Integral exercise for \(\frac{1}{{(1+sin(x))}^{2}}\)

Exercise:

Integrate, $\int{\frac{dx}{{{\left( 1+\sin x \right)}^{2}}}}$

Solution: we have $\frac{1}{{{\left( 1+\sin x \right)}^{2}}}\times \frac{{{\left( 1-\sin x \right)}^{2}}}{{{\left( 1-\sin x \right)}^{2}}}=\frac{{{\left( 1-\sin x \right)}^{2}}}{{{\left[ \left( 1+\sin x \right)\left( 1-\sin x \right) \right]}^{2}}}=\frac{{{\left( 1-\sin x \right)}^{2}}}{{{\left( 1-{{\sin }^{2}}x \right)}^{2}}}$

 $=\frac{{{\left( 1-\sin x \right)}^{2}}}{{{\left( {{\cos }^{2}}x \right)}^{2}}}=\frac{1-2\sin x+{{\sin }^{2}}x}{{{\cos }^{4}}x}=\frac{1-2\sin x+1-{{\cos }^{2}}x}{{{\cos }^{4}}x}=\frac{2-2\sin x-{{\cos }^{2}}x}{{{\cos }^{4}}x}$

$=\frac{2}{{{\cos }^{4}}x}-2\frac{\sin x}{\cos x}\times \frac{1}{{{\cos }^{3}}x}-\frac{1}{{{\cos }^{2}}x}=2{{\sec }^{4}}x-2\tan x{{\sec }^{3}}x-{{\sec }^{2}}x$

So $\int{\frac{dx}{{{\left( 1+\sin x \right)}^{2}}}=2\int{{{\sec }^{4}}x\,dx}-2\int{\tan x{{\sec }^{3}}x\,dx-\int{{{\sec }^{2}}x\,dx}}}$

Let ${{A}_{1}}=\int{{{\sec }^{4}}x\,dx}\,\,,\,\,{{A}_{2}}=\int{\tan x{{\sec }^{3}}x\,dx}\,\,\,\And \,\,{{A}_{3}}=\int{{{\sec }^{2}}x\,dx}$

In ${{A}_{2}}$ we have $\int{\tan x{{\sec }^{3}}x\,dx}=\int{\tan x\sec x{{\sec }^{2}}x\,dx}$

Take $u=\sec x\Leftrightarrow du=\tan x\sec x\,dx$

So ${{A}_{2}}=\int{\tan x{{\sec }^{3}}xdx}=\int{{{u}^{2}}du=\frac{1}{3}{{u}^{3}}+c=\frac{1}{3}{{\sec }^{3}}x+c}$

Take $u=\tan x\Leftrightarrow du={{\sec }^{2}}x\,dx$ so ${{A}_{3}}=\int{{{\sec }^{2}}xdx}=\int{d\left( \tan x \right)=\tan x+c}$

${{A}_{1}}=\int{{{\sec }^{4}}xdx}=\int{{{\sec }^{2}}x{{\sec }^{2}}xdx}$ but ${{\sec }^{2}}x=1+{{\tan }^{2}}x$

So ${{A}_{1}}=\int{\left( 1+{{\tan }^{2}}x \right){{\sec }^{2}}x}\,dx$ take $w=\tan x\Rightarrow dw={{\sec }^{2}}xdx$

Thus ${{A}_{1}}=\int{\left( 1+{{u}^{2}} \right)du}=u+\frac{1}{3}{{u}^{3}}+c=\tan x+\frac{1}{3}{{\tan }^{3}}x+c$

$\therefore \int{\frac{dx}{{{\left( 1+\sin x \right)}^{2}}}=\tan x+\frac{2}{3}{{\tan }^{3}}x-\frac{2}{3}{{\sec }^{3}}x+c}$