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Question Number 142404 by mohammad17 last updated on 31/May/21

Commented by mohammad17 last updated on 31/May/21

$$helpmesirplease$$

Answered by Ar Brandon last updated on 31/May/21

$$\mathcal{I}=\int \frac{\mathrm{dx}}{\sec {\mathrm{xtan}}^2\mathrm{x}}=\int \frac{{\sec }^2\mathrm{x}-{\tan }^2\mathrm{x}}{\sec {\mathrm{xtan}}^2\mathrm{x}}\mathrm{dx}$$$$=\int \frac{\cos \mathrm{x}}{{\sin }^2\mathrm{x}}\mathrm{dx}-\int \mathrm{cosxdx}=-\frac{1}{\sin \mathrm{x}}-\sin \mathrm{x}+\mathrm{C}$$

Answered by Ar Brandon last updated on 31/May/21

$$\mathrm{J}=\int {\cot }^3{\mathrm{xcsc}}^3\mathrm{xdx}$$$$=\int \left({\cot }^2{\mathrm{xcsc}}^2\mathrm{x}\right)\left(\csc \mathrm{xcot}\mathrm{x}\right)\mathrm{dx}$$$$=-\int ({\csc }^4\mathrm{x}-{\csc }^2\mathrm{x})\mathrm{d}\left(\csc \mathrm{x}\right)$$$$=\frac{{\csc }^3\mathrm{x}}{3}-\frac{{\csc }^5\mathrm{x}}{5}+\mathrm{C}$$

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