Question
Evaluate the following integrals:
$\int\text{e}^{2\text{x}}\Big(\frac{1-\sin2\text{x}}{1-\cos2\text{x}}\Big)\text{dx}$
$\int\text{e}^{2\text{x}}\Big(\frac{1-\sin2\text{x}}{1-\cos2\text{x}}\Big)\text{dx}$
✓
Answer
WE have,
$\text{I}=\int\text{e}^{2\text{x}}\Big(\frac{1-\sin2\text{x}}{1-\cos2\text{x}}\Big)\text{dx}$
$=\int\text{e}^{2\text{x}}\Big(\frac{1-2\sin\text{x}\cos\text{x}}{2\sin^2\text{x}}\Big)\text{dx}$
Put $\text{t}=2\text{x}.$ Then $\text{dt}=2\text{dx}$
Therefore,
$\text{I}=\frac{1}{2}\int\text{e}^{\text{t}}\bigg(\frac{1-2\sin\frac{\text{t}}{2}\cos\frac{\text{t}}{2}}{2\sin^2\frac{\text{t}}{2}}\bigg)\text{dt}$
$=\frac{1}{4}\int\text{e}^{\text{t}}\bigg(\frac{1-2\sin\frac{\text{t}}{2}\cos\frac{\text{t}}{2}}{\sin^2\frac{\text{t}}{2}}\bigg)\text{dt}$
$=\frac{1}4{\int\text{e}^{\text{t}}}\bigg(\frac{1}{\sin^2\frac{\text{t}}{2}}-\frac{2\sin\frac{\text{t}}{2}\cos\frac{\text{t}}{2}}{\sin^2\frac{\text{t}}{2}}\bigg)\text{dt}$
$=\frac{1}{4}\int\text{e}^{\text{t}}\big(\text{cosec}^2\frac{\text{t}}{2}-2\cot\frac{\text{t}}{2}\big)\text{dt}$
$=-\frac{1}{4}\int\text{e}^{\text{t}}\big(2\cot\frac{\text{t}}{2}-\text{cosec}^2\frac{\text{t}}{2}\big)\text{dt}$
Consider, $\text{f(x)}=2\cot\frac{\text{t}}{2},$ then $\text{f}'\text{(x)}=-\text{cosec}^2\frac{\text{t}}{2}$
Thus, the given integrand is of the from $\text{e}^{\text{x}}\big[\text{f(x)}+\text{f'}\text{(x)}\big].$
Therefore, $\text{I}=-\frac{1}{4}\big(2\cot\frac{\text{t}}{2}\big)\text{e}^{\text{t}}+\text{C}$
$=-\frac{1}{4}\big(2\cot\frac{2\text{x}}{2}\big)\text{e}^{2\text{x}}+\text{C}$
Hence, $\int\text{e}^{2\text{x}}\Big(\frac{1-\sin2\text{x}}{1-\cos2\text{x}}\Big)\text{dx}=-\frac{1}{2}(\cot\text{x})\text{e}^{2\text{x}}+\text{C}$
$\text{I}=\int\text{e}^{2\text{x}}\Big(\frac{1-\sin2\text{x}}{1-\cos2\text{x}}\Big)\text{dx}$
$=\int\text{e}^{2\text{x}}\Big(\frac{1-2\sin\text{x}\cos\text{x}}{2\sin^2\text{x}}\Big)\text{dx}$
Put $\text{t}=2\text{x}.$ Then $\text{dt}=2\text{dx}$
Therefore,
$\text{I}=\frac{1}{2}\int\text{e}^{\text{t}}\bigg(\frac{1-2\sin\frac{\text{t}}{2}\cos\frac{\text{t}}{2}}{2\sin^2\frac{\text{t}}{2}}\bigg)\text{dt}$
$=\frac{1}{4}\int\text{e}^{\text{t}}\bigg(\frac{1-2\sin\frac{\text{t}}{2}\cos\frac{\text{t}}{2}}{\sin^2\frac{\text{t}}{2}}\bigg)\text{dt}$
$=\frac{1}4{\int\text{e}^{\text{t}}}\bigg(\frac{1}{\sin^2\frac{\text{t}}{2}}-\frac{2\sin\frac{\text{t}}{2}\cos\frac{\text{t}}{2}}{\sin^2\frac{\text{t}}{2}}\bigg)\text{dt}$
$=\frac{1}{4}\int\text{e}^{\text{t}}\big(\text{cosec}^2\frac{\text{t}}{2}-2\cot\frac{\text{t}}{2}\big)\text{dt}$
$=-\frac{1}{4}\int\text{e}^{\text{t}}\big(2\cot\frac{\text{t}}{2}-\text{cosec}^2\frac{\text{t}}{2}\big)\text{dt}$
Consider, $\text{f(x)}=2\cot\frac{\text{t}}{2},$ then $\text{f}'\text{(x)}=-\text{cosec}^2\frac{\text{t}}{2}$
Thus, the given integrand is of the from $\text{e}^{\text{x}}\big[\text{f(x)}+\text{f'}\text{(x)}\big].$
Therefore, $\text{I}=-\frac{1}{4}\big(2\cot\frac{\text{t}}{2}\big)\text{e}^{\text{t}}+\text{C}$
$=-\frac{1}{4}\big(2\cot\frac{2\text{x}}{2}\big)\text{e}^{2\text{x}}+\text{C}$
Hence, $\int\text{e}^{2\text{x}}\Big(\frac{1-\sin2\text{x}}{1-\cos2\text{x}}\Big)\text{dx}=-\frac{1}{2}(\cot\text{x})\text{e}^{2\text{x}}+\text{C}$
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