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Question Number 133688 by mnjuly1970 last updated on 23/Feb/21

$$\text{You can't use 'macro parameter character \#' in math mode}$$$$find:$$$$\mathit{\varphi }={\int }_0^{\frac{\pi }{2}}\frac{dx}{{sin}^{10}\left(x\right)+{cos}^{10}\left(x\right)}=???$$$$$$

Answered by Ar Brandon last updated on 23/Feb/21

$$\varphi ={\int }_0^{\frac{\pi }{2}}\frac{\mathrm{d}x}{{\sin }^{10}x+{\cos }^{10}x}={\int }_0^{\frac{\pi }{2}}\frac{{\sec }^{10}x}{{\tan }^{10}x+1}\mathrm{d}x={\int }_0^{\mathrm{\infty }}\frac{{({\mathrm{t}}^2+1)}^4}{{\mathrm{t}}^{10}+1}\mathrm{dt}$$$$={\int }_0^{\mathrm{\infty }}\left\{\frac{{\mathrm{t}}^8+{4\mathrm{t}}^6+{6\mathrm{t}}^4+{4\mathrm{t}}^2+1}{{\mathrm{t}}^{10}+1}\right\}\mathrm{dt},\mathrm{u}={\mathrm{t}}^{10}\Rightarrow \mathrm{du}={10\mathrm{t}}^9\mathrm{dt}={10\mathrm{u}}^{\frac{9}{10}}\mathrm{dt}$$$$\varphi =\frac{1}{10}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{u}}^{-\frac{1}{10}}}{\mathrm{u}+1}\mathrm{du}+\frac{2}{5}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{u}}^{-\frac{3}{10}}}{\mathrm{u}+1}\mathrm{du}+\frac{3}{5}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{u}}^{-\frac{5}{10}}}{\mathrm{u}+1}\mathrm{du}+\frac{2}{5}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{u}}^{-\frac{7}{10}}}{\mathrm{u}+1}\mathrm{du}+\frac{1}{10}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{u}}^{-\frac{9}{10}}}{\mathrm{u}+1}\mathrm{du}$$$$=\frac{1}{10}\beta (\frac{9}{10},\frac{1}{10})+\frac{2}{5}\beta (\frac{7}{10},\frac{3}{10})+\frac{3}{5}\beta (\frac{1}{2},\frac{1}{2})+\frac{2}{5}\beta (\frac{3}{10},\frac{7}{10})+\frac{1}{10}\beta (\frac{1}{10},\frac{9}{10})$$$$=\frac{\pi }{5\sin \left(\frac{\pi }{10}\right)}+\frac{4\pi }{5\sin \left(\frac{3\pi }{10}\right)}+\frac{3\pi }{5\sin \left(\frac{\pi }{2}\right)}$$

Commented by Dwaipayan Shikari last updated on 23/Feb/21

$$\frac{\pi }{5}(\frac{4}{\sqrt{5}-1}+\frac{16}{\sqrt{5}+1}+3)=\frac{\pi }{5}(\sqrt{5}+1+4\sqrt{5}-4+3)$$$$=\sqrt{5}\pi$$

Commented by mnjuly1970 last updated on 23/Feb/21

$$thanksalotmasterbrandon$$$$niceverynicesolution...$$

Commented by Ar Brandon last updated on 23/Feb/21

My pleasure Sir

Answered by Ajetunmobi last updated on 23/Feb/21

$$\mathbf{i}\mathbf{have}\mathbf{did}\mathbf{this}\mathbf{on}\mathbf{facebook}$$

Answered by Ajetunmobi last updated on 23/Feb/21

Commented by Ar Brandon last updated on 23/Feb/21

Ambiguity on your name Sir. Ajetunmobi or Ajeutunmobi ?��

Commented by Ajetunmobi last updated on 24/Feb/21

$$itis\mathbf{Ajetunmobi}\mathbf{Abdulqoyyum}$$$$\mathbf{thank}\mathbf{u}\mathbf{that}\mathbf{was}\mathbf{a}\mathbf{mistake}$$

Answered by Ajetunmobi last updated on 23/Feb/21

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