Trig-integral

Firstly using the concept of geometric series we get :

$\sum_{n=0}^{\infty} \sin^{2n}x = \sum_{n=0}^{\infty} {(\sin^{2}x)}^{n} = \frac{1}{1-\sin^{2}x}=\frac{1}{\cos^{2}x} = \sec^{2}x$

Now our integral becomes :

$I=\int_{\frac{5\pi}{8}}^{\frac{7\pi}{8}} \sqrt{\sec^{2}x} dx=\int _{ \frac { 5\pi }{ 8 } }^{ \frac { 7\pi }{ 8 } }{ \left| sec(x) \right| dx }$

Put $x=\pi-t$ to get our integral as :

$I=\int _{ \frac { \pi }{ 8 } }^{ \frac { 3\pi }{ 8 } }{ \left| sec(t) \right| dt }$

Since $sec(x)>0$ in the interval hence :

$I=\int _{ \frac { \pi }{ 8 } }^{ \frac { 3\pi }{ 8 } }{ sec(t)dt }$

$I=(ln(sec(t)+tan(t)))\overset { \frac { 3\pi }{ 8 } }{ \underset { \frac { \pi }{ 8 } }{ | } }$

$I=1.211691197011.....$