Mixed Exercise between Geometric series and improper integral $\sum\limits_{i=1}^{\infty }{\int_{0}^{\frac{\pi }{4}}{{{\sin }^{i}}x\,dx}}$

Exercise:

Compute, $\sum\limits_{i=1}^{\infty }{\int_{0}^{\frac{\pi }{4}}{{{\sin }^{i}}x\,dx}}$

Solution: we have $\sum\limits_{i=1}^{\infty }{\int_{0}^{\frac{\pi }{4}}{{{\sin }^{i}}x\,dx}=\int_{0}^{\frac{\pi }{4}}{\sin x\,dx}+\int_{0}^{\frac{\pi }{4}}{{{\sin }^{2}}x\,dx}+....}$

So $\sum\limits_{i=1}^{\infty }{\int_{0}^{\frac{\pi }{4}}{{{\sin }^{i}}x\,dx}=\int_{0}^{\frac{\pi }{4}}{\sum\limits_{i=1}^{\infty }{{{\sin }^{i}}x\,dx}}}$

Notice that $0\le x\le \frac{\pi }{4}\Leftrightarrow 0\le \sin x\le \frac{\sqrt{2}}{2}<1\Leftrightarrow \left| \sin x \right|<1$

Notice that $\sum\limits_{i=1}^{\infty }{{{t}^{i}}}$ is a Geometric series with common ratio is $t$

Hence $\sum\limits_{i=1}^{\infty }{{{t}^{i}}=\frac{t}{1-t}}$ where $\left| t \right|<1$ thus $\sum\limits_{i=1}^{\infty }{{{\sin }^{i}}x=\frac{\sin x}{1-\sin x}}$ with $\left| \sin x \right|<1$

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

                 $=\sec x\tan x+{{\sec }^{2}}x-1$

So $\int_{0}^{\frac{\pi }{4}}{\sum\limits_{i=1}^{\infty }{{{\sin }^{i}}x\,dx}=\int_{0}^{\frac{\pi }{4}}{\left( \sec x\tan x+{{\sec }^{2}}x-1 \right)dx}}=\int_{0}^{\frac{\pi }{4}}{\sec x\tan x}\,dx+\int_{0}^{\frac{\pi }{4}}{{{\sec }^{2}}x}\,dx-\int_{0}^{\frac{\pi }{4}}{dx}$

                        $$=\int_{0}^{\frac{\pi }{4}}{\sec x\tan x\,dx}+\int_{0}^{\frac{\pi }{4}}{d\left( \tan x \right)-\left[ x \right]_{0}^{\frac{\pi }{4}}}=\int_{0}^{\frac{\pi }{4}}{\sec x\tan x}\,dx+\left[ \tan x \right]_{0}^{\frac{\pi }{4}}-\frac{\pi }{4}$$

                       $=\int_{0}^{\frac{\pi }{4}}{\sec x\tan x+1-\frac{\pi }{4}}=\int_{0}^{\frac{\pi }{4}}{\frac{\sin x}{{{\cos }^{2}}x}dx}+1-\frac{\pi }{4}=-\int_{0}^{\frac{\pi }{4}}{\frac{d\left( \cos x \right)}{{{\cos }^{2}}x}+1-\frac{\pi }{4}}$

                       $=\left[ \sec x \right]_{0}^{\frac{\pi }{4}}+1-\frac{\pi }{4}=\sqrt{2}-1+1-\frac{\pi }{4}=\sqrt{2}-\frac{\pi }{4}$