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Question Number 30258 by mondodotto@gmail.com last updated on 19/Feb/18

Answered by $@ty@m last updated on 19/Feb/18

$$Let\sec x-1=z$$$$\tan {}^{2}xdx=dz$$$$dx=\frac{dz}{\sec {}^{2}x-1}=\frac{dz}{z(z+2)}$$$$I=\int \frac{dx}{z^2(z+2)}$$$$Let\frac{1}{z^2(z+2)}=\frac{A}{z^2}+\frac{B}{z+2}+\frac{C}{z}$$$$\Rightarrow A(z+2)+{Bz}^2+Cz(z+2)\equiv 1$$$$\Rightarrow A=\frac{1}{2},B=\frac{1}{4}\mathrm{\&}$$$$A+B-C=1\Rightarrow C=\frac{-1}{4}$$$$\therefore I=\int \frac{dz}{2z^2}+\int \frac{dz}{4(z+2)}-\int \frac{dz}{4z}$$$$=\frac{1}{2}\left(\frac{-1}{z}\right)+\frac{1}{4}\ln (z+2)-\frac{1}{4}\ln z+C$$$$=\frac{-1}{2(\sec x-1)}+\frac{1}{4}\ln (\sec x+1)-\frac{1}{4}\ln (\sec x-1)+C$$

Commented by rahul 19 last updated on 19/Feb/18

$$\mathrm{try}\mathrm{my}\mathrm{ques}.\mathrm{then}:)$$

Commented by $@ty@m last updated on 19/Feb/18

$$Notaschallangingasclaimed.$$$$:($$

Answered by ajfour last updated on 19/Feb/18

$$\int \frac{dx}{\sec x-1}=\int \frac{\sec x+1}{\sec {}^{2}x-1}dx$$$$=\int \frac{(\sec x+1)}{\tan {}^{2}x}dx$$$$=\int \mathrm{cosec}x\cot xdx+\int \cot {}^{2}xdx$$$$=-\mathrm{cosec}x+c_1+\int (\mathrm{cosec}{}^{2}x-1)dx$$$$=-\mathrm{cosec}x-\cot x-x-c.$$

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