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Question
Evaluate the following integrals:
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Solution
\[\text{ We have,} \]
\[I = \int e^{2x} \left( \frac{1 - \sin2x}{1 - \cos2x} \right)dx\]
\[ = \int e^{2x} \left( \frac{1 - 2 sinx \cos x}{2 \sin^2 x} \right)dx\]
\[\text{ Put t }= 2x . \text{ Then dt} = 2dx\]
\[\text{ Therefore }, \]
\[I = \frac{1}{2}\int e^t \left( \frac{1 - 2 \sin\frac{t}{2} \cos\frac{t}{2}}{2 \sin^2 \frac{t}{2}} \right)dt\]
\[ = \frac{1}{4}\int e^t \left( \frac{1 - 2 \sin\frac{t}{2} \cos\frac{t}{2}}{\sin^2 \frac{t}{2}} \right)dt\]
\[ = \frac{1}{4}\int e^t \left( \frac{1}{\sin^2 \frac{t}{2}} - \frac{2 \sin\frac{t}{2}\cos\frac{t}{2}}{\sin^2 \frac{t}{2}} \right)dt\]
\[ = \frac{1}{4}\int e^t \left( {cosec}^2 \frac{t}{2} - 2\cot\frac{t}{2} \right)dt\]
\[ = - \frac{1}{4}\int e^t \left( 2\cot\frac{t}{2} - {cosec}^2 \frac{t}{2} \right)dt\]
\[\text{ Consider, }f\left( x \right) = 2\cot\frac{t}{2}, \text{ then f}^ \left( x \right) = - {cosec}^2 \frac{t}{2}\]
\[ \text{Thus, the given integrand is of the form} \text{ e}^x \left[ f \left( x \right) + f^{ '} \left( x \right) \right] . \]
\[\text{ Therefore, I }= - \frac{1}{4}\left( 2\cot\frac{t}{2} \right) e^t + c\]
\[ = - \frac{1}{4}\left( 2\cot\frac{2x}{2} \right) e^{2x} + c\]
\[\text{ Hence, }\int e^{2x} \left( \frac{1 - \sin2x}{1 - \cos2x} \right)dx = - \frac{1}{2}\left( \cot x \right) e^{2x} + c\]
