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Question Number 116998 by TANMAY PANACEA last updated on 08/Oct/20

$${\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}\frac{{sin}^{2}x}{1+2^x}dx$$

Commented by Dwaipayan Shikari last updated on 08/Oct/20

$${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{{sin}^{2}x}{1+2^x}dx={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{{sin}^{2}x}{1+2^{-x}}dx$$$$2I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{sin}^{2}xdx$$$$2I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{1-cos2x}{2}$$$$2I=\frac{\pi }{2}-\frac{1}{4}{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}sin2x$$$$I=\frac{\pi }{4}$$$$$$

Commented by TANMAY PANACEA last updated on 08/Oct/20

$$thankyou$$

Answered by TANMAY PANACEA last updated on 08/Oct/20

$${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$$$${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{{sin}^{2}x}{1+2^{-\mathit{x}}}\mathit{d}\mathit{x}$$$$2\mathit{I}={\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}{sin}^{2}x(\frac{1}{1+2^x}+\frac{2^x}{2^x+1})dx$$$$2I={\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}\frac{1-cos2x}{2}dx$$$$I=\frac{1}{4}\mid x-\frac{sin2x}{2}{\mid }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}=\frac{1}{4}\left(\pi \right)$$

Answered by mathmax by abdo last updated on 08/Oct/20

$$\mathrm{let}{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{{\sin }^2\mathrm{x}}{1+2^{\mathrm{x}}}\mathrm{dx}=\mathrm{I}\mathrm{changement}\mathrm{x}=-\mathrm{t}\mathrm{give}$$$$\mathrm{I}={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{{\sin }^2\mathrm{t}}{1+2^{-\mathrm{t}}}\mathrm{dt}\Rightarrow 2\mathrm{I}={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{{\sin }^2\mathrm{x}}{1+2^{\mathrm{x}}}\mathrm{dx}+{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{{\sin }^2\mathrm{x}}{1+2^{-\mathrm{x}}}\mathrm{dx}$$$$={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}(\frac{1}{1+2^{\mathrm{x}}}+\frac{1}{1+2^{-\mathrm{x}}}){\sin }^2\mathrm{x}\mathrm{dx}={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\left(\frac{2+2^{\mathrm{x}}+2^{-\mathrm{x}}}{1+2^{-\mathrm{x}}+2^{\mathrm{x}}+1}\right){\sin }^2\mathrm{x}\mathrm{dx}$$$$={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{1-\cos \left(2\mathrm{x}\right)}{2}\mathrm{dx}=\frac{\pi }{2}-\frac{1}{4}{\left[\sin \left(2\mathrm{x}\right)\right]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}=\frac{\pi }{2}\Rightarrow 2\mathrm{I}=\frac{\pi }{2}$$$$\Rightarrow ★\mathrm{I}=\frac{\pi }{4}★$$

Commented by TANMAY PANACEA last updated on 08/Oct/20

$$thankyousir$$

Commented by Bird last updated on 09/Oct/20

$$youarewelcome$$

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