c\frac{x}{1+(log \; x)^2} + c

Question 13: Find the value of 2sin  xsin(xπ/4)  dx\sqrt{2} \int \frac{sin \; x}{sin(x – \pi/4)}\; dx

Solution:

2sin  xsin(xπ/4)  dx\sqrt{2} \int \frac{sin \; x}{sin(x – \pi/4)}\; dx

= 2sin  (xπ/4+π/4)sin(xπ/4)  dx\sqrt{2} \int \frac{sin \; (x – \pi/4 + \pi/4)}{sin(x – \pi/4)}\; dx

= 2[cos(π/4)+cot(xπ/4)sinπ/4]  dx\sqrt{2} \int [cos(\pi/4) + cot(x – \pi/4)sin \pi/4]\; dx

= dx+cot(xπ/4)  dx\int dx + \int cot(x – \pi/4) \; dx

= x + ln|sin(x – π/4)| + c

Also Visit :-  Integral Calculus Previous Year Questions With Solutions

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