First slide

Methods of integration

Question

$\int e^{x} (\frac{1 + \sin x}{1 + \cos x}) dx$ is equal to

Easy

A

$e^{x} \sec \frac{x}{2} + C$
B

$\frac{e^{x}}{1 + \cos x} + C$
C

$e^{x} \tan \frac{x}{2} + C$
D

None of these

Solution

$\begin{matrix} \int e^{x} (\frac{1}{1 + \cos x} + \frac{\sin x}{1 + \cos x}) dx \\ = \int e^{x} (\frac{1}{2 \cos^{2} \frac{x}{2}} + \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos^{2} \frac{x}{2}}) dx \\ (∵ \cos 2 x = 2 \cos^{2} x - 1 and \sin 2 x = 2 \sin xcos x) \\ = \int e^{x} (\frac{\sec^{2} \frac{x}{2}}{2} + \tan \frac{x}{2}) dx \\ = \int e^{x} (\tan \frac{x}{2} + \frac{1}{2} \sec^{2} \frac{x}{2}) dx \end{matrix}$
Let,
$\begin{array}{l} f (x) = \tan \frac{x}{2} \Rightarrow f^{'} (x) = \frac{\sec^{2} \frac{x}{2}}{2} \\ \begin{aligned} ∴ \int e^{x} (\tan \frac{x}{2} + \frac{\sec^{2} \frac{x}{2}}{2}) dx = e^{x} \tan \frac{x}{2} + C \\ [∵ \int e^{x} [f (x) + f^{'} (x)] dx = e^{x} f (x)] \end{aligned} \end{array}$

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