pi-2-pi-2-sin-2-x-1-2-x-dx-

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

\int_{\frac{- π}{2}}^{\frac{π}{2}} \frac{{s i n}^{2} x}{1 + 2^{x}} d x

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

\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{{s i n}^{2} x}{1 + 2^{x}} d x = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{{s i n}^{2} x}{1 + 2^{- x}} d x

2 I = \int_{- \frac{π}{2}}^{\frac{π}{2}} {s i n}^{2} x d x

2 I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{1 - c o s 2 x}{2}

2 I = \frac{π}{2} - \frac{1}{4} \int_{- \frac{π}{2}}^{\frac{π}{2}} s i n 2 x

I = \frac{π}{4}

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

t h a n k y o u

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

\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x

\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{{s i n}^{2} x}{1 + 2^{- \boldsymbol x}} \boldsymbol d x

2 \boldsymbol I = \int_{\frac{- π}{2}}^{\frac{π}{2}} {s i n}^{2} x (\frac{1}{1 + 2^{x}} + \frac{2^{x}}{2^{x} + 1}) d x

2 I = \int_{\frac{- π}{2}}^{\frac{π}{2}} \frac{1 - c o s 2 x}{2} d x

I = \frac{1}{4} ∣ x - \frac{s i n 2 x}{2} ∣_{\frac{- π}{2}}^{\frac{π}{2}} = \frac{1}{4} (π)

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

let \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{\sin^{2} x}{1 + 2^{x}} dx = I changement x = - t give

I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{\sin^{2} t}{1 + 2^{- t}} dt \Rightarrow 2 I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{\sin^{2} x}{1 + 2^{x}} dx + \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{\sin^{2} x}{1 + 2^{- x}} dx

= \int_{- \frac{π}{2}}^{\frac{π}{2}} (\frac{1}{1 + 2^{x}} + \frac{1}{1 + 2^{- x}}) \sin^{2} x dx = \int_{- \frac{π}{2}}^{\frac{π}{2}} (\frac{2 + 2^{x} + 2^{- x}}{1 + 2^{- x} + 2^{x} + 1}) \sin^{2} x dx

= \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{1 - \cos (2 x)}{2} dx = \frac{π}{2} - \frac{1}{4} {[\sin (2 x)]}_{- \frac{π}{2}}^{\frac{π}{2}} = \frac{π}{2} \Rightarrow 2 I = \frac{π}{2}

\Rightarrow ★ I = \frac{π}{4} ★

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

t h a n k y o u s i r

t h a n k y o u s i r

Commented by Bird last updated on 09/Oct/20

y o u a r e w e l c o m e

y o u a r e w e l c o m e

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