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Integrals of Particular Function - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Some Special Integration, Application of Special Integral Formula (Part 1) is considered one of the most asked concept.

  • 27 Questions around this concept.

Solve by difficulty

  Evaluate the integral of \int \frac{dx}{\sqrt{x^{2}+49}}:

Find the integral of\int \frac{dx}{9-x^{2}}

The integral value of \int\frac{x\ dx}{\sqrt{x^{4}+1}}

Concepts Covered - 3

Some Special Integration

\\\mathrm{1.\;\;\int \frac{d x}{x^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C}\\\\\mathrm{\;\;\;\;\;\;put\;\;x=a\tan\theta,\;then\;\;dx=a\sec^2\theta\;d\theta}\\\mathrm{\;\;\;\;\;\;\therefore \;\;\int \frac{d x}{x^{2}+a^{2}}=\int\frac{a\sec^2\theta}{a^2+a^2\tan^2\theta}\;d\theta}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\int\frac{a\sec^2\theta}{a^2\left (1+\tan^2\theta \right )}\;d\theta}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{1}{a}\int d\theta=\frac{1}{a}\theta+C=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C}

 

\\\mathrm{2.\;\;\int \frac{d x}{x^{2}-a^{2}}=\frac{1}{2a} \log\left | \frac{x-a}{x+a} \right |+C}\\\\\text{\;\;\;\;\;\;\;we can rewrite above integral as}\\\mathrm{\;\;\;\;\;\;\int \frac{d x}{x^{2}-a^{2}}=\frac{1}{2a}\int\left ( \frac{1}{x-a}-\frac{1}{x+a} \right )dx}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{1}{2a}\left ( \log|x-a|-\log|x+a| \right )+c}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{1}{2a} \log\left | \frac{x-a}{x+a} \right |+C}

 

\\\mathrm{3.\;\;\int \frac{d x}{a^{2}-x^{2}}=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+C}\\\\\mathrm{4.\;\;\int \frac{d x}{\sqrt{a^{2}-x^{2}}}=\sin ^{-1}\left(\frac{x}{a}\right)+C}\\\\\mathrm{5.\;\;\int \frac{d x}{\sqrt{a^{2}+x^{2}}}=\log |x+\sqrt{x^{2}+a^{2}}|+C}\\\\\mathrm{6.\;\;\int \frac{d x}{\sqrt{x^{2}-a^{2}}}=\log |x+\sqrt{x^{2}-a^{2}}|+C}

\\\\\mathrm{7.\;\;\int \sqrt{a^{2}-x^{2}} d x=\frac{1}{2} x \sqrt{a^{2}-x^{2}}+\frac{1}{2} a^{2} \sin ^{-1}\left(\frac{x}{a}\right)+C}\\\\\mathrm{8.\;\;\int \sqrt{a^{2}+x^{2}} d x =\frac{1}{2} x \sqrt{a^{2}+x^{2}}+\frac{1}{2} a^{2} \log |x+\sqrt{a^{2}+x^{2}}|+C}\\\\\mathrm{9.\;\;\int \sqrt{x^{2}-a^{2}} d x=\frac{1}{2} x \sqrt{x^{2}-a^{2}}-\frac{1}{2} a^{2} \log |x+\sqrt{x^{2}-a^{2}}|+C}

 

Following are some important substitutions useful in evaluating integrals.

\begin{array}{c|| c } \mathbf { Expression } & {\mathbf { Substitution }} \\\\ \hline \\a^{2}+x^{2} & {x=a \tan \theta} {\text { or }} {x=a \cot \theta} \\ \\\hline \\a^{2}-x^{2} & {x=a \sin \theta} {\text { or } x=a \cos \theta}\\ \\ \hline \\x^{2}-a^{2} & {x=a \sec \theta} {\text { or } x=a \csc \theta} \\\\ \hline\\ \sqrt{\frac{a-x}{a+x}} {\text { or } \sqrt{\frac{a+x}{a-x}}} & {x=a \cos 2 \theta}\\ \\\hline\end{array}

Application of Special Integral Formula (Part 1)

Integration of the type 

\\ {\text { (i) } \int \frac{(p x+q) \,dx}{a x^{2}+b x+c} }\\\\ \text { (ii) } \int \frac{(p x+q)}{\sqrt{a x^{2}+b x+c}} \,dx\\\\ {\text { (iii) } \int(p x+q) \sqrt{a x^{2}+b x+c} \,\,dx}\\

\\\mathrm{Express\;the\;linear\;factor\;\mathit{px+q}\;in\;terms\;of\;the\;derivative\;of}\\\mathrm{quadratic\;factor\;\mathit{ax^2+bx+c}}\\\text{i.e.}\\\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;{p x+q=\lambda.\frac{ d}{d x}\left(a x^{2}+b x+c\right)+\mu}}\\\\\mathrm{\Rightarrow \;\;\;\;\;\;\;\;{p x+q=\lambda(2 a x+b)+\mu}}\\\\\mathrm{{\text {Find } \lambda \text { and } \mu \text { and replace }(p x+q) \text { by }} \\ {\lambda(2 a x+b)+\mu}}

Application of Special Integral Formula (Part 2)

Integration of the type

\\\mathrm{1.\;\;\int \frac{a x^{2}+b x+c}{\left(p x^{2}+q x+r\right)} \,\,d x}\\\\\mathrm{2.\;\;\int \frac{\left(a x^{2}+b x+c\right)}{\sqrt{p x^{2}+q x+r}} \,\,d x}\\\\\mathrm{3.\;\;\int\left(a x^{2}+b x+c\right) \sqrt{p x^{2}+q x+r} \,\,dx}

\\\text{Substitute,}\\\\\mathrm{a x^{2}+b x+c=\lambda\left(p x^{2}+q x+r\right)+\mu\left\{\frac{d}{d x}\left(p x^{2}+q x+r\right)\right\}+\gamma}

Find λ, μ and γ. These integrations reduces to integration of three independent functions.

 

Integration of the form \mathbf{\int \frac{k(x)}{ax^2+bx+c}\;dx}

(here, k(x) is a polynomial of degree greater than 2)

To solve this type of integration, divide the numerator by the denominator and express the intagral as 

\mathrm{Q(x)+\frac{R(x)}{ax^2+bx+c}} 

Here, R(x) is a linear function of x.

Study it with Videos

Some Special Integration
Application of Special Integral Formula (Part 1)
Application of Special Integral Formula (Part 2)

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Books

Reference Books

Some Special Integration

Mathematics for Joint Entrance Examination JEE (Advanced) : Calculus

Page No. : 7.14

Line : 1

Application of Special Integral Formula (Part 1)

Mathematics for Joint Entrance Examination JEE (Advanced) : Calculus

Page No. : 7.19

Line : 1

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