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Question Number 200586 by Calculusboy last updated on 20/Nov/23

Answered by Frix last updated on 21/Nov/23

$$\underset{0}{\stackrel{\pi }{\int }}\frac{x}{2+{\tan }^{2}x}dx=\underset{0}{\stackrel{\pi }{\int }}xdx-\underset{0}{\stackrel{\pi }{\int }}\frac{x}{1+{\cos }^{2}x}$$$$I_1=\underset{0}{\stackrel{\pi }{\int }}xdx=\frac{{\pi }^2}{2}$$$$I_2=\underset{0}{\stackrel{\pi }{\int }}\frac{x}{1+{\cos }^{2}x}dx\stackrel{t=2x}{=}\frac{1}{2}\underset{0}{\stackrel{2\pi }{\int }}\frac{t}{3+\cos t}dt\stackrel{u=2\pi -t}{=}$$$$=\frac{1}{2}\underset{2\pi }{\stackrel{0}{\int }}\frac{u-2\pi }{3+\cos u}du=\frac{1}{2}\underset{0}{\stackrel{2\pi }{\int }}\frac{2\pi -u}{3+\cos u}du$$$$\Rightarrow$$$$\left[\mathrm{Renaming}\mathrm{variables}\right]$$$$I_2=\frac{1}{2}(\frac{1}{2}\underset{0}{\stackrel{2\pi }{\int }}\frac{v}{3+\cos v}dv+\frac{1}{2}\underset{0}{\stackrel{2\pi }{\int }}\frac{2\pi -v}{3+\cos v}dv)=$$$$=\frac{\pi }{2}\underset{0}{\stackrel{2\pi }{\int }}\frac{dv}{3+\cos v}=\pi \underset{0}{\stackrel{\pi }{\int }}\frac{dv}{3+\cos v}\stackrel{w=\tan \frac{v}{2}}{=}$$$$=\pi \underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{dw}{w^2+2}=\frac{\pi }{\sqrt{2}}{\left[{\tan }^{-1}\frac{w}{\sqrt{2}}\right]}_0^{\mathrm{\infty }}=\frac{\sqrt{2}{\pi }^2}{4}$$$$\Rightarrow$$$$\underset{0}{\stackrel{\pi }{\int }}\frac{x}{2+{\tan }^{2}x}dx=I_1-I_2=\frac{(2-\sqrt{2}){\pi }^2}{4}$$

Commented by Calculusboy last updated on 21/Nov/23

$$\mathit{t}\mathit{h}\mathit{a}\mathit{n}\mathit{k}\mathit{s}\mathit{s}\mathit{i}\mathit{r}$$

Commented by Frix last updated on 21/Nov/23

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