# Basic Examples On Integration Set-3

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Example – 8

Evaluate \begin{align}\int {\frac{{ax + b}}{{cx + d}}dx}\end{align}  .

Solution: As in Ex-2 and Ex-7, we try to express the numerator in terms of the denominator, so that a cancellation is possible and we can get rid of the variable term $$\left( {cx + d} \right)$$ in the denominator:

\begin{align}& \qquad \qquad \qquad ax+b=a\left( x+\frac{b}{a} \right) \\ & \\ & \qquad \qquad \qquad \quad \quad \ \ =\frac{a}{c}\left( cx+\frac{bc}{a} \right) \\ & \\ & \qquad \qquad \qquad \quad \quad \ \ =\frac{a}{c}\left( cx+d+\frac{bc}{a}-d \right) \\ & \\ & \qquad \qquad \qquad \quad \quad \ \ =\frac{a}{c}\left( cx+d \right)+\left( b-\frac{ad}{c} \right) \\ & \\ & \ \ \Rightarrow \quad \int{\frac{ax+b}{cx+d}dx=\int{\frac{\frac{a}{c}\left( cx+d \right)+\left( b-\frac{ad}{c} \right)}{cx+d}dx}} \\ & \\ & \qquad \qquad \qquad \quad \quad \ \ =\int{\left\{ \frac{a}{c}+\frac{\left( b-\frac{ad}{c} \right)}{cx+d} \right\}}dx \\ & \\ & \qquad \qquad \qquad \quad \quad \ \ =\frac{ax}{c}+\frac{1}{c}\left( b-\frac{ad}{c} \right)\ln \left( cx+d \right)+C \\ & \\ & \qquad \qquad \qquad \quad \quad \ \ =\frac{ax}{c}+\left( \frac{bc-ad}{{{c}^{2}}} \right)\ln \left( cx+d \right)+C \\ \end{align}

Example - 9

Evaluate  \begin{align}\int {\frac{{\sin 4x}}{{\sin x}}dx} \end{align}.

Solution: The expression can be simplified by a straight forward expansion of the numerator :

\begin{align}& \int{\frac{\sin 4x}{\sin x}dx}=\int{\frac{2\sin 2x\cos 2x}{\sin x}dx} \\ & \qquad \qquad \quad =4\int{\frac{\sin x\cos x\cos 2x}{\sin x}dx} \\ & \qquad \qquad \quad =4\int{\cos x\cos 2xdx} \\ & \qquad \qquad \quad =2\int{\left\{ \cos 3x+\cos x \right\}dx} \\ & \qquad \qquad \quad =\frac{2}{3}\sin 3x+2\sin x+C \\ \end{align}

Example - 10

Evaluate the following integrals:

 (a) $$\int {{{\cos }^3}x\,\,dx}$$ (b) $$\int {{{\cos }^4}x\,\,dx}$$ (c) $$\int {\sin 2x\cos 4x\cos 5x\,\,dx}$$ (d) $$\int {{{\sin }^3}x{{\cos }^3}x\,\,dx}$$

Solution: (a) We know the integral of  $$\cos x$$ ; we must express the cubic cos term $$\left( {co{s^3}x} \right)$$ in terms of linear cos terms; this can be done using the triple angle formula:

\begin{align}& \qquad \qquad \,\,\,\cos 3x=4{{\cos }^{3}}x-3\cos x \\& \;\Rightarrow \qquad \;\;\;{{\cos }^{3}}x=\frac{1}{4}\left\{ 3\cos x+\cos 3x \right\} \\ & \Rightarrow \quad \int{{{\cos }^{3}}xdx=\frac{3}{4}\sin x+\frac{1}{12}\sin 3x+C} \\ \end{align}

(b) Here again, we need to express the fourth degree cos term $$\left( {{{\cos }^4}x} \right)$$ in terms of linear cos terms:

\begin{align} & \qquad \quad \quad \ \ {{\cos }^{4}}x={{\left( {{\cos }^{2}}x \right)}^{2}} \\ & \qquad \qquad \qquad \quad ={{\left( \frac{1+\cos 2x}{2} \right)}^{2}} \\ & \qquad \qquad \qquad \quad =\frac{1}{4}+\frac{{{\cos }^{2}}2x}{4}+\frac{1}{2}\cos 2x \\ & \qquad \qquad \qquad \quad =\frac{1}{4}+\frac{\left( 1+\cos 4x \right)}{8}+\frac{1}{2}\cos 2x \\ & \qquad \qquad \qquad \quad =\frac{3}{8}+\frac{1}{2}\cos 2x+\frac{1}{8}\cos 4x \\ & \Rightarrow \quad \int{{{\cos }^{4}}xdx=\frac{3x}{8}+\frac{\sin 2x}{4}+\frac{\sin 4x}{32}+C} \\ \end{align}

(c) \begin{align}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\sin 2x\cos 4x\cos 5x = \frac{1}{2}\left( {2\sin 2x\cos 4x} \right)\cos 5x\end{align}

\begin{align}& = \frac{1}{2}\left( {\sin 6x - \sin 2x} \right)\cos 5x\\ &= \frac{1}{4}\left\{ {\left( {2\sin 6x\cos 5x} \right) - \left( {2\sin 2x\cos 5x} \right)} \right\}\\&= \frac{1}{4}\left\{ {\sin 11x + \sin x - \sin 7x + \sin 3x} \right\}\\ &= \frac{1}{4}\left\{ {\sin x + \sin 3x - \sin 7x + \sin 11x} \right\}\end{align}

$\Rightarrow \qquad \,\int {\sin 2x\cos 4x\cos 5x\,\,dx = \frac{{ - \cos x}}{4} - \frac{{\cos 3x}}{{12}} + \frac{{\cos 7x}}{{28}} - \frac{{\cos 11x}}{{44}} + C}$

(d) \begin{align}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\;\;\;{\sin ^3}x\;{\cos ^3}x = {\left( {\sin x\cos x} \right)^3}\end{align}

\begin{align}&= \frac{{{{\left( {2\sin x\cos x} \right)}^3}}}{8}\\& = \frac{{{{\sin }^3}2x}}{8}\\& = \frac{1}{8}\left\{ {\frac{{3\sin 2x - \sin 6x}}{4}} \right\}\qquad\left( {{\rm{Triple \;angle \;formula}}} \right)\\& = \frac{3}{{32}}\sin 2x - \frac{1}{{32}}\sin 6x\end{align}

$\Rightarrow \qquad \int {{{\sin }^3}x{{\cos }^3}x\,\,dx = \frac{{ - 3}}{{64}}\cos 2x + \frac{1}{{192}}\cos 6x + C}$