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Question Number 200418 by mnjuly1970 last updated on 18/Nov/23

$$$$$$\mathrm{calculus}\left(\mathrm{I}\right)$$$$\mathrm{I}f,\mathrm{I}={\int }_0^{\pi }\frac{x}{1+{\sin }^2\left(x\right)}\mathrm{d}x=a\zeta \left(2\right)$$$$\Rightarrow a=?$$$$where,\zeta \left(s\right)=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{n^s}$$$$$$

Answered by witcher3 last updated on 18/Nov/23

$$\mathrm{I}={\int }_0^{\pi }\frac{\pi -\mathrm{x}}{1+{\sin }^2\left(\mathrm{x}\right)}\mathrm{dx}=\pi {\int }_0^{\pi }\frac{\mathrm{dx}}{1+{\sin }^2\left(\mathrm{x}\right)}-\mathrm{I}$$$$2\mathrm{I}=\pi {\int }_0^{\pi }\frac{\mathrm{dx}}{{\cos }^2\left(\mathrm{x}\right)+{\sin }^2\left(\mathrm{x}\right)+{\sin }^2\left(\mathrm{x}\right)}$$$$=\pi {\int }_0^{\frac{\pi }{2}}\frac{\mathrm{dx}}{{\cos }^2\left(\mathrm{x}\right)+{2\sin }^2\left(\mathrm{x}\right)}+\pi {\int }_{\frac{\pi }{2}}^{\pi }\frac{\mathrm{dx}}{{\cos }^2\left(\mathrm{x}\right)+{2\sin }^2\left(\mathrm{x}\right)}\mathrm{dx}$$$$\mathrm{t}=\pi -\mathrm{x}..\mathrm{in}2\mathrm{nd}$$$$=2\pi \left\{{\int }_0^{\frac{\pi }{2}}\frac{\mathrm{dx}}{{\cos }^2\left(\mathrm{x}\right)+{2\sin }^2\left(\mathrm{x}\right)}\right\}$$$$=\pi {\int }_0^{\frac{\pi }{2}}\frac{1}{{\cos }^2\left(\mathrm{x}\right)}.\frac{\mathrm{dx}}{1+{(\sqrt{2}.\mathrm{tg}\left(\mathrm{x}\right))}^2}$$$$=2\pi {\left[\frac{1}{\sqrt{2}}{\tan }^{-1}\left(\sqrt{2}\tan \left(\mathrm{x}\right)\right)\right]}_0^{\frac{\pi }{2}}$$$$=\frac{{\pi }^2}{\sqrt{2}}$$$$\mathrm{I}=\frac{{\pi }^2}{2\sqrt{2}}=\frac{6}{2\sqrt{2}}\zeta \left(2\right)\therefore \mathrm{a}=\frac{3}{\sqrt{2}}$$$$$$$$$$

Commented by mnjuly1970 last updated on 18/Nov/23

$$thanksalotsir...$$

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