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Question Number 197637 by SLVR last updated on 25/Sep/23

Commented by SLVR last updated on 25/Sep/23

$$kindlyhelpmesir$$

Commented by SLVR last updated on 25/Sep/23

$$sir..imean\underset{0}{\stackrel{\pi }{\int }}\frac{x}{1-sinxcosx}dx=?$$$$itwasaskedtoproveas$$$$\frac{5{\pi }^2}{6\sqrt{3}}...kindlyhelpme$$

Commented by SLVR last updated on 26/Sep/23

$$kindlyhelpme$$

Answered by witcher3 last updated on 27/Sep/23

$$\mathrm{I}={\int }_0^{\pi }\frac{\mathrm{x}}{1-\sin \left(\mathrm{x}\right)\cos \left(\mathrm{x}\right)}\mathrm{dx}$$$$={\int }_0^{\pi }\frac{\pi -\mathrm{x}}{1+\sin \left(\mathrm{x}\right)\cos \left(\mathrm{x}\right)}$$$$2\mathrm{I}={\int }_0^{\pi }\frac{\pi }{1+\sin \left(\mathrm{x}\right)\cos \left(\mathrm{x}\right)}\mathrm{dx}+{\int }_0^{\pi }\mathrm{x}\frac{\sin \left(2\mathrm{x}\right)}{1-{\sin }^2\left(\mathrm{x}\right){\cos }^2\left(\mathrm{x}\right)}\mathrm{dx}$$$${\int }_0^{\pi }\frac{\pi }{1+\sin \left(\mathrm{x}\right)\cos \left(\mathrm{x}\right)}\mathrm{dx}$$$$=\pi {\int }_0^{\frac{\pi }{2}}\frac{\mathrm{dx}}{1+\sin \left(\mathrm{x}\right)\cos \left(\mathrm{x}\right)}+\frac{\mathrm{dx}}{1-\sin \left(\mathrm{x}\right)\cos \left(\mathrm{x}\right)}$$$$\pi {\int }_0^{\frac{\pi }{2}}(1+{\tan }^2\left(\mathrm{x}\right))(\frac{\mathrm{dx}}{1+{\tan }^2\left(\mathrm{x}\right)+\tan \left(\mathrm{x}\right)}+\frac{\mathrm{dx}}{1-\tan \left(\mathrm{x}\right)+{\tan }^2\left(\mathrm{x}\right)})$$$$=\pi {\int }_0^{\mathrm{\infty }}(\frac{1}{({\mathrm{y}}^2+\mathrm{y}+1)}+\frac{1}{{\mathrm{y}}^2-\mathrm{y}+1})\mathrm{dy}$$$$=\pi {[\frac{2}{\sqrt{3}}{\tan }^{-1}((\mathrm{y}+\frac{1}{2}).\frac{2}{\sqrt{3}})+\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2}{\sqrt{3}}(\mathrm{y}-\frac{1}{2})\right)]}_0^{\mathrm{\infty }}$$$$\pi \left(\frac{2}{\sqrt{3}}(\frac{\pi }{2}+\frac{\pi }{2})\right)=\frac{2}{\sqrt{3}}{\pi }^2$$$$$$$${\int }_0^{\pi }\frac{\mathrm{xsin}\left(2\mathrm{x}\right)}{1-{\sin }^2\left(\mathrm{x}\right){\cos }^2\left(\mathrm{x}\right)}$$$$\int \frac{\sin \left(2\mathrm{x}\right)}{1-{\sin }^2\left(\mathrm{x}\right){\cos }^2\left(\mathrm{x}\right)}$$$$=4\int \frac{\sin \left(2\mathrm{x}\right)}{4-{\sin }^2\left(2\mathrm{x}\right)}$$$$=2\int \frac{\sin \left(\mathrm{y}\right)}{4-(1-{\cos }^2\left(\mathrm{y}\right))}=2\int \frac{\sin \left(\mathrm{y}\right)}{3+{\cos }^2\left(\mathrm{y}\right)}$$$$=-2\int \frac{\mathrm{d}\left(\cos \left(\mathrm{y}\right)\right)}{3+{\cos }^2\left(\mathrm{y}\right)}$$$$=-\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{\cos \left(\mathrm{y}\right)}{\sqrt{3}}\right)+\mathrm{c}$$$${\int }_0^{\pi }\frac{\mathrm{xsin}\left(2\mathrm{x}\right)}{3+{\cos }^2(2\mathrm{x}}=[{}_0^{\pi }\mathrm{x}.\frac{-2}{\sqrt{3}}{\tan }^{-1}\left(\frac{\cos \left(2\mathrm{x}\right)}{\sqrt{3}}\right)]$$$$+\frac{2}{\sqrt{3}}{\int }_0^{\pi }{\tan }^{-1}\left(\frac{\cos \left(2\mathrm{x}\right)}{\sqrt{3}}\right)\mathrm{dx}$$$$=-\frac{{\pi }^2}{3\sqrt{3}}+\frac{2}{\sqrt{3}}{\int }_0^{\frac{\pi }{2}}{\tan }^{-1}\left(\frac{\cos \left(2\mathrm{x}\right)}{\sqrt{3}}\right)+{\tan }^{-1}\left(\frac{\cos \left(2(\frac{\pi }{2}+\mathrm{x})\right)}{\sqrt{3}}\right)\mathrm{dx}$$$$=-\frac{{\pi }^2}{3\sqrt{3}}+\frac{2}{\sqrt{3}}{\int }_0^{\frac{\pi }{2}}{\tan }^{-1}\left(\frac{\cos \left(2\mathrm{x}\right)}{\sqrt{3}}\right)+{\tan }^{-1}(-\frac{\cos \left(2\mathrm{x}\right)}{\sqrt{3}}))\mathrm{dx}=-\frac{{\pi }^2}{3\sqrt{3}}$$$$2\mathrm{I}=\frac{2{\pi }^2}{\sqrt{3}}-\frac{{\pi }^2}{3\sqrt{3}}\iff \mathrm{I}=\frac{5{\pi }^2}{6\sqrt{3}}$$$$$$$$$$$$$$

Commented by witcher3 last updated on 05/Oct/23

$$\mathrm{i}\mathrm{will}\mathrm{Try}\mathrm{to}\mathrm{find}\mathrm{a}\mathrm{batter}\mathrm{Way}$$$$$$

Commented by SLVR last updated on 29/Sep/23

$$sir...itisclumsy...forme..$$$$kindly..giveanyanotherway$$$$ifpossible$$

Commented by SLVR last updated on 05/Oct/23

$$Sokindofyousir$$

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