intlimits_0^pi {frac{{sin left( {n + frac{1}{ | vedclass

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

Question

(C) We know that $2\sin \frac{x}{2} \left( {\frac{1}{2} + \cos x + \cos 2x + \dots + \cos nx} \right) = \sin \frac{x}{2} + \sum_{k=1}^n 2\sin \frac{x}

Vedclass · Accepted Answer

$\pi $

Vedclass · Answer

(C) We know that $2\sin \frac{x}{2} \left( {\frac{1}{2} + \cos x + \cos 2x + \dots + \cos nx} \right) = \sin \frac{x}{2} + \sum_{k=1}^n 2\sin \frac{x}{2} \cos kx$. Using the identity $2\sin A \cos B = \sin(A+B) + \sin(A-B)$,we get: $2\sin \frac{x}{2} \cos kx = \sin \left( k + \frac{1}{2} \right)x - \sin \left( k - \frac{1}{2} \right)x$. Summing this from $k=1$ to $n$,we get a telescoping series: $\sin \frac{x}{2} + \left( \sin \frac{3x}{2} - \sin \frac{x}{2} \right) + \left( \sin \frac{5x}{2} - \sin \frac{3x}{2} \right) + \dots + \left( \sin \left( n + \frac{1}{2} \right)x - \sin \left( n - \frac{1}{2} \right)x \right) = \sin \left( n + \frac{1}{2} \right)x$. Thus,$\frac{\sin \left( n + \frac{1}{2} \right)x}{\sin \frac{x}{2}} = 1 + 2\cos x + 2\cos 2x + \dots + 2\cos nx$. Now,integrate from $0$ to $\pi$: $\int_0^\pi \left( 1 + 2\sum_{k=1}^n \cos kx \right) dx = \left[ x + 2\sum_{k=1}^n \frac{\sin kx}{k} \right]_0^\pi$. Since $\sin(k\pi) = 0$ for all integers $k$,the integral evaluates to $\pi - 0 = \pi$.

$\int\limits_0^\pi {\frac{{\sin \left( {n + \frac{1}{2}} \right)x}}{{\sin \frac{x}{2}}}} \,dx$,$(n \in N)$ equals

Similar Questions

$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \left( \frac{x+\frac{\pi}{4}}{2-\cos 2x} \right) dx$ is equal to

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

$\int_0^{2a} f(x) dx - \int_a^{2a} f(x) dx =$

If $f(x)$ is an odd function of $x,$ then $\int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {f(\cos x)\,dx} $ is equal to

$\int_0^1 \frac{\log x}{\sqrt{1 - x^2}} \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module