If int_{0}^{100 pi} frac{sin ^{2} x}{e^{left( | vedclass

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

Question

(B) Let $I = \int_{0}^{100 \pi} \frac{\sin ^{2} x}{e^{\left(\frac{x}{\pi}-\left[\frac{x}{\pi}\right]\right)}} d x$. Since the function $f(x) = \frac{\

Vedclass · Accepted Answer

$200(1-e^{-1})$

Vedclass · Answer

(B) Let $I = \int_{0}^{100 \pi} \frac{\sin ^{2} x}{e^{\left(\frac{x}{\pi}-\left[\frac{x}{\pi}\right]\right)}} d x$. Since the function $f(x) = \frac{\sin^2 x}{e^{\left(\frac{x}{\pi}-\left[\frac{x}{\pi}\right]\right)}}$ is periodic with period $\pi$,we have $I = 100 \int_{0}^{\pi} \frac{\sin^2 x}{e^{x/\pi}} dx$. $I = 100 \int_{0}^{\pi} e^{-x/\pi} \left(\frac{1-\cos 2x}{2}\right) dx = 50 \left[ \int_{0}^{\pi} e^{-x/\pi} dx - \int_{0}^{\pi} e^{-x/\pi} \cos 2x dx \right]$. Let $I_1 = \int_{0}^{\pi} e^{-x/\pi} dx = [-\pi e^{-x/\pi}]_0^{\pi} = \pi(1-e^{-1})$. Let $I_2 = \int_{0}^{\pi} e^{-x/\pi} \cos 2x dx$. Using integration by parts,$I_2 = [-\pi e^{-x/\pi} \cos 2x]_0^{\pi} - \int_0^{\pi} \pi e^{-x/\pi} (2 \sin 2x) dx = \pi(1-e^{-1}) - 2\pi \int_0^{\pi} e^{-x/\pi} \sin 2x dx$. Integrating the second part again,$\int_0^{\pi} e^{-x/\pi} \sin 2x dx = [-\pi e^{-x/\pi} \sin 2x]_0^{\pi} - \int_0^{\pi} \pi e^{-x/\pi} (2 \cos 2x) dx = 2\pi I_2$. Thus,$I_2 = \pi(1-e^{-1}) - 2\pi(2\pi I_2) = \pi(1-e^{-1}) - 4\pi^2 I_2$. $I_2(1+4\pi^2) = \pi(1-e^{-1}) \implies I_2 = \frac{\pi(1-e^{-1})}{1+4\pi^2}$. Substituting back,$I = 50 [\pi(1-e^{-1}) - \frac{\pi(1-e^{-1})}{1+4\pi^2}] = 50 \pi(1-e^{-1}) [1 - \frac{1}{1+4\pi^2}] = 50 \pi(1-e^{-1}) [\frac{4\pi^2}{1+4\pi^2}] = \frac{200(1-e^{-1})\pi^3}{1+4\pi^2}$. Comparing with $\frac{\alpha \pi^3}{1+4\pi^2}$,we get $\alpha = 200(1-e^{-1})$.

If $\int_{0}^{100 \pi} \frac{\sin ^{2} x}{e^{\left(\frac{x}{\pi}-\left[\frac{x}{\pi}\right]\right)}} d x=\frac{\alpha \pi^{3}}{1+4 \pi^{2}}, \alpha \in R$,where $[x]$ is the greatest integer less than or equal to $x$,then the value of $\alpha$ is :

Similar Questions

Statement $-1$: The value of the integral $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1 + \sqrt{\tan x}} = \frac{\pi}{6}$.
Statement $-2$: $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx$.

$\int_0^{\pi} x f(\sin x) \, dx$ is equal to

By using the properties of definite integrals,evaluate the integral $\int_{0}^{2} x \sqrt{2-x} \, dx$.

$\int_0^{\frac{\pi}{2}} \frac{\cos x \, dx}{\sqrt{1+\cos x \sin x}} = $

$\int_{-\pi}^{\pi} \frac{\sin^4 x}{\sin^4 x + \cos^4 x} \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module