Question

Evaluate the following integrals:

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

Answer 1

$$\text{ We have,}$$

$$I=\int e^{2x}\left(\frac{1-\sin 2x}{1-\cos 2x}\right)dx$$

$$=\int e^{2x}\left(\frac{1-2sinx\cos x}{2{\sin }^{2}x}\right)dx$$

$$\text{ Put t }=2x.\text{ Then dt}=2dx$$

$$\text{ Therefore },$$

$$I=\frac{1}{2}\int e^t\left(\frac{1-2\sin \frac{t}{2}\cos \frac{t}{2}}{2{\sin }^2\frac{t}{2}}\right)dt$$

$$=\frac{1}{4}\int e^t\left(\frac{1-2\sin \frac{t}{2}\cos \frac{t}{2}}{{\sin }^2\frac{t}{2}}\right)dt$$

$$=\frac{1}{4}\int e^t(\frac{1}{{\sin }^2\frac{t}{2}}-\frac{2\sin \frac{t}{2}\cos \frac{t}{2}}{{\sin }^2\frac{t}{2}})dt$$

$$=\frac{1}{4}\int e^t({cosec}^2\frac{t}{2}-2\cot \frac{t}{2})dt$$

$$=-\frac{1}{4}\int e^t(2\cot \frac{t}{2}-{cosec}^2\frac{t}{2})dt$$

$$\text{ Consider, }f\left(x\right)=2\cot \frac{t}{2},{\text{ then f}}^{\left(x\right)}=-{cosec}^2\frac{t}{2}$$

$$\text{Thus, the given integrand is of the form}{\text{ e}}^x[f\left(x\right)+f^{{}^{\prime }}\left(x\right)].$$

$$\text{ Therefore, I }=-\frac{1}{4}(2\cot \frac{t}{2})e^t+c$$

$$=-\frac{1}{4}(2\cot \frac{2x}{2})e^{2x}+c$$

$$\text{ Hence, }\int e^{2x}\left(\frac{1-\sin 2x}{1-\cos 2x}\right)dx=-\frac{1}{2}(\cot x)e^{2x}+c$$