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Question Number 116427 by bemath last updated on 04/Oct/20

$$\int {(\sec \mathrm{x}-\tan \mathrm{x})}^2\mathrm{dx}=?$$

Answered by john santu last updated on 04/Oct/20

$$\int {\left(\frac{1-\sin x}{\cos x}\right)}^{2}dx=\int \frac{1-2\sin x+\sin {}^{2}x}{\cos {}^{2}x}dx$$$$=\int \sec {}^{2}xdx+\int \frac{2d\left(\cos x\right)}{\cos {}^{2}x}+\int \tan {}^{2}xdx$$$$=\int 2\sec {}^{2}xdx+\frac{2}{\cos x}-x+c$$$$=2\tan x+2\sec x-x+c$$

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