Exercise :
Integrate, $\int{\frac{{{\sin }^{4}}x}{{{\sin }^{6}}x+{{\cos }^{6}}x}dx}$
Solution: we have $\frac{{{\sin }^{4}}x}{{{\sin }^{6}}x+{{\cos }^{6}}x}$ so divided by ${{\cos }^{6}}x$ to get :
$\frac{{{\sin }^{4}}x}{{{\sin }^{6}}x+{{\cos }^{6}}x}=\frac{\frac{{{\sin }^{4}}x}{{{\cos }^{4}}x}\times \frac{1}{{{\cos }^{2}}x}}{{{\tan }^{6}}x+1}=\frac{{{\tan }^{4}}x{{\sec }^{2}}x}{{{\tan }^{6}}x+1}$
So $\int{\frac{{{\sin }^{4}}x}{{{\sin }^{6}}x+{{\cos }^{6}}x}dx}=\int{\frac{{{\tan }^{4}}x{{\sec }^{2}}x}{{{\tan }^{6}}x+1}dx}$
Let $u=\tan x\Leftrightarrow du={{\sec }^{2}}xdx$
So $\int{\frac{{{\tan }^{4}}x{{\sec }^{2}}x}{{{\tan }^{6}}x+1}dx}=\int{\frac{{{u}^{4}}du}{{{u}^{6}}+1}}$
As we know that ${{u}^{6}}+1={{\left( {{u}^{2}} \right)}^{3}}+1=\left( {{u}^{2}}+1 \right)\left( {{u}^{4}}-{{u}^{2}}+1 \right)$
But ${{u}^{4}}-{{u}^{2}}+1={{u}^{4}}-3{{u}^{2}}+2{{u}^{2}}+1=\left( {{u}^{4}}+2{{u}^{2}}+1 \right)-3{{u}^{2}}={{\left( {{u}^{2}}+1 \right)}^{2}}-3{{u}^{2}}$
$=\left( {{u}^{2}}+\sqrt{3}u+1 \right)\left( {{u}^{2}}-\sqrt{3}u+1 \right)$
So $\frac{{{u}^{4}}}{{{u}^{6}}+1}=\frac{{{u}^{4}}}{\left( {{u}^{2}}+1 \right)\left( {{u}^{2}}+\sqrt{3}u+1 \right)\left( {{u}^{2}}-\sqrt{3}u+1 \right)}$
Thus $\frac{{{u}^{4}}}{{{u}^{6}}+1}=\frac{3u-\sqrt{3}}{6\sqrt{3}\left( -{{u}^{2}}+\sqrt{3}u-1 \right)}+\frac{1}{3\left( {{u}^{2}}+1 \right)}+\frac{-\sqrt{3}u-1}{6\left( {{u}^{2}}+\sqrt{3}u+1 \right)}$
Hence $\frac{1}{3}\int{\frac{du}{{{u}^{2}}+1}}=\frac{1}{3}\arctan \left( u \right)+c$
Now Let’s Compute the integral for the first term as follows :
$\int{\frac{3u-\sqrt{3}}{6\sqrt{3}\left( -{{u}^{2}}+\sqrt{3}u-1 \right)}du}=\frac{1}{6\sqrt{3}}\int{\frac{3u-\sqrt{3}}{-{{u}^{2}}-1+\sqrt{3}u}du}$
But $-{{u}^{2}}-1+\sqrt{3}u=-\left( {{u}^{2}}+1-\sqrt{3}u \right)=-\left( {{u}^{2}}-2\frac{\sqrt{3}}{2}u+\frac{3}{4}-\frac{3}{4}+1 \right)=-{{\left( u-\frac{\sqrt{3}}{2} \right)}^{2}}-\frac{1}{4}$
Let $z=u-\frac{\sqrt{3}}{2}\Leftrightarrow dz=du$
So $\frac{1}{6\sqrt{3}}\int{\frac{3u-\sqrt{3}}{-{{u}^{2}}-1+\sqrt{3}}}=\frac{1}{6\sqrt{3}}\int{\frac{3u-\sqrt{3}}{-{{z}^{2}}-\frac{1}{4}}dz}=\frac{4}{6\sqrt{3}}\int{\frac{6z+\sqrt{3}}{8{{z}^{2}}+2}dz}$
$=\frac{4}{6\sqrt{3}}\int{\frac{6z}{8{{z}^{2}}+2}dz}+\frac{4}{6\sqrt{3}}\int{\frac{\sqrt{3}}{8{{z}^{2}}+2}dz}$
Let $q=8{{z}^{2}}+2\Leftrightarrow dq=16zdz\Leftrightarrow \frac{dq}{16}=zdz$
So $\int{\frac{6zdz}{8{{z}^{2}}+2}}=\int{\frac{\frac{6dq}{16}}{q}=\frac{6}{16}\int{\frac{dq}{q}=\frac{3}{8}\ln \left| q \right|+c}=\frac{3}{8}\ln \left| 8{{z}^{2}}+2 \right|+c}=\frac{3}{8}\ln \left( {{\left( u-\frac{\sqrt{3}}{2} \right)}^{2}}+2 \right)+c$
And $\int{\frac{\sqrt{3}}{8{{z}^{2}}+2}dz}=\frac{\sqrt{3}}{2}\int{\frac{dz}{4{{z}^{2}}+1}}=\frac{\sqrt{3}}{2}\arctan \left( 2z \right)+c=\frac{\sqrt{3}}{2}\arctan \left( 2\left( u-\frac{\sqrt{3}}{2} \right) \right)+c$
Now we need to compute the integral of the last term as follows :
$\int{\frac{-\sqrt{3}u-1}{6\left( {{u}^{2}}+\sqrt{3}u+1 \right)}du=\frac{1}{6}\int{\frac{-\sqrt{3}u-1}{{{u}^{2}}+\sqrt{3}u+1}du}}=\frac{1}{6}\int{\frac{-\sqrt{3}u-1}{{{\left( u-\frac{\sqrt{3}}{2} \right)}^{2}}+\frac{1}{4}}du}$
Let $y=u-\frac{\sqrt{3}}{2}\Leftrightarrow dy=du\,\,\And \,\,u=y+\frac{\sqrt{3}}{2}$
So $\int{\frac{-\sqrt{3}u-1}{{{\left( u-\frac{\sqrt{3}}{2} \right)}^{2}}+\frac{1}{4}}du}=-4\int{\frac{2\sqrt{3}y-1}{8{{y}^{2}}+2}dy=-4\int{\frac{2\sqrt{3}ydy}{8{{y}^{2}}+2}+4\int{\frac{dy}{8{{y}^{2}}+1}}}}$
Let $U=8{{y}^{2}}+2\Leftrightarrow dU=16ydy\Leftrightarrow \frac{dU}{16}=ydy$
$\Rightarrow \int{\frac{2\sqrt{3}ydy}{8{{y}^{2}}+2}}=\int{\frac{\frac{2\sqrt{3}}{16}dU}{U}}=\frac{\sqrt{3}}{8}\ln \left| U \right|+c=\frac{\sqrt{3}}{8}\ln \left( 8{{y}^{2}}+2 \right)+c=\frac{\sqrt{3}}{8}\ln \left( 8{{\left( u-\frac{\sqrt{3}}{2} \right)}^{2}}+2 \right)+c$
And $\int{\frac{dy}{8{{y}^{2}}+2}=\frac{1}{2}\int{\frac{dy}{4{{y}^{2}}+1}}=\frac{1}{2}\arctan \left( 2y \right)+c=\frac{1}{2}\arctan \left( 2\left( u-\frac{\sqrt{3}}{2} \right) \right)+c}$
By Backward substitution we get :
$\int{\frac{{{\sin }^{4}}x}{{{\cos }^{6}}x+{{\sin }^{6}}x}dx}=\frac{1}{12}\left( 4x-2\arctan \left( \sqrt{3}-2\tan x \right)+2\arctan \left( 2\tan x+\sqrt{3} \right) \right.$
$+\sqrt{3}\log \left( \sqrt{3}-3\sin x\cos x \right)-\sqrt{3}\log \left. \left( 3\sin x\cos x+\sqrt{3} \right) \right)+c$
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Remark 1:
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Here another way to integrate the terms of the partial fractions (first term )
Exercise:
Integrate, $\int{\frac{\sqrt{3}-3x}{6\sqrt{3}\left( -{{x}^{2}}+\sqrt{3}x-1 \right)}dx}$
Solution: we have $\int{\frac{\sqrt{3}-3x}{6\sqrt{3}\left( -{{x}^{2}}+\sqrt{3}x-1 \right)}dx}=\frac{1}{6\sqrt{3}}\int{\frac{\sqrt{3}-2x-x}{-{{x}^{2}}+\sqrt{3}x-1}dx}$
$=\frac{1}{6\sqrt{3}}\int{\frac{\sqrt{3}-2x}{-{{x}^{2}}+\sqrt{3}x-1}dx}-\frac{1}{6\sqrt{3}}\int{\frac{x}{-{{x}^{2}}+\sqrt{3}x-1}dx}$
$=\frac{1}{6\sqrt{3}}\int{\frac{d\left( -{{x}^{2}}+\sqrt{3}x-1 \right)}{-{{x}^{2}}+\sqrt{3}x-1}+\frac{1}{6\sqrt{3}}\int{\frac{-\frac{2x}{2}+\sqrt{3}-\sqrt{3}}{-{{x}^{2}}+\sqrt{3}x-1}dx}}$
$=\frac{1}{6\sqrt{3}}\ln \left| -{{x}^{2}}+\sqrt{3}x-1 \right|+\frac{1}{12\sqrt{3}}\ln \left| -{{x}^{2}}+\sqrt{3}x-1 \right|-\frac{1}{6}\int{\frac{dx}{-{{x}^{2}}+\sqrt{3}x-1}dx}$
But $\int{\frac{dx}{-{{x}^{2}}+\sqrt{3}x-1}=\int{\frac{dx}{-\left( {{x}^{2}}-\sqrt{3}x+1 \right)}=-\int{\frac{dx}{{{x}^{2}}-\sqrt{3}x+\frac{3}{4}-\frac{3}{4}+1}=-\int{\frac{dx}{{{\left( x-\frac{\sqrt{3}}{2} \right)}^{2}}+\frac{1}{4}}}}}}$
Let $u=x-\frac{\sqrt{3}}{2}\Leftrightarrow du=dx$
So $-\int{\frac{dx}{{{\left( x-\frac{\sqrt{3}}{2} \right)}^{2}}+\frac{1}{4}}=-\int{\frac{du}{{{u}^{2}}+\frac{1}{4}}=-\int{\frac{du}{\frac{4{{u}^{2}}+1}{4}}=-4\int{\frac{du}{{{\left( 2u \right)}^{2}}+1}=-4\arctan \left( 2u \right)+c}}}}$
So $\int{\frac{\sqrt{3}-3x}{6\sqrt{3}\left( -{{x}^{2}}+\sqrt{3}x-1 \right)}dx}=\frac{1}{6\sqrt{3}}\ln \left| -{{x}^{2}}+\sqrt{3}x-1 \right|+\frac{1}{12\sqrt{3}}\ln \left| -{{x}^{2}}+\sqrt{3}x-1 \right|+\frac{2}{3}\arctan \left( 2x-\sqrt{3} \right)+c$
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