First slide

Introduction to integration

Question

$Evaluate \int \frac{1 - \tan x}{1 + \tan x} d x$

Easy

A

$\log |\cos x + \sin x| + C$
B

$\log |\cos x - \sin x| + C$
C

$\log |\tan x + \sin x| + C$
D

None of these

Solution

$\begin{array}{l} \int \frac{1 - \tan x}{1 + \tan x} d x = \int \frac{\cos x - \sin x}{\cos x + \sin x} d x \\ = \int \frac{\frac{d}{d x} (\cos x + \sin x)}{\cos x + \sin x} d x \\ = \log |\cos x + \sin x| + C \end{array}$

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