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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left[x^2 + \log \left(\frac{\pi-x}{\pi+x}\right)\right] \cos x \, dx$. We can split the integral a

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left[x^2 + \log \left(\frac{\pi-x}{\pi+x}\right)\right] \cos x \, dx$. We can split the integral as $I = I_1 + I_2$,where $I_1 = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \cos x \, dx$ and $I_2 = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{\pi-x}{\pi+x}\right) \cos x \, dx$. Since $f(x) = \log \left(\frac{\pi-x}{\pi+x}\right) \cos x$ is an odd function because $f(-x) = \log \left(\frac{\pi+x}{\pi-x}\right) \cos(-x) = -\log \left(\frac{\pi-x}{\pi+x}\right) \cos x = -f(x)$,we have $I_2 = 0$. Now,$I_1 = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \cos x \, dx = 2 \int_{0}^{\frac{\pi}{2}} x^2 \cos x \, dx$ (since $x^2 \cos x$ is an even function). Using integration by parts: $\int x^2 \cos x \, dx = x^2 \sin x - \int 2x \sin x \, dx = x^2 \sin x - 2 \left[x(-\cos x) - \int 1(-\cos x) \, dx\right] = x^2 \sin x + 2x \cos x - 2 \sin x$. Evaluating from $0$ to $\frac{\pi}{2}$: $2 \left[x^2 \sin x + 2x \cos x - 2 \sin x\right]_{0}^{\frac{\pi}{2}} = 2 \left[\left(\frac{\pi^2}{4} \cdot 1 + 2 \cdot \frac{\pi}{2} \cdot 0 - 2 \cdot 1\right) - (0 + 0 - 0)\right] = 2 \left(\frac{\pi^2}{4} - 2\right) = \frac{\pi^2}{2} - 4$. Thus,$I = \frac{\pi^2}{2} - 4$.

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

Similar Questions

The value of $\int_{-\pi/2}^{\pi/2} (3\sin x + \sin^3 x) \, dx$ is

$\int_{0}^{\pi} \frac{x \cos x \sin x}{\cos^{3} x + \cos x} dx = $

$\int_{-1}^{1} \log \left( \frac{1+x}{1-x} \right) \, dx = $

The value of $\int_{-1}^1 \frac{(1+\sqrt{|x|-x}) e^x+(\sqrt{|x|-x}) e^{-x}}{e^x+e^{-x}} d x$ is equal to

If $g(x) = \int_0^x \cos^4 t \,dt$, then $g(x+\pi)$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module