Let I_n = int_0^{pi / 2} x^n cos x , dx,where | vedclass

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

Question

(A) We have $I_n = \int_0^{\pi / 2} x^n \cos x \, dx$. Using integration by parts,$I_n = [x^n \sin x]_0^{\pi / 2} - \int_0^{\pi / 2} n x^{n-1} \sin x

Vedclass · Accepted Answer

$e^{\pi / 2} - 1 - \frac{\pi}{2}$

Vedclass · Answer

(A) We have $I_n = \int_0^{\pi / 2} x^n \cos x \, dx$. Using integration by parts,$I_n = [x^n \sin x]_0^{\pi / 2} - \int_0^{\pi / 2} n x^{n-1} \sin x \, dx$. $I_n = (\frac{\pi}{2})^n - n \int_0^{\pi / 2} x^{n-1} \sin x \, dx$. Applying integration by parts again to the integral,$\int_0^{\pi / 2} x^{n-1} \sin x \, dx = [x^{n-1} (-\cos x)]_0^{\pi / 2} - \int_0^{\pi / 2} (n-1) x^{n-2} (-\cos x) \, dx = 0 + (n-1) I_{n-2}$. Substituting this back,$I_n = (\frac{\pi}{2})^n - n(n-1) I_{n-2}$. Thus,$I_n + n(n-1) I_{n-2} = (\frac{\pi}{2})^n$. Dividing by $n!$,we get $\frac{I_n}{n!} + \frac{I_{n-2}}{(n-2)!} = \frac{(\pi/2)^n}{n!}$. Summing from $n=2$ to $\infty$,$\sum_{n=2}^{\infty} \frac{(\pi/2)^n}{n!} = (\sum_{n=0}^{\infty} \frac{(\pi/2)^n}{n!}) - \frac{(\pi/2)^0}{0!} - \frac{(\pi/2)^1}{1!} = e^{\pi / 2} - 1 - \frac{\pi}{2}$.

Let $I_n = \int_0^{\pi / 2} x^n \cos x \, dx$,where $n$ is a non-negative integer. Then,$\sum_{n=2}^{\infty} \left( \frac{I_n}{n!} + \frac{I_{n-2}}{(n-2)!} \right)$ equals

Similar Questions

Let $I = \int_{\pi / 4}^{\pi / 3} \frac{\sin x}{x} dx$. Then

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

If $I_n = \int_{0}^{\pi} \frac{\sin(nx)}{\sin(x)} dx$,then the value of $\sum_{n=1}^{10} I_n$ is-

$\int_{ - 3}^3 {\frac{{{x^2}\sin 2x}}{{{x^2} + 1}}\,dx = } $

$\int_0^\pi \frac{x \tan x}{\sec x + \cos x} \,dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module