Integral Exercise asked by Imad Zak in the math group

Exercise:

Integrate, $\int{\frac{1-\sin x}{1+\sin x}dx}$  then deduce $\int_{0}^{\pi }{\frac{1-\sin x}{1+\sin x}dx=4-\pi }$

Solution: we have $\frac{1-\sin x}{1+\sin x}=\frac{1}{1+\sin x}-\frac{\sin x+1-1}{1+\sin x}=\frac{1}{1+\sin x}-1+\frac{1}{1+\sin x}=\frac{2}{1+\sin x}-1$

So $\int{\frac{1-\sin x}{1+\sin x}\,dx}=\int{\frac{2}{1+\sin x}dx}-\int{\,dx}=\int{\frac{2}{1+\sin x}\,dx}-x+c$

 we know that $\sin x=\frac{2\tan \left( \frac{x}{2} \right)}{1+{{\tan }^{2}}\left( \frac{x}{2} \right)}$ and put $u=\tan \left( \frac{x}{2} \right)$ so we get $\sin x=\frac{2u}{1+{{u}^{2}}}$

so $1+\sin x=1+\frac{2u}{1+{{u}^{2}}}=\frac{1+{{u}^{2}}+2u}{1+{{u}^{2}}}=\frac{{{\left( 1+u \right)}^{2}}}{1+{{u}^{2}}}$

 Taking $u=\tan \left( \frac{x}{2} \right)$ $\Rightarrow du=\frac{1}{2}{{\sec }^{2}}\left( \frac{x}{2} \right)dx$ and we now that ${{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta $

Thus $2du=\left( 1+{{\tan }^{2}}\frac{x}{2} \right)dx\Leftrightarrow 2du=\left( 1+{{u}^{2}} \right)dx$

So \[\int{\frac{2}{1+\sin x}dx}=2\int{\frac{1+{{u}^{2}}}{{{\left( 1+u \right)}^{2}}}\times \frac{2}{1+{{u}^{2}}}du=4\int{\frac{1}{{{\left( 1+u \right)}^{2}}}du}}\]

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

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

But $\underset{x\to 0}{\mathop{\lim }}\,\left( -\frac{4}{1+\tan \left( \frac{x}{2} \right)}-x \right)=-4$ & $\underset{x\to \pi }{\mathop{\lim }}\,\left( -\frac{4}{1+\tan \left( \frac{x}{2} \right)}-x \right)=-\pi $

Thus $\int_{0}^{\pi }{\frac{1-\sin x}{1+\sin x}\,dx}=-\pi -\left( -4 \right)=4-\pi $