# Basic Examples On Integration-set-2

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## Basic integration examples

Example - 4

Evaluate \begin{align}\int {\frac{x}{{\sqrt {x + a} + \sqrt {x + b} }}dx} \end{align}.

Solution : The form of the expression in the denominator clearly hints that a reduction is possible by rationalization which would lead to a constant term in the denominator:

\begin{align}& \int{\frac{x}{\sqrt{x+a}+\sqrt{x+b}}dx}=\int{\frac{x\left\{ \sqrt{x+a}-\sqrt{x+b} \right\}}{\left( x+a \right)-\left( x+b \right)}}dx \\& \qquad \qquad \qquad \qquad \qquad =\frac{1}{\left( a-b \right)}\int{\left\{ x\sqrt{x+a}-x\sqrt{x+b} \right\}dx} \\ & \qquad \qquad \qquad \qquad \qquad =\frac{1}{a-b}\int{\left\{ \left( x+a-a \right)\sqrt{x+a}-\left( x+b-b \right)\sqrt{x+b} \right\}dx} \\ & \qquad \qquad \qquad \qquad \qquad =\frac{1}{a-b}\int{\left\{ {{\left( x+a \right)}^{{}^{3}\!\!\diagup\!\!{}_{2}\;}}-{{\left( x+b \right)}^{{}^{3}\!\!\diagup\!\!{}_{2}\;}}-a{{\left( x+a \right)}^{{}^{1}\!\!\diagup\!\!{}_{2}\;}}+b{{\left( x+b \right)}^{{}^{1}\!\!\diagup\!\!{}_{2}\;}} \right\}dx} \\ & \qquad \qquad \qquad \qquad \qquad =\frac{1}{a-b}\left\{ \frac{{{\left( x+a \right)}^{{}^{5}\!\!\diagup\!\!{}_{2}\;}}}{{}^{5}\!\!\diagup\!\!{}_{2}\;}-\frac{{{\left( x+b \right)}^{{}^{5}\!\!\diagup\!\!{}_{2}\;}}}{{}^{5}\!\!\diagup\!\!{}_{2}\;}-\frac{a{{\left( x+a \right)}^{{}^{3}\!\!\diagup\!\!{}_{2}\;}}}{{}^{3}\!\!\diagup\!\!{}_{2}\;}+\frac{b{{\left( x+b \right)}^{{}^{3}\!\!\diagup\!\!{}_{2}\;}}}{{}^{3}\!\!\diagup\!\!{}_{2}\;} \right\}+C \\ \end{align}

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The first simplification by rationalization led to an expression which involved two terms of the form $$x\sqrt {x + a} ;$$ to integrate these terms, we wrote the x outside the root as $$\left( {x + a-{\text{ }}a} \right)$$ so that a final expression is obtained which contains only terms of the form $${\left( {x + k} \right)^n};$$ these could then be integrated easily.

Example - 5

Evaluate \begin{align}\int {\frac{{1 + \cos 4x}}{{\cot x - \tan x}}dx} \end{align} .

Solution: We simply both the numerator and the denominator separately:

\begin{align}& \int{\frac{1+\cos 4x}{\cot x-\tan x}dx}=\int{\frac{2{{\cos }^{2}}2x}{\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}}dx} \\ & \qquad \qquad \qquad \qquad =\int{\frac{2\sin x\cos x{{\cos }^{2}}2x}{{{\cos }^{2}}x-{{\sin }^{2}}x}dx} \\ & \qquad \qquad \qquad \qquad =\int{\frac{\sin 2x\cdot {{\cos }^{2}}2x}{\cos 2x}dx} \\ & \qquad \qquad \qquad \qquad =\int{\sin 2x\cos 2x}dx \\ & \qquad \qquad \qquad \qquad =\frac{1}{2}\int{\sin 4xdx} \\ & \qquad \qquad \qquad \qquad =-\frac{1}{8}\cos 4x+C \\ \end{align}

Example - 6

Evaluate \begin{align}\int {\tan x\tan 2x\tan 3x} \,dx\end{align}.

Solution: Notice that $$3x = 2x + x,$$ so that

\begin{align} &\qquad\qquad\qquad\qquad \tan 3x=\tan \left( 2x+x \right) \\ &\qquad\;\;\qquad\qquad\qquad\qquad =\frac{\tan 2x+\tan x}{1-\tan 2x\tan x} \\ & \Rightarrow \quad \tan x\tan 2x\tan 3x=\tan 3x-\tan 2x-\tan x \\ \end{align}

The required integral is now easy to evaluate :

\begin{align}& \int{\tan x\tan 2x\tan 3x}dx=\int{\left\{ \tan 3x-\tan 2x-\tan x \right\}}dx \\ &\qquad\qquad\qquad\qquad\qquad =\frac{1}{3}\ln \left| \sec 3x \right|-\frac{1}{2}\ln \left| \sec 2x \right|-\ln \left| \sec x \right|+C \\ \end{align}

Example - 7

Evaluate \begin{align}\int {\frac{{\sin \left( {x - a} \right)}}{{\sin \left( {x - b} \right)}}dx} \end{align} .

Solution: Taking cue from Example-2, our aim should be to somehow get rid of the variable term $$sin\left(\text{ }x-\text{ }b \right)$$ in the denominator; to do this, we write the numerator$$\sin \left( x-a \right)\,\,as\,\,\sin \left\{ \left( x-b \right)-\left( a-b \right) \right\}:$$

\begin{align}& \int{\frac{\sin \left( x-a \right)}{\sin \left( x-b \right)}dx=}\int{\frac{\sin \left\{ \left( x-b \right)-\left( a-b \right) \right\}}{\sin \left( x-b \right)}dx} \\ & \qquad \qquad \qquad \quad =\int{\frac{\sin \left( x-b \right)\cos \left( a-b \right)-\cos \left( x-b \right)\sin \left( a-b \right)}{\sin \left( x-b \right)}dx} \\ & \qquad \qquad \qquad \quad =\int{\left\{ \cos \left( a-b \right)-\sin \left( a-b \right)\cot \left( x-b \right) \right\}dx} \\ & \qquad \qquad \qquad \quad =x\cos \left( a-b \right)-\sin \left( a-b \right)\ln \left| \sin \left( x-b \right) \right|+C \\ \end{align}