I have trouble while integrating $\tfrac{1}{1+\sin x}$ from $0$ to $\pi$

Question

I am getting trouble while integrating $$ \int_0^\pi \! \frac{1}{1 + \sin x} \, \mathrm{d}x $$ by substituting $u$ as $\sin x$.

Answer 1

Hint: \begin{align} \frac{1}{1 + \sin x} &= \frac{1 - \sin x}{(1 + \sin x)(1 - \sin x)} \\[2pt] &= \frac{1 - \sin x}{1 - \sin^2 x} \\[2pt] &= \frac{1 - \sin x}{\cos^2 x} \\[2pt] &= \frac{1}{\cos^2 x} - \frac{\sin x}{\cos^2 x} \\[2pt] &= \sec^2 x - \sec x \tan x \end{align}

Answer 2

I would recommend you to notice that

\begin{align*} \int_{0}^{\pi}\frac{1}{1 + \sin(x)}\mathrm{d}x & = \int_{0}^{\pi}\frac{1}{1 + 2\sin(x/2)\cos(x/2)}\mathrm{d}x\\\\ & = \int_{0}^{\pi}\frac{1}{(\cos(x/2) + \sin(x/2))^{2}}\mathrm{d}x\\\\ & = \int_{0}^{\pi}\frac{1}{\cos^{2}(x/2)(1 + \tan(x/2))^{2}}\mathrm{d}x\\\\ & = \int_{0}^{\pi}\frac{\sec^{2}(x/2)}{(1 + \tan(x/2))^{2}}\mathrm{d}x\\\\ & = 2\int_{0}^{\infty}\frac{1}{(1 + u)^{2}}\mathrm{d}u = 2 \end{align*}

Answer 3

Exploit $\displaystyle \frac1{1-x}=\sum_{n=0}^\infty x^n$ (when $|x|<1$) to rewrite the integral in terms of the beta function, then utilize the power series derived in this answer:

$$\begin{align*} \int_0^\pi \frac{dx}{1+\sin x} &= \sum_{n=0}^\infty (-1)^n \int_0^\pi \sin^n x \, dx \\ &= \Gamma\left(\frac12\right) \sum_{n=0}^\infty (-1)^n \frac{\Gamma\left(\frac n2+\frac12\right)}{\Gamma\left(\frac n2+1\right)} \\ &= \sqrt\pi \sum_{n=0}^\infty \left(\frac{\Gamma\left(n+\frac12\right)}{\Gamma(n+1)} - \frac{\Gamma(n+1)}{\Gamma\left(n+\frac32\right)}\right) \\ &= 2 \lim_{x\to1^-} \frac{\frac\pi2 - \arcsin x}{\sqrt{1-x^2}} \end{align*}$$

Answer 4

Let $I$ denote the integral, and observe that, by symmetry,

$$I=2\int_0^{\frac{\pi}{2}}\frac{\mathrm{d}x}{1+\sin x}.$$

Furthermore, if $u=\frac{x}{2}+\frac{\pi}{4}$, then

\begin{align*} \frac{1}{1+\sin x} &=\frac{1}{1+\sin\left(2u-\frac{\pi}{2}\right)} \\ &=\frac{1}{1-\cos2u} \\ &=\frac{1}{2\sin^2u} \\ &=\frac{1}{2}\frac{\sin^2 u+\cos^2u}{\sin^2u} \\ &=\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}u}\left(\frac{-\cos u}{\sin u}\right) \\ &=-\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}u}\left(\frac{1}{\tan u}\right) \end{align*}

by recognizing the quotient rule, so

$$I=-2\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\mathrm{d}}{\mathrm{d}u}\left(\frac{1}{\tan u}\right)~\mathrm{d}u=\frac{2}{\tan\frac{\pi}{4}}-\lim_{u\uparrow\frac{\pi}{2}}\frac{2}{\tan u}=2$$

by the FTC.