Integral exercise of $\frac{1}{\sin^{2}{x}+\tan^{2}{x}}$

Exercise:

Integrate, $\int{\frac{dx}{{{\sin }^{2}}x+{{\tan }^{2}}x}}$

Solution: we have $\frac{1}{{{\sin }^{2}}x+{{\tan }^{2}}x}\times \frac{{{\sec }^{2}}x}{{{\sec }^{2}}x}=\frac{{{\sec }^{2}}x}{\frac{{{\sin }^{2}}x}{{{\cos }^{2}}x}+{{\tan }^{2}}x{{\sec }^{2}}x}=\frac{{{\sec }^{2}}x}{{{\tan }^{2}}x\left( 1+{{\sec }^{2}}x \right)}=\frac{{{\sec }^{2}}x}{{{\tan }^{4}}x+2{{\tan }^{2}}x}$

So $\int{\frac{dx}{{{\sin }^{2}}x+{{\tan }^{2}}x}=\int{\frac{{{\sec }^{2}}x}{{{\tan }^{4}}x+2{{\tan }^{2}}x}dx}}$

Put $u=\tan x\Leftrightarrow du={{\sec }^{2}}xdx$

So $\int{\frac{{{\sec }^{2}}xdx}{{{\tan }^{4}}x+2{{\tan }^{2}}x}=\int{\frac{du}{{{u}^{4}}+2{{u}^{2}}}}}=\int{\frac{du}{{{u}^{2}}\left( {{u}^{2}}+2 \right)}}$

But $\frac{1}{{{u}^{2}}\left( {{u}^{2}}+2 \right)}=\frac{1}{2{{u}^{2}}}-\frac{1}{2\left( {{u}^{2}}+2 \right)}$  So $\int{\frac{du}{{{u}^{2}}\left( {{u}^{2}}+2 \right)}}=\frac{1}{2}\int{\frac{1}{{{u}^{2}}}du-\frac{1}{2}\int{\frac{du}{{{u}^{2}}+2}}}$

But $\int{\frac{du}{{{u}^{2}}}}=\int{{{u}^{-2}}du}=-{{u}^{-1}}+c=\frac{-1}{u}+c$

And $\int{\frac{du}{{{u}^{2}}+2}=\int{\frac{du}{2\left( {{\left( \frac{u}{\sqrt{2}} \right)}^{2}}+1 \right)}}}$ Take $v=\frac{u}{\sqrt{2}}\Leftrightarrow dv=\frac{1}{\sqrt{2}}du\Leftrightarrow \sqrt{2}dv=du$

So $\int{\frac{du}{{{u}^{2}}+2}=\frac{1}{2}\int{\frac{\sqrt{2}dv}{{{v}^{2}}+1}=\frac{\sqrt{2}}{2}\arctan \left( v \right)+c}=\frac{\sqrt{2}}{2}\arctan \left( \frac{u}{\sqrt{2}} \right)+c}$

Thus $\int{\frac{du}{{{u}^{4}}+2{{u}^{2}}}=\frac{-1}{2u}-\frac{\sqrt{2}}{4}\arctan \left( \frac{u}{\sqrt{2}} \right)+c}$

$\therefore \int{\frac{dx}{{{\sin }^{2}}x+{{\tan }^{2}}x}=-\frac{1}{2\tan x}-\frac{\sqrt{2}}{4}\arctan \left( \frac{\tan x}{\sqrt{2}} \right)+c}$