Question

Evaluate the following integrals:

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int {\mathrm{e}}^{2x}\left(\frac{1-\sin 2\mathrm{x}}{1-\cos 2\mathrm{x}}\right)\mathrm{d}x \\ =\int {\mathrm{e}}^{2x}\left(\frac{1-2\mathrm{sinx}\cos x}{2{\sin }^2\mathrm{x}}\right)\mathrm{d}x \\ \mathrm{Put}t=2x.\mathrm{Then}\mathrm{d}t=2\mathrm{d}x \\ \mathrm{Therefore}, \\ I=\frac{1}{2}\int {\mathrm{e}}^t\left(\frac{1-2\sin {\displaystyle \frac{t}{2}}\cos {\displaystyle \frac{t}{2}}}{2{\sin }^2{\displaystyle \frac{t}{2}}}\right)\mathrm{d}t \\ =\frac{1}{4}\int {\mathrm{e}}^t\left(\frac{1-2\sin {\displaystyle \frac{t}{2}}\cos {\displaystyle \frac{t}{2}}}{{\sin }^2{\displaystyle \frac{t}{2}}}\right)\mathrm{d}t \\ =\frac{1}{4}\int {\mathrm{e}}^t\left(\frac{1}{{\sin }^2{\displaystyle \frac{t}{2}}}-\frac{2\sin {\displaystyle \frac{t}{2}}\cos {\displaystyle \frac{t}{2}}}{{\sin }^2{\displaystyle \frac{t}{2}}}\right)\mathrm{d}t \\ =\frac{1}{4}\int {\mathrm{e}}^t\left({\mathrm{cosec}}^2\frac{t}{2}-2\cot \frac{t}{2}\right)\mathrm{d}t \\ =-\frac{1}{4}\int {\mathrm{e}}^t\left(2\cot \frac{t}{2}-{\mathrm{cosec}}^2\frac{t}{2}\right)\mathrm{d}t \\ \mathrm{Consider},f\left(x\right)=2\cot \frac{t}{2},\mathrm{then}{f}^{\text{'}}\left(x\right)=-{\mathrm{cosec}}^2\frac{t}{2} \\ \mathrm{Thus},\mathrm{the}\mathrm{given}\mathrm{integrand}\mathrm{is}\mathrm{of}\mathrm{the}\mathrm{form}{\mathrm{e}}^x\left[f\left(x\right)+f^{\text{'}}\left(x\right)\right]. \\ \mathrm{Therefore},I=-\frac{1}{4}\left(2\cot \frac{t}{2}\right){\mathrm{e}}^t+c \\ =-\frac{1}{4}\left(2\cot \frac{2x}{2}\right){\mathrm{e}}^{2x}+c \\ \mathrm{Hence},\int {\mathrm{e}}^{2x}\left(\frac{1-\sin 2\mathrm{x}}{1-\cos 2\mathrm{x}}\right)\mathrm{d}x=-\frac{1}{2}\left(\cot x\right){\mathrm{e}}^{2x}+c \end{gathered}$$