Question

Integrate $\int \frac{{\sin }^2\left(\mathrm{x}\right)}{1-\cos \left(\mathrm{x}\right)}\mathrm{dx}$.

Answer 1

Compute the given integration.

In the question, an integration $\int \frac{{\sin }^2\left(\mathrm{x}\right)}{1-\cos \left(\mathrm{x}\right)}\mathrm{dx}$ is given.

Assume that, $I=\int \frac{{\sin }^2\left(\mathrm{x}\right)}{1-\cos \left(\mathrm{x}\right)}\mathrm{dx}$.

$$\begin{gathered} I=\int \frac{{\sin }^2\left(\mathrm{x}\right)}{1-\cos \left(\mathrm{x}\right)}\mathrm{dx}\left[\because {\sin }^2\left(\mathrm{x}\right)+{\cos }^2\left(\mathrm{x}\right)=1\right] \\ \Rightarrow I=\int \frac{1-{\cos }^2\left(\mathrm{x}\right)}{1-\cos \left(\mathrm{x}\right)}\mathrm{dx}\left[\because 1-{\cos }^2\left(\mathrm{x}\right)=\left(1-\cos \left(\mathrm{x}\right)\right)\left(1+\cos \left(\mathrm{x}\right)\right)\right] \\ \Rightarrow I=\int \left(1+\cos \left(\mathrm{x}\right)\right)\mathrm{dx} \\ \Rightarrow I=\int \mathrm{dx}+\int \cos \left(\mathrm{x}\right)\mathrm{dx}\left[\because \int \cos \left(\mathrm{x}\right)\mathrm{dx}=\sin \left(\mathrm{x}\right),\int \mathrm{dx}=\mathrm{x}\right] \\ \Rightarrow I=\mathrm{x}+\sin \left(\mathrm{x}\right)+\mathrm{C} \end{gathered}$$

Hence $\int \frac{{\sin }^2\left(\mathrm{x}\right)}{1-\cos \left(\mathrm{x}\right)}\mathrm{dx}=x+\sin \left(x\right)+C$.