Check if an integral is convergent and calculate its value I have improper integral $$\int_0^\pi \frac{\cos(x)}{(1-2\sin(x))^{1/3}} dx$$
I have to check if it is convergent. If yes, then i have to evaluate it.
I think it is convergent. I can make substitution $t=\sin(x)$. But what next?
 A: First, since it is an improper integral you need to split it into the sum of three parts
$$\lim_{a\rightarrow\frac{5\pi}6}\lim_{b\rightarrow\frac{\pi}6}\int_a^\pi \frac{\cos(x)}{(1-2\sin(x))^{1/3}} dx + \int_b^a\frac{\cos(x)}{(1-2\sin(x))^{1/3}} dx + \int_0^b\frac{\cos(x)}{(1-2\sin(x))^{1/3}} dx$$
If you make the substitution $t = 1-2\sin(x)$ you will get $dt = -2\cos(x) dx$.
The boundaries become $t(\pi) = 1-2\sin(\pi) = 1$, $t(\frac{5\pi}6) = 1-2\sin(\frac{5\pi}6) = 0$, $t(\frac{\pi}6) = 1-2\sin(\frac{\pi}6) = 0$, and $t(0) = 1-2\sin(0) = 1$
Use this to simplify your equation to $$\int_0^1\frac{-1}{2t^{1/3}} dt + \int_0^0\frac{-1}{2t^{1/3}} dt + \int_1^0\frac{-1}{2t^{1/3}} dt$$
We will use $$\int_0^0\frac{-1}{2t^{1/3}} dt = 0$$ and $\int_1^0\frac{-1}{2t^{1/3}} dt = -\int_0^1\frac{-1}{2t^{1/3}} dt$
So the sum totals to $0$
A: Formally, we can state
\begin{align}
\int_{0}^{\pi}\frac{\cos x}{(1-2\sin x)^{1/3}}dx&=\int_0^{\pi/2}\frac{\cos x}{(1-2\sin x)^{1/3}}dx+\int_{\pi/2}^{\pi}\frac{\cos x}{(1-2\sin x)^{1/3}}dx\\
&=\int_0^{\pi/2}\frac{\cos x}{(1-2\sin x)^{1/3}}dx+\int_{y=\pi/2}^{y=0}\frac{\cos (\pi-y)}{[1-2\sin (\pi-y)]^{1/3}}d(\pi-y)\\
\end{align}
Where $y=\pi-x$, then
\begin{align}
\int_{0}^{\pi}\frac{\cos x}{(1-2\sin x)^{1/3}}dx&=\int_0^{\pi/2}\frac{\cos x}{(1-2\sin x)^{1/3}}dx-\int_{\pi/2}^{0}\frac{-\cos y}{(1-2\sin y)^{1/3}}dy\\
&=\int_0^{\pi/2}\frac{\cos x}{(1-2\sin x)^{1/3}}dx+\int_{\pi/2}^0\frac{\cos y}{(1-2\sin y)^{1/3}}dy\\
&=\int_0^{\pi/2}\frac{\cos x}{(1-2\sin x)^{1/3}}dx-\int_0^{\pi/2}\frac{\cos y}{(1-2\sin y)^{1/3}}dy\\
&=0
\end{align}
This is true if $$\int_0^{\pi/2}\frac{\cos x}{(1-2\sin x)^{1/3}}dx$$ converges.

By using $t=\sin x,\;\;dt=\cos x dx$ we have
$$\int\frac{\cos x}{(1-2\sin x)^{1/3}}dx=\int\frac{dt}{(1-2t)^{1/3}}=-\frac{3}{4}(1-2t)^{2/3}+C=-\frac{3}{4}(1-2\sin x)^{2/3}+C$$
Then
$$\int_0^{\pi/6}\frac{\cos x}{(1-2\sin x)^{1/3}}dx=\lim_{b\to\frac{\pi}{6}^-}\int_0^{b}\frac{\cos x}{(1-2\sin x)^{1/3}}dx=\lim_{b\to\frac{\pi}{6}^-}-\frac{3}{4}(1-2\sin b)^{2/3}+\frac{3}{4}=\frac{3}{4},$$
$$\int_{\pi/6}^{\pi/2}\frac{\cos x}{(1-2\sin x)^{1/3}}dx=\lim_{a\to\frac{\pi}{6}^+}\int_a^{\pi/2}\frac{\cos x}{(1-2\sin x)^{1/3}}dx=\frac{3}{4}+\lim_{a\to\frac{\pi}{6}^+}\frac{3}{4}(1-2\sin a)^{2/3}=\frac{3}{4}$$
Then, the integral converges, also
$$\int_0^{\pi/2}\frac{\cos x}{(1-2\sin x)^{1/3}}dx=\frac{3}{2}$$
Therefore
$$\int_{0}^{\pi}\frac{\cos x}{(1-2\sin x)^{1/3}}dx=0$$
