The value of intlimits_{frac{-pi}{2}}^{frac{p | vedclass

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Question

(A) Let $I = \int\limits_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2 x}{1 + (2017)^x} \, dx$. Using the property $\int_{a}^{b} f(x) \, dx = \int_{a}

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$\frac{\pi}{4}$

Vedclass · Answer

(A) Let $I = \int\limits_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2 x}{1 + (2017)^x} \, dx$. Using the property $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$,we have $I = \int\limits_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2(-x)}{1 + (2017)^{-x}} \, dx$. Since $\sin^2(-x) = \sin^2 x$,we get $I = \int\limits_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2 x}{1 + \frac{1}{(2017)^x}} \, dx = \int\limits_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{(2017)^x \sin^2 x}{(2017)^x + 1} \, dx$. Adding the two expressions for $I$: $2I = \int\limits_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2 x + (2017)^x \sin^2 x}{1 + (2017)^x} \, dx = \int\limits_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \, dx$. Since $\sin^2 x$ is an even function,$2I = 2 \int\limits_{0}^{\frac{\pi}{2}} \sin^2 x \, dx$. $I = \int\limits_{0}^{\frac{\pi}{2}} \sin^2 x \, dx = \int\limits_{0}^{\frac{\pi}{2}} \frac{1 - \cos 2x}{2} \, dx = \left[ \frac{x}{2} - \frac{\sin 2x}{4} \right]_{0}^{\frac{\pi}{2}} = \frac{\pi}{4} - 0 = \frac{\pi}{4}$.

The value of $\int\limits_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^2 x}{1 + (2017)^x} \, dx$ is

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