Question

Prove that:
$\underset{-\pi /4}{\overset{\pi /4}{\int }}\frac{dx}{1+\sin x}=$

Answer 1

Apply $u$ - substitution: $u=\tan \left(\frac{x}{2}\right)$

$=\int \frac{2}{1+u^2+2u}du$

Take the constant out:$\int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx$

$=2\cdot \int \frac{1}{1+u^2+2u}du$

$Completethesquare:$1 + u ^ { 2 } + 2 u : ( u + 1 ) ^ { 2 }$

$=2\cdot \int \frac{1}{(u+1{)}^2}du$

Apply u- substitution: $v=u+1$

$=2\cdot \int \frac{1}{v^2}dv$

$=2\cdot \int v^{-2}dv$

Apply the Power Rule: $\int x^{a}dx=\frac{x^{a+1}}{a+1},a\ne -1$

$=2\cdot \frac{v^{-2+1}}{-2+1}$

Substitute back $v=u+1,u=\tan \left(\frac{x}{2}\right)$

$=2\cdot \frac{{(\tan \left(\frac{x}{2}\right)+1)}^{-2+1}}{-2+1}$

Simplify $2\cdot \frac{{(\tan \left(\frac{x}{2}\right)+1)}^{-2+1}}{-2+1}:-\frac{2}{\tan \left(\frac{x}{2}\right)+1}$

$=-\frac{2}{\tan \left(\frac{x}{2}\right)+1}$

Compute the boundaries: ${\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{1}{1+\sin \left(x\right)}dx=-\sqrt{2}-(-2-\sqrt{2})$

$=-\sqrt{2}-(-2-\sqrt{2})$ =2