integrate x.tanx.(secx)^2

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$$\begin{gathered} \mathrm{Let} \mathrm{I}=\int \mathrm{xtanx} {\sec }^2\mathrm{x} \mathrm{dx} \\ \mathrm{Let} \mathrm{tanx}=\mathrm{t}\Rightarrow {\sec }^2\mathrm{dx}=\mathrm{dt}, \mathrm{so} \mathrm{integra}\left(1\right) \mathrm{becomes}, \\ \mathrm{I}=\int \mathrm{t} {\tan }^{-1}\mathrm{t} \mathrm{dt} \\ ={\tan }^{-1}\mathrm{t}\int \mathrm{t} \mathrm{dt}-\int \frac{\mathrm{d}}{\mathrm{dt}}\left({\tan }^{-1}\mathrm{t}\right)\left(\int \mathrm{t} \mathrm{dt}\right)\mathrm{dt} \\ ={\tan }^{-1}\mathrm{t}\left(\frac{{\mathrm{t}}^2}{2}\right)-\int \frac{1}{1+{\mathrm{t}}^2}\left(\frac{{\mathrm{t}}^2}{2}\right)\mathrm{dt} \\ ={\tan }^{-1}\mathrm{t}\left(\frac{{\mathrm{t}}^2}{2}\right)-\frac{1}{2}\int \frac{{\mathrm{t}}^2+1-1}{{\mathrm{t}}^2+1}\mathrm{dt} \\ ={\tan }^{-1}\mathrm{t}\left(\frac{{\mathrm{t}}^2}{2}\right)-\frac{1}{2}\left(\int 1\mathrm{dt}-\int \frac{1}{{\mathrm{t}}^2+1}\mathrm{dt}\right) \\ ={\tan }^{-1}\mathrm{t}\left(\frac{{\mathrm{t}}^2}{2}\right)-\frac{1}{2}\left(\mathrm{t}-{\tan }^{-1}\mathrm{t}\right)+\mathrm{C} \\ ={\tan }^{-1}\mathrm{t}\left(\frac{{\mathrm{t}}^2}{2}\right)-\frac{1}{2}+\frac{{\tan }^{-1}\mathrm{t}}{2}+\mathrm{C} \\ ={\tan }^{-1}\mathrm{t}\left(\frac{{\mathrm{t}}^2+1}{2}\right)-\frac{1}{2}+\mathrm{C} \\ ={\tan }^{-1}\left(\mathrm{tanx}\right)\left(\frac{{\tan }^2\mathrm{x}+1}{2}\right)-\frac{1}{2}+\mathrm{C} \\ =\mathrm{x}\left(\frac{{\mathrm{sce}}^2\mathrm{x}}{2}\right)-\frac{1}{2}+\mathrm{C} \\ \Rightarrow \mathbf{\int }\mathbf{xtanx}\mathbf{ }{\mathbf{sec}}^{\mathbf{2}}\mathbf{x}\mathbf{ }\mathbf{dx}\mathbf{=}\mathbf{x}\left(\frac{{\mathbf{sce}}^{\mathbf{2}}\mathbf{x}}{\mathbf{2}}\right)\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{+}\mathbf{C} \end{gathered}$$

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