JEE Previous year questions on Indefinite integrals allow students 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 be a great reference tool for students 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.
JEE Main Indefinite Integration Past Year Questions With Solutions
Question 1: EvaluateÂ
\(\begin{array}{l}\int_{{}}^{{}}{\frac{x-1}{{{(x+1)}^{3}}}{{e}^{x}}\ dx}\end{array} \)
Solution:
\(\begin{array}{l}\int \frac{x-1}{(x+1)^3} \; e^x dx\end{array} \)
\(\begin{array}{l}=\int_{{}}^{{}}{{{e}^{x}}\left( \frac{(x+1)}{{{(x+1)}^{3}}}-\frac{2}{{{(x+1)}^{3}}} \right)\,dx}\end{array} \)
\(\begin{array}{l}=\int_{{}}^{{}}{{{e}^{x}}\left( \frac{1}{{{(x+1)}^{2}}}-\frac{2}{{{(x+1)}^{3}}} \right)\,dx}\end{array} \)
\(\begin{array}{l}\text{Using the formula}\frac{d}{dx}\left( \frac{1}{{{(x+1)}^{2}}} \right) =- \frac{2}{(x+1)^3},\ \text{we get;}\end{array} \)
\(\begin{array}{l}=\frac{{{e}^{x}}}{{{(x+1)}^{2}}}+c\end{array} \)
Question 2: If
\(\begin{array}{l}\int_{{}}^{{}}{\frac{{{e}^{x}}(1+\sin x)dx}{1+\cos x}={{e}^{x}}f(x)+c},\end{array} \)
then find f(x).
Solution:
\(\begin{array}{l}I=\int_{{}}^{{}}{{{e}^{x}}\left( \frac{1+\sin x}{1+\cos x} \right)\,dx}=\int_{{}}^{{}}{{{e}^{x}}\left[ \frac{1+2\sin (x/2)\,\cos (x/2)}{2{{\cos }^{2}}(x/2)} \right]dx}\\ I=\int_{{}}^{{}}{{{e}^{x}}\left[ \frac{1}{2}{{\sec }^{2}}(x/2)+\tan (x/2) \right]\,dx}={{e}^{x}}.\tan (x/2)+c \end{array} \)
Because,
\(\begin{array}{l}\int_{{}}^{{}}{{{e}^{x}}[f(x)+{f}'(x)\,]dx={{e}^{x}}.\,f(x)+c\}}\end{array} \)
Question 3: Evaluate: ∫x sin2x dx
Solution:
\(\begin{array}{l}\int_{{}}^{{}}{x{{\sin }^{2}}x\,dx}=\int_{{}}^{{}}{x\,.\,\frac{(1-\cos 2x)}{2}\,dx}\\ =\frac{1}{2}\left[ \int_{{}}^{{}}{x\,dx}-\int_{{}}^{{}}{x\,.\,\cos 2x\,dx} \right]\\ =\frac{{{x}^{2}}}{4}-\frac{x}{4}\sin 2x-\frac{1}{8}\cos 2x+c.\end{array} \)
Question 4: Evaluate: ∫x2 sin 2x dx
Solution:
\(\begin{array}{l}I=\int_{{}}^{{}}{{{x}^{2}}\sin 2x\,dx}=\frac{-{{x}^{2}}\cos 2x}{2}+\int_{{}}^{{}}{\frac{2x\cos 2x}{2}\,dx}+c \\ =-\frac{{{x}^{2}}\cos 2x}{2}+\frac{x\sin 2x}{2}+\frac{\cos 2x}{4}+c.\end{array} \)
Question 5: Evaluate ∫ex sin x dx
Solution:
\(\begin{array}{l}I=\int_{{}}^{{}}{{{e}^{x}}\sin x\,dx}={{e}^{x}}\sin x-\int_{{}}^{{}}{{{e}^{x}}\cos x\,dx+c}\\ ={{e}^{x}}\sin x-{{e}^{x}}\cos x-\int_{{}}^{{}}{{{e}^{x}}\sin x\,dx+c}\\ 2I={{e}^{x}}(\sin x-\cos x)+c\\ I=\frac{1}{2}{{e}^{x}}(\sin x-\cos x)+c.\end{array} \)
Question 6: Evaluate ∫cos√x dx
Solution:
Put
\(\begin{array}{l}\sqrt{x}=t\Rightarrow \frac{1}{2\sqrt{x}}\,dx=dt\Rightarrow dx=2t\,dt,\end{array} \)
It reduces to
\(\begin{array}{l}\int_{{}}^{{}}{2t\,.\cos t\,dt}=2\left[ t\,.\,\sin t-\int_{{}}^{{}}{\sin t\,dt} \right]\\ =2t\sin t+2\cos t\\ =2[\sqrt{x}\sin \sqrt{x}+\cos \sqrt{x}]+c.\end{array} \)
Question 7:Â Simplify
\(\begin{array}{l}\int_{{}}^{{}}{\frac{x{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}\ }dx\end{array} \)
Solution:
Putting
\(\begin{array}{l}{{\sin }^{-1}}x=t\Rightarrow \frac{1}{\sqrt{1-{{x}^{2}}}}\,dx=dt,\end{array} \)
We get
\(\begin{array}{l}\int_{{}}^{{}}{\frac{x{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}\,dx=}\int_{{}}^{{}}{t\sin t\,dt=-t\cos t+\sin t+c}\\ =-{{\sin }^{-1}}x\cos ({{\sin }^{-1}}x)+\sin ({{\sin }^{-1}}x)+c\\ =x-{{\sin }^{-1}}x\sqrt{1-{{x}^{2}}}+c.\end{array} \)
Question 8: EvaluateÂ
\(\begin{array}{l}\int_{{}}^{{}}{\frac{dx}{\sin x+\cos x}}\end{array} \)
Solution:
\(\begin{array}{l}\int_{{}}^{{}}{\frac{dx}{\sin x+\cos x}}=\frac{1}{\sqrt{2}}\int_{{}}^{{}}{\frac{dx}{\sin x\cos \frac{\pi }{4}+\cos x\sin \frac{\pi }{4}}} \\ =\frac{1}{\sqrt{2}}\int_{{}}^{{}}{\text{cosec }\left( x+\frac{\pi }{4} \right)\,dx\\=\frac{1}{\sqrt{2}}\log \tan \left( \frac{\pi }{8}+\frac{x}{2} \right)}+c.\end{array} \)
Question 9: EvaluateÂ
\(\begin{array}{l}\int{\sqrt{{{x}^{2}}+{{a}^{2}}}\,\,dx}\end{array} \)
Solution:
\(\begin{array}{l}I=\int{\sqrt{{{x}^{2}}+{{a}^{2}}}\,dx}\\ =\int{\sqrt{{{x}^{2}}+{{a}^{2}}}\,.1\,dx}\\ =\sqrt{{{x}^{2}}+{{a}^{2}}}\int{1dx-\int{\left[ \frac{d}{dx}\left( \sqrt{{{x}^{2}}+{{a}^{2}}} \right)\int{1\,dx} \right]\,dx}}\\ =x\sqrt{{{x}^{2}}+{{a}^{2}}}-\int{\left[ \frac{2x}{2\sqrt{{{x}^{2}}+{{a}^{2}}}}x \right]}\,dx\\ =x\sqrt{{{x}^{2}}+{{a}^{2}}}-\int{\left[ \frac{{{x}^{2}}+{{a}^{2}}-{{a}^{2}}}{\sqrt{{{x}^{2}}+{{a}^{2}}}} \right]}dx\\ =x\sqrt{{{x}^{2}}+{{a}^{2}}}-\int{\left[ \sqrt{{{x}^{2}}+{{a}^{2}}}-\frac{{{a}^{2}}}{\sqrt{{{x}^{2}}+{{a}^{2}}}} \right]\,}dx \\ =x\sqrt{{{x}^{2}}+{{a}^{2}}}-\int{\sqrt{{{x}^{2}}+{{a}^{2}}}dx+{{a}^{2}}\int{\frac{dx}{\sqrt{{{x}^{2}}+{{a}^{2}}}}}} \\ =\sqrt{{{x}^{2}}+{{a}^{2}}}-I+{{a}^{2}}\log \left[ x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right]+C\\ 2I=x\sqrt{{{x}^{2}}+{{a}^{2}}+}{{a}^{2}}\log \left[ x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right]+C\\ I=\frac{x}{2}\sqrt{{{x}^{2}}+{{a}^{2}}}+\frac{{{a}^{2}}}{2}\log \left[ x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right]+C.\end{array} \)
Question 10: Evaluate
\(\begin{array}{l}\int{\frac{{{(x+1)}^{2}}\,\,dx}{x({{x}^{2}}+1)}}\end{array} \)
Solution:
\(\begin{array}{l}\int{\frac{{{(x+1)}^{2}}\,\,dx}{x({{x}^{2}}+1)}}\end{array} \)
\(\begin{array}{l}=\int{\frac{{{x}^{2}}+1+2x}{x({{x}^{2}}+1)}dx}\end{array} \)
\(\begin{array}{l}=\int{\frac{{{x}^{2}}+1}{x({{x}^{2}}+1)}dx+2\int{\frac{x}{x({{x}^{2}}+1)}dx}}\end{array} \)
\(\begin{array}{l}=\int{\frac{dx}{x}+2\int{\frac{dx}{{{x}^{2}}+1}}}={{\log }_{e}}x+2{{\tan }^{-1}}x+c\end{array} \)
Question 11: Find the value of A and B if
\(\begin{array}{l}\int \frac{sinx}{sin(x – \alpha)} \; \; dx = Ax + B \; log \; sin(x – \alpha) + C\end{array} \)
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
\(\begin{array}{l}\Rightarrow \int \frac{sin (\alpha + m)}{sin \; m} \; \; dm = sin \alpha \int cot \; m \; dm + cos \alpha \int dm\end{array} \)
= cos α(x – α) + sin α ln|sin m| + c
Therefore, by comparing, we have;
A = cos α and B = sin α
Question 12: Solve
\(\begin{array}{l}\int (\frac{(log \; x – 1)}{1 + (log \; x)^2})^2\; dx\end{array} \)
Solution:
\(\begin{array}{l}\int (\frac{(log \; x – 1)}{1 + (log \; x)^2})^2\; dx\end{array} \)
\(\begin{array}{l}=\int \frac{(log \; x – 1)^2}{(1 + (log \; x)^2)^2}\; dx\end{array} \)
\(\begin{array}{l}=\int[\frac{1}{1+(logx)^2} – \frac{2logx}{1+(logx)^2}]\end{array} \)
Put log x = t so dx = et dt
\(\begin{array}{l}=\int[\frac{e^t}{1+t^2} – \frac{2te^t}{(1+t^2)^2}] \; dt\end{array} \)
\(\begin{array}{l}=\int e^t[\frac{1}{1+t^2} – \frac{2t}{(1+t^2)^2}] \; dt\end{array} \)
\(\begin{array}{l}=\frac{e^t}{1+t^2} + c\end{array} \)
\(\begin{array}{l}=\frac{x}{1+(log \; x)^2} + c\end{array} \)
Question 13: Find the value of
\(\begin{array}{l}\sqrt{2} \int \frac{sin \; x}{sin(x – \pi/4)}\; dx\end{array} \)
Solution:
\(\begin{array}{l}\sqrt{2} \int \frac{sin \; x}{sin(x – \pi/4)}\; dx\end{array} \)
\(\begin{array}{l}=\sqrt{2} \int \frac{sin \; (x – \pi/4 + \pi/4)}{sin(x – \pi/4)}\; dx\end{array} \)
\(\begin{array}{l}=\sqrt{2} \int [cos(\pi/4) + cot(x – \pi/4)sin \pi/4]\; dx\end{array} \)
= x + ln|sin(x – Ï€/4)| + c
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