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Question Number 205873 by universe last updated on 01/Apr/24

$${\int }_0^{\pi }\frac{1}{{\pi }^2}\frac{x}{\sqrt{1+{\sin }^{3}x}}[(3\pi \cos x+4\sin x){\sin }^{2}x+4]dx$$

Answered by Berbere last updated on 02/Apr/24

$$x\to \pi -x;let\mathrm{\Omega }=integral$$$$\mathrm{\Omega }={\int }_0^{\pi }\frac{1}{{\pi }^2}.\frac{3\pi xcos\left(x\right){sin}^2\left(x\right)}{\sqrt{1+{sin}^3\left(x\right)}}dx+\frac{1}{{\pi }^2}{\int }_0^{\pi }\frac{4(1+{sin}^3\left(x\right))x}{\sqrt{1+{sin}^3\left(x\right)}}dx$$$$=\frac{1}{\pi }{\int }_0^{\pi }\frac{3cos\left(x\right){sin}^2\left(x\right)}{\sqrt{1+{sin}^3\left(x\right)}}.x+4{\int }_0^{\pi }\sqrt{1+{sin}^3\left(x\right)}xdx$$$$=\frac{1}{\pi }.A+4B$$$$B;x\to \pi -x;B=\frac{1}{{\pi }^2}{\int }_0^{\pi }\sqrt{1+{sin}^3\left(x\right)}(\pi -x)dx\Rightarrow 2B=\frac{1}{\pi }{\int }_0^{\pi }\sqrt{1+{sin}^3\left(x\right)}$$$$B=\frac{1}{2\pi }{\int }_0^{\pi }\sqrt{1+{sin}^3\left(x\right)}dx$$$$A;\{\begin{array}{l}{u}^{\prime }=\frac{3cos\left(x\right){sin}^2\left(x\right)}{\sqrt{1+{sin}^3\left(x\right)}};u=2\sqrt{1+{sin}^3\left(x\right)}\\ v=x\Rightarrow v^{\prime }=1\end{array}$$$$A={\left[2x\sqrt{1+{sin}^3\left(x\right)}\right]}_0^{\pi }-2{\int }_0^{\pi }\sqrt{1+{sin}^3\left(x\right)}dx$$$$=2\pi -2{\int }_0^{\pi }\sqrt{1+{sin}^3\left(x\right)}dx$$$$\mathrm{\Omega }=\frac{1}{\pi }(2\pi -2{\int }_0^{\pi }\sqrt{1+{sin}^3\left(x\right)}dx)+4\left(\frac{1}{2\pi }{\int }_0^{\pi }\sqrt{1+{sin}^3\left(x\right)}dx\right)$$$$=2$$$$$$

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