First slide

Introduction to integration

Question

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

Easy

A

$\log | \tan x | + C$
B

$e^{x} \tan x / 2 + C$
C

$e^{x} \cot x + C$
D

$\sin \log x + C$

Solution

$\begin{array}{l} \int \frac{e^{x} (1 + \sin x)}{1 + \cos x} d x \\ = \int \frac{e^{x} [1 + 2 \sin (x / 2) \cos (x / 2)]}{2 \cos^{2} (x / 2)} d x \\ = \int e^{x} (\frac{1}{2} \sec^{2} \frac{x}{2} + \tan \frac{x}{2}) d x \\ = \frac{1}{2} \int e^{x} \sec^{2} \frac{x}{2} d x + \int e^{x} \tan \frac{x}{2} d x \end{array}$
$\begin{array}{l} = \frac{1}{2} \int e^{x} \sec^{2} \frac{x}{2} d x + e^{x} \tan \frac{x}{2} - \frac{1}{2} \int e^{x} \sec^{2} \frac{x}{2} + C \\ [integrating by parts] \\ = e^{x} \tan \frac{x}{2} + C \end{array}$

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