Evaluate the following integrals:

Solution:

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

We have,

$\cos 2 x=1-2 \sin ^{2} x$

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

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

$=\int \mathrm{e}^{2 \mathrm{x}}\left(\frac{\operatorname{cosec}^{2} \mathrm{x}}{2}-\frac{2 \sin \mathrm{x} \cos \mathrm{x}}{2 \sin ^{2} \mathrm{x}}\right) \mathrm{dx}$

$=\int \mathrm{e}^{2 \mathrm{x}}\left(\frac{\operatorname{cosec}^{2} \mathrm{x}}{2}-\frac{\cos \mathrm{x}}{\sin \mathrm{x}}\right) \mathrm{dx}$

$=\int \mathrm{e}^{2 \mathrm{x}}\left(\frac{\operatorname{cosec}^{2} \mathrm{x}}{2}-\operatorname{cotx}\right) \mathrm{dx}$

Using integration by parts,

$=\frac{1}{2} \int \mathrm{e}^{2 \mathrm{x}} \operatorname{cosec}^{2} \mathrm{xdx}-\int \mathrm{e}^{2 \mathrm{x}} \cot \mathrm{x} \mathrm{dx}$

That is,

$\mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}$

$I_{1}=\frac{1}{2} \int e^{2 x} \operatorname{cosec}^{2} x d x$

$I_{2}=\int e^{2 x} \operatorname{cotxdx}$

Consider

$\mathrm{I}_{1}=\frac{1}{2} \int \mathrm{e}^{2 \mathrm{x}} \operatorname{cosec}^{2} \mathrm{xdx}$

take $\mathrm{e}^{2 \mathrm{x}}$ as first function and $\operatorname{cosec}^{2} \mathrm{x}$ as second function

$u=e^{2 x} ; d u=2 e^{2 x} d x$

$\int \operatorname{cosec}^{2} x d x=\int d v$

Let $v=-\cot x$

$\mathrm{I}_{1}=\frac{1}{2}\left[\mathrm{e}^{2 \mathrm{x}}(-\cot \mathrm{x})-\int(-\cot \mathrm{x}) 2 \mathrm{e}^{2 \mathrm{x}} \mathrm{dx}\right]$

$\mathrm{I}_{1}=\frac{1}{2}\left[\mathrm{e}^{2 \mathrm{x}}(-\cot \mathrm{x})-2 \int \operatorname{cotx} \mathrm{e}^{2 \mathrm{x}} \mathrm{dx}\right]$

$\mathrm{I}_{1}=\frac{1}{2}\left(\mathrm{e}^{2 \mathrm{x}}(-\cot \mathrm{x})\right)+\int \operatorname{cotx} \mathrm{e}^{2 \mathrm{x}} \mathrm{dx}$

Thus,

$I=\frac{1}{2}\left(e^{2 x}(-\cot x)\right)+\int \operatorname{cotx} e^{2 x} d x-\int e^{2 x} \operatorname{cotx} d x$

$I=\frac{1}{2}\left[e^{2 x}(-\cot x)\right]+c$