Question

$\underset{-\pi /4}{\overset{\pi /4}{\int }}\frac{dx}{1+\sin x}$ equals

Answer 1

Let , ${\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{1}{1+\sin x}dx$

We know that , ${\int }_{-a}^{a}f\left(x\right)dx={\int }_0^{2a}f\left(x\right)dx$

${\int }_0^{2\times \frac{\pi }{4}}\frac{1}{1+\sin x}dx={\int }_0^{\frac{\pi }{2}}\frac{1}{1+\sin x}dx$

$I={\int }_0^{\frac{\pi }{2}}\frac{1}{1+\cos (\frac{\pi }{2}-x)}dx$

$\because \cos A=2{\cos }^2\frac{A}{2}-1$

$={\int }_0^{\frac{\pi }{2}}\frac{1}{1+2{\cos }^2\frac{(\frac{\pi }{2}-x)}{2}-1}dx$

$={\int }_o^{\frac{\pi }{2}}\frac{1}{2{\cos }^2\frac{(\frac{\pi }{2}-x)}{2}}dx$

$=={\int }_0^{\frac{\pi }{2}}\frac{1}{2{\cos }^2(\frac{\pi }{4}-\frac{x}{2})}dx$

$=\frac{1}{2}{\int }_o^{\frac{\pi }{2}}{\sec }^2(\frac{\pi }{4}-\frac{x}{2})dx$

$={{\left[\frac{1}{2}\tan (\frac{\pi }{4}-\frac{x}{2})\right]}^{\frac{\pi }{2}}}_0+C$

$=\frac{1}{2}\tan (\frac{\pi }{4}-\frac{\pi }{2})-\frac{1}{2}\tan (\frac{\pi }{4}-\frac{0}{2})+C$

$=\frac{1}{2}\tan (-\frac{\pi }{4})-\frac{1}{2}\tan \frac{\pi }{4}+C$

$=-\frac{1}{2}\tan \frac{\pi }{4}-\frac{1}{2}\tan \frac{\pi }{4}+C$

$=-\frac{1}{2}\times 1-\frac{1}{2}\times 1+C$

$=-\frac{1}{4}$