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Question Number 166320 by bobhans last updated on 18/Feb/22

$$\int \frac{\mathrm{dx}}{\tan {}^2\mathrm{x}+\sin {}^2\mathrm{x}}=?$$

Answered by cortano1 last updated on 18/Feb/22

$$\mathrm{Y}=\int \frac{\mathrm{dx}}{\tan {}^2\mathrm{x}+\sin {}^2\mathrm{x}}=?$$$$\mathrm{Y}=\int \frac{\sec {}^2\mathrm{x}}{\sec {}^2\mathrm{x}(\tan {}^2\mathrm{x}+\sin {}^2\mathrm{x})}\mathrm{dx}$$$$\mathrm{Y}=\int \frac{\sec {}^2\mathrm{x}}{\sec {}^2\mathrm{x}\tan {}^2\mathrm{x}+\tan {}^2\mathrm{x}}\mathrm{dx}$$$$\mathrm{Y}=\int \frac{\sec {}^2\mathrm{x}}{\tan {}^2\mathrm{x}(\sec {}^2\mathrm{x}+1)}\mathrm{dx}$$$$\mathrm{Y}=\int \frac{\mathrm{d}\left(\tan \mathrm{x}\right)}{\tan {}^2\mathrm{x}(2+\tan {}^2\mathrm{x})}$$$$\mathrm{Y}=\int \frac{\mathrm{dy}}{{\mathrm{y}}^2(2+{\mathrm{y}}^2)};[\mathrm{y}=\tan \mathrm{x}]$$$$\mathrm{Y}=\int \frac{1}{2}(\frac{1}{{\mathrm{y}}^2}-\frac{1}{2+{\mathrm{y}}^2})\mathrm{dy}$$$$\mathrm{Y}=\frac{1}{2}\int ({\mathrm{y}}^{-2}-\frac{1}{{\left(\sqrt{2}\right)}^2+{\mathrm{y}}^2})\mathrm{dy}$$$$\mathrm{Y}=-\frac{1}{2\mathrm{y}}-\frac{1}{2}.\frac{1}{\sqrt{2}}\arctan \left(\frac{\mathrm{y}}{\sqrt{2}}\right)+\mathrm{c}$$$$\mathrm{Y}=-\frac{1}{2\tan \mathrm{x}}-\frac{\sqrt{2}}{4}\arctan \left(\frac{\tan \mathrm{x}}{\sqrt{2}}\right)+\mathrm{c}$$

Commented by peter frank last updated on 19/Feb/22

$$\mathrm{thank}\mathrm{you}$$

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