Is there another way to solve this integral?

My way to solve this integral. I wonder is there another way to solve it as it's very long for me.

$$\int_{0}^{\pi}\frac{1-\sin (x)}{\sin (x)+1}dx$$

Let $$u=\tan (\frac{x}{2})$$ $$du=\frac{1}{2}\sec ^2(\frac{x}{2})dx $$

By Weierstrass Substitution

$$\sin (x)=\frac{2u}{u^2+1}$$

$$\cos (x)=\frac{1-u^2}{u^2+1}$$

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

$$=\int_{0}^{\infty }\frac{2(1-\frac{2u}{u^2+1})}{(u^2+1)(\frac{2u}{u^2+1}+1)}du$$

$$=\int_{0}^{\infty }\frac{2(u-1)^2}{u^4+2u^3+2u^2+2u+1}du $$

$$=2\int_{0}^{\infty }\frac{(u-1)^2}{u^4+2u^3+2u^2+2u+1}du $$

$$=2\int_{0}^{\infty }\frac{(u-1)^2}{(u+1)^2(u^2+1)}du $$

$$=2\int_{0}^{\infty }(\frac{2}{(u+1)^2}-\frac{1}{u^2+1})du $$

$$=-2\int_{0}^{\infty }\frac{1}{u^2+1}du+4\int_{0}^{\infty}\frac{1}{(u+1)^2}du $$

$$\lim_{b\rightarrow \infty }\left | (-2\tan^{-1}(u)) \right |_{0}^{b}+4\int_{0}^{\infty}\frac{1}{(u+1)^2}du$$

$$=(\lim_{b\rightarrow \infty}-2\tan^{-1}(b))+4\int_{0}^{\infty}\frac{1}{(u+1)^2}du$$

$$=-\pi+4\int_{0}^{\infty}\frac{1}{(u+1)^2}du$$

Let $$s=u+1$$

$$ds=du$$

$$=-\pi+4\int_{1}^{\infty}\frac{1}{s^2}ds$$

$$=-\pi+\lim_{b\rightarrow \infty}\left | (-\frac{4}{s}) \right |_{1}^{b}$$

$$=-\pi+(\lim_{b\rightarrow \infty} -\frac{4}{b}) +4$$

$$=4-\pi$$

$$\approx 0.85841$$

Solution 1:

Substitute $x=\pi/2-2t$ so the integral becomes $$ -2\int_{\pi/4}^{-\pi/4}\frac{1-\cos 2t}{1+\cos 2t}\,dt= 2\int_{-\pi/4}^{\pi/4}\frac{1-\cos^2t}{\cos^2t}\,dt =2\Bigl[\tan t - t\Bigr]_{-\pi/4}^{\pi/4} $$

Solution 2:

First use partial fractions to get rid of the sine in the numerator: $$ \int_0^{\pi} \frac{1-\sin{x}-1+1}{1+\sin{x}} \, dx = \int_0^{\pi} \left( \frac{2}{1+\sin{x}} - 1 \right) \, dx = -\pi + \int_0^{\pi} \frac{2 \, dx}{1+\sin{x}}. $$ We have the identity $$ 1+\sin{x} = 2\sin^2{\left( \frac{x}{2} + \frac{\pi}{4} \right)} $$ (from $\cos{2\theta}=1-2\sin^2{\theta}$ and $\sin{\theta}=-\cos{(\theta+\pi/2)}$), so the remaining integral is $$ \int_0^{\pi} \csc^2{\left( \frac{x}{2} + \frac{\pi}{4} \right)} \, dx = \left[ -2 \cot{\left( \frac{x}{2} + \frac{\pi}{4} \right)} \right]_0^{\pi} = -2(-1-1) = 4 $$

Solution 3:

Multiply numerator and denominator by $1 - \sin x$. So that $\displaystyle\int_0^\pi \dfrac{1-\sin x}{1+\sin x} \cdot \dfrac{1-\sin x}{1-\sin x}dx $

$$ = \displaystyle\int_0^\pi \dfrac{1-2\sin x+ \sin^2x}{\cos^2x}dx = \int_0^\pi \sec^2x - 2\dfrac{\sin x}{\cos^2x} + \tan^2x dx$$

We know that $\tan^2x + 1 = \sec^2x$. So the integral would look like: $$= \int_0^\pi 2\sec^2x -1 - 2\dfrac{\sin x}{\cos^2x}dx$$

We know the integral of $\sec^2x$ is just $\tan x$, and the last integrand can be solved by letting $u=\cos x$.

Solution 4:

Hint: Note that by symmetry our integral is twice the integral from $0$ to $\pi/2$. Then by symmetry replace $\sin x$ by $\cos x$. Then use the identities $\cos x=2\cos^2 (x/2)-1=1-2\sin^2(x/2)$.