The value of the integral int_{-pi / 2}^{pi / | vedclass

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(B) Let $I = \int_{-\pi / 2}^{\pi / 2} \left(x^2 + \ln \frac{\pi+x}{\pi-x}\right) \cos x \, dx$.We can split this into two integrals: $I = \int_{-\pi

Vedclass · Accepted Answer

$\frac{\pi^2}{2}-4$

Vedclass · Answer

(B) Let $I = \int_{-\pi / 2}^{\pi / 2} \left(x^2 + \ln \frac{\pi+x}{\pi-x}\right) \cos x \, dx$.We can split this into two integrals: $I = \int_{-\pi / 2}^{\pi / 2} x^2 \cos x \, dx + \int_{-\pi / 2}^{\pi / 2} \ln \left(\frac{\pi+x}{\pi-x}\right) \cos x \, dx$.Let $f(x) = \ln \left(\frac{\pi+x}{\pi-x}\right) \cos x$. Since $\ln \left(\frac{\pi+x}{\pi-x}\right)$ is an odd function and $\cos x$ is an even function,their product is an odd function. Thus,$\int_{-\pi / 2}^{\pi / 2} f(x) \, dx = 0$.Now,$I = \int_{-\pi / 2}^{\pi / 2} x^2 \cos x \, dx$. Since $x^2 \cos x$ is an even function,$I = 2 \int_0^{\pi / 2} x^2 \cos x \, dx$.Using integration by parts: $\int x^2 \cos x \, dx = x^2 \sin x - \int 2x \sin x \, dx = x^2 \sin x - 2 \left( -x \cos x + \int \cos x \, dx \right) = x^2 \sin x + 2x \cos x - 2 \sin x$.Evaluating from $0$ to $\pi / 2$: $2 \left[ (x^2 \sin x + 2x \cos x - 2 \sin x) \right]_0^{\pi / 2} = 2 \left[ ((\pi/2)^2 \cdot 1 + 0 - 2(1)) - (0 + 0 - 0) \right] = 2 \left( \frac{\pi^2}{4} - 2 \right) = \frac{\pi^2}{2} - 4$.

The value of the integral $\int_{-\pi / 2}^{\pi / 2}\left(x^2+\ln \frac{\pi+x}{\pi-x}\right) \cos x \, dx$ is

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