Question Number 205873 by universe last updated on 01/Apr/24
![∫_0 ^π (1/π^2 ) (x/( (√(1+sin^3 x ))))[(3πcosx+4sinx)sin^2 x+4]dx](Q205873.png)
$$\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{1}}{\pi^{\mathrm{2}} }\:\frac{{x}}{\:\sqrt{\mathrm{1}+\mathrm{sin}^{\mathrm{3}} {x}\:}}\left[\left(\mathrm{3}\pi\mathrm{cos}{x}+\mathrm{4sin}{x}\right)\mathrm{sin}^{\mathrm{2}} {x}+\mathrm{4}\right]{dx}\:\:\: \\ $$
Answered by Berbere last updated on 02/Apr/24
![x→π−x;let Ω=integral Ω=∫_0 ^π (1/π^2 ).((3πxcos(x)sin^2 (x))/( (√(1+sin^3 (x)))))dx+(1/π^2 )∫_0 ^π ((4(1+sin^3 (x))x)/( (√(1+sin^3 (x)))))dx =(1/π)∫_0 ^π ((3cos(x)sin^2 (x))/( (√(1+sin^3 (x))))).x+4∫_0 ^π (√(1+sin^3 (x)))xdx =(1/π).A+4B B;x→π−x;B=(1/π^2 )∫_0 ^π (√(1+sin^3 (x)))(π−x)dx⇒2B=(1/π)∫_0 ^π (√(1+sin^3 (x))) B=(1/(2π))∫_0 ^π (√(1+sin^3 (x)))dx A; { ((u′=((3cos(x)sin^2 (x))/( (√(1+sin^3 (x)))));u=2(√(1+sin^3 (x))))),((v=x⇒v′=1)) :} A=[2x(√(1+sin^3 (x)))]_0 ^π −2∫_0 ^π (√(1+sin^3 (x)))dx =2π−2∫_0 ^π (√(1+sin^3 (x)))dx Ω=(1/π)(2π−2∫_0 ^π (√(1+sin^3 (x)))dx)+4((1/(2π))∫_0 ^π (√(1+sin^3 (x)))dx) =2](Q205891.png)
$${x}\rightarrow\pi−{x};{let}\:\Omega={integral} \\ $$$$\Omega=\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{1}}{\pi^{\mathrm{2}} }.\frac{\mathrm{3}\pi{xcos}\left({x}\right){sin}^{\mathrm{2}} \left({x}\right)}{\:\sqrt{\mathrm{1}+{sin}^{\mathrm{3}} \left({x}\right)}}{dx}+\frac{\mathrm{1}}{\pi^{\mathrm{2}} }\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{4}\left(\mathrm{1}+{sin}^{\mathrm{3}} \left({x}\right)\right){x}}{\:\sqrt{\mathrm{1}+{sin}^{\mathrm{3}} \left({x}\right)}}{dx} \\ $$$$=\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{3}{cos}\left({x}\right){sin}^{\mathrm{2}} \left({x}\right)}{\:\sqrt{\mathrm{1}+{sin}^{\mathrm{3}} \left({x}\right)}}.{x}+\mathrm{4}\int_{\mathrm{0}} ^{\pi} \sqrt{\mathrm{1}+{sin}^{\mathrm{3}} \left({x}\right)}{xdx} \\ $$$$=\frac{\mathrm{1}}{\pi}.{A}+\mathrm{4}{B} \\ $$$${B};{x}\rightarrow\pi−{x};{B}=\frac{\mathrm{1}}{\pi^{\mathrm{2}} }\int_{\mathrm{0}} ^{\pi} \sqrt{\mathrm{1}+{sin}^{\mathrm{3}} \left({x}\right)}\left(\pi−{x}\right){dx}\Rightarrow\mathrm{2}{B}=\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\pi} \sqrt{\mathrm{1}+{sin}^{\mathrm{3}} \left({x}\right)} \\ $$$${B}=\frac{\mathrm{1}}{\mathrm{2}\pi}\int_{\mathrm{0}} ^{\pi} \sqrt{\mathrm{1}+{sin}^{\mathrm{3}} \left({x}\right)}{dx} \\ $$$${A};\begin{cases}{{u}'=\frac{\mathrm{3}{cos}\left({x}\right){sin}^{\mathrm{2}} \left({x}\right)}{\:\sqrt{\mathrm{1}+{sin}^{\mathrm{3}} \left({x}\right)}};{u}=\mathrm{2}\sqrt{\mathrm{1}+{sin}^{\mathrm{3}} \left({x}\right)}}\\{{v}={x}\Rightarrow{v}'=\mathrm{1}}\end{cases} \\ $$$${A}=\left[\mathrm{2}{x}\sqrt{\mathrm{1}+{sin}^{\mathrm{3}} \left({x}\right)}\right]_{\mathrm{0}} ^{\pi} −\mathrm{2}\int_{\mathrm{0}} ^{\pi} \sqrt{\mathrm{1}+{sin}^{\mathrm{3}} \left({x}\right)}{dx} \\ $$$$=\mathrm{2}\pi−\mathrm{2}\int_{\mathrm{0}} ^{\pi} \sqrt{\mathrm{1}+{sin}^{\mathrm{3}} \left({x}\right)}{dx} \\ $$$$\Omega=\frac{\mathrm{1}}{\pi}\left(\mathrm{2}\pi−\mathrm{2}\int_{\mathrm{0}} ^{\pi} \sqrt{\mathrm{1}+{sin}^{\mathrm{3}} \left({x}\right)}{dx}\right)+\mathrm{4}\left(\frac{\mathrm{1}}{\mathrm{2}\pi}\int_{\mathrm{0}} ^{\pi} \sqrt{\mathrm{1}+{sin}^{\mathrm{3}} \left({x}\right)}{dx}\right) \\ $$$$=\mathrm{2} \\ $$$$ \\ $$