First slide

Introduction to integration

Question

If $\int (1 - \tan 3 x)^{2} d x = \frac{1}{2} [\tan 3 x + \log f (x)] + C$ then $f (x)$ is given by

Easy

A

$\cos 3 x$
B

$\cos^{2} 3 x$
C

$\sin^{2} 3 x$
D

$\cos^{3} 3 x$

Solution

$\begin{array}{l} \int (1 - \tan 3 x)^{2} d x = \int (1 + \tan^{2} 3 x - 2 \tan 3 x) d x \\ = \int (\sec^{2} 3 x - 2 \tan 3 x) d x \\ = \frac{\tan 3 x}{3} - \frac{2}{3} \log | \sec 3 x | + C \\ = \frac{1}{3} [\tan 3 x + \log \cos^{2} 3 x] + C \end{array}$

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