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Question Number 116231 by bemath last updated on 02/Oct/20

$$\int \sec \mathrm{x}\tan \mathrm{x}\sqrt{\tan {}^2\mathrm{x}-3}\mathrm{dx}?$$

Answered by john santu last updated on 02/Oct/20

$$\int \sec x\tan x\sqrt{\sec {}^{2}x-4}dx$$$$[letu=\sec x\to du=\sec x\tan xdx]$$$$\int \sqrt{u^2-4}du=\frac{u}{2}\sqrt{u^2-4}-2\ln \mid u+\sqrt{u^2-4}\mid +c$$$$=\frac{\sec x\sqrt{\sec {}^{2}x-4}}{2}-2\ln \mid \sec x+\sqrt{\sec {}^{2}x-4}\mid +c$$

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