Question Number 205873 by universe last updated on 01/Apr/24
$$\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{1}}{\pi^{\mathrm{2}} }\:\frac{{x}}{\:\sqrt{\mathrm{1}+\mathrm{sin}^{\mathrm{3}} {x}\:}}\left[\left(\mathrm{3}\pi\mathrm{cos}{x}+\mathrm{4sin}{x}\right)\mathrm{sin}^{\mathrm{2}} {x}+\mathrm{4}\right]{dx}\:\:\: \\ $$
Answered by Berbere last updated on 02/Apr/24
$${x}\rightarrow\pi−{x};{let}\:\Omega={integral} \\ $$$$\Omega=\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{1}}{\pi^{\mathrm{2}} }.\frac{\mathrm{3}\pi{xcos}\left({x}\right){sin}^{\mathrm{2}} \left({x}\right)}{\:\sqrt{\mathrm{1}+{sin}^{\mathrm{3}} \left({x}\right)}}{dx}+\frac{\mathrm{1}}{\pi^{\mathrm{2}} }\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{4}\left(\mathrm{1}+{sin}^{\mathrm{3}} \left({x}\right)\right){x}}{\:\sqrt{\mathrm{1}+{sin}^{\mathrm{3}} \left({x}\right)}}{dx} \\ $$$$=\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{3}{cos}\left({x}\right){sin}^{\mathrm{2}} \left({x}\right)}{\:\sqrt{\mathrm{1}+{sin}^{\mathrm{3}} \left({x}\right)}}.{x}+\mathrm{4}\int_{\mathrm{0}} ^{\pi} \sqrt{\mathrm{1}+{sin}^{\mathrm{3}} \left({x}\right)}{xdx} \\ $$$$=\frac{\mathrm{1}}{\pi}.{A}+\mathrm{4}{B} \\ $$$${B};{x}\rightarrow\pi−{x};{B}=\frac{\mathrm{1}}{\pi^{\mathrm{2}} }\int_{\mathrm{0}} ^{\pi} \sqrt{\mathrm{1}+{sin}^{\mathrm{3}} \left({x}\right)}\left(\pi−{x}\right){dx}\Rightarrow\mathrm{2}{B}=\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\pi} \sqrt{\mathrm{1}+{sin}^{\mathrm{3}} \left({x}\right)} \\ $$$${B}=\frac{\mathrm{1}}{\mathrm{2}\pi}\int_{\mathrm{0}} ^{\pi} \sqrt{\mathrm{1}+{sin}^{\mathrm{3}} \left({x}\right)}{dx} \\ $$$${A};\begin{cases}{{u}'=\frac{\mathrm{3}{cos}\left({x}\right){sin}^{\mathrm{2}} \left({x}\right)}{\:\sqrt{\mathrm{1}+{sin}^{\mathrm{3}} \left({x}\right)}};{u}=\mathrm{2}\sqrt{\mathrm{1}+{sin}^{\mathrm{3}} \left({x}\right)}}\\{{v}={x}\Rightarrow{v}'=\mathrm{1}}\end{cases} \\ $$$${A}=\left[\mathrm{2}{x}\sqrt{\mathrm{1}+{sin}^{\mathrm{3}} \left({x}\right)}\right]_{\mathrm{0}} ^{\pi} −\mathrm{2}\int_{\mathrm{0}} ^{\pi} \sqrt{\mathrm{1}+{sin}^{\mathrm{3}} \left({x}\right)}{dx} \\ $$$$=\mathrm{2}\pi−\mathrm{2}\int_{\mathrm{0}} ^{\pi} \sqrt{\mathrm{1}+{sin}^{\mathrm{3}} \left({x}\right)}{dx} \\ $$$$\Omega=\frac{\mathrm{1}}{\pi}\left(\mathrm{2}\pi−\mathrm{2}\int_{\mathrm{0}} ^{\pi} \sqrt{\mathrm{1}+{sin}^{\mathrm{3}} \left({x}\right)}{dx}\right)+\mathrm{4}\left(\frac{\mathrm{1}}{\mathrm{2}\pi}\int_{\mathrm{0}} ^{\pi} \sqrt{\mathrm{1}+{sin}^{\mathrm{3}} \left({x}\right)}{dx}\right) \\ $$$$=\mathrm{2} \\ $$$$ \\ $$