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Question Number 93959 by mashallah last updated on 16/May/20

$$\int (\tan 3\mathrm{x}+\sec 3\mathrm{x})\mathrm{dx}=$$

Answered by john santu last updated on 16/May/20

$$\int \frac{\sin 3\mathrm{x}}{\cos 3\mathrm{x}}\mathrm{dx}=-\frac{1}{3}\int \frac{\mathrm{d}\left(\cos 3\mathrm{x}\right)}{\cos 3\mathrm{x}}$$$$=-\frac{1}{3}\ln \mid \cos 3\mathrm{x}\mid$$$$\int \sec 3\mathrm{x}\mathrm{dx}=\frac{1}{3}\int \sec 3\mathrm{x}\mathrm{d}\left(3\mathrm{x}\right)$$$$=\frac{1}{3}\ln \mid \sec 3\mathrm{x}+\tan 3\mathrm{x}\mid$$$$\therefore =-\frac{1}{3}\ln \mid \cos 3\mathrm{x}\mid +\frac{1}{3}\ln \mid \sec 3\mathrm{x}+\tan 3\mathrm{x}\mid +\mathrm{c}$$$$=\frac{1}{3}\ln \mid \frac{\sec 3\mathrm{x}+\tan 3\mathrm{x}}{\cos 3\mathrm{x}}\mid +\mathrm{c}$$$$$$

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