How to take the integral of a function with trigonometry?

Question

Tell me please how to take this integral?

$$\int{\left(1-\frac{1}{\sin x}\right)}^{\frac{3}{2}}dx$$

I tried making replacements and using various trigonometric formulas but nothing worked.

Answer 1

The integrand is defined for $\sin x<0$ (assuming we're working in real numbers). Then you can make the tangent half-angle substitution $t=\tan\frac x2$ and note $t<0$. Then $\sin x=\frac{2t}{1+t^2}$, $dx=\frac{2}{1+t^2}\,dt$, and the integral becomes $$I=\int\left(\frac{-(t-1)^2}{2t}\right)^{3/2}\frac{2}{1+t^2}\,dt=\int\left(-\frac{1}{2t}\right)^{3/2}\frac{2(1-t)^3}{1+t^2}\,dt$$ Next, you can substitute $u=\sqrt{-\dfrac{1}{2t}}\Rightarrow t=-\dfrac{1}{2u^2}\Rightarrow dt=\dfrac{du}{u^3}$ to arrive at $$I=\int\frac{\left(2u^2+1\right)^3}{u^2\left(4u^4+1\right)}\,du=\int\left(2+\frac{1}{u^2}+\frac{2}{2u^2-2u+1}+\frac{2}{2u^2+2u+1}\right)du $$ Can you continue the integration from here?