# Evaluate the Following Integrals: ∫ E 2 X ( 1 − Sin 2 X 1 − Cos 2 X ) D X - Mathematics

Sum

Evaluate the following integrals:

$\int e^{2x} \left( \frac{1 - \sin2x}{1 - \cos2x} \right)dx$

#### Solution

$\text{ We have,}$

$I = \int e^{2x} \left( \frac{1 - \sin2x}{1 - \cos2x} \right)dx$

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

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

$\text{ Therefore },$

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

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

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

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

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

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

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

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

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

$\text{ Hence, }\int e^{2x} \left( \frac{1 - \sin2x}{1 - \cos2x} \right)dx = - \frac{1}{2}\left( \cot x \right) e^{2x} + c$

Concept: Evaluation of Simple Integrals of the Following Types and Problems
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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Exercise 19.26 | Q 24 | Page 143