If (n - m) is odd and |m| ne |n|, then int_0^ | vedclass

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

Question

(A) Let $I = \int_0^\pi \cos mx \sin nx \,dx$. Using the trigonometric identity $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$,we have $\cos mx \sin nx = \

Vedclass · Accepted Answer

$\frac{2n}{n^2 - m^2}$

Vedclass · Answer

(A) Let $I = \int_0^\pi \cos mx \sin nx \,dx$. Using the trigonometric identity $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$,we have $\cos mx \sin nx = \frac{1}{2} [\sin(n+m)x - \sin(m-n)x] = \frac{1}{2} [\sin(n+m)x + \sin(n-m)x]$. Integrating this,we get: $I = \frac{1}{2} \int_0^\pi [\sin(n+m)x + \sin(n-m)x] \,dx$ $I = \frac{1}{2} \left[ -\frac{\cos(n+m)x}{n+m} - \frac{\cos(n-m)x}{n-m} \right]_0^\pi$ $I = -\frac{1}{2} \left[ \left( \frac{\cos(n+m)\pi}{n+m} + \frac{\cos(n-m)\pi}{n-m} \right) - \left( \frac{1}{n+m} + \frac{1}{n-m} \right) \right]$ Since $(n-m)$ is odd,$(n+m)$ is also odd. Thus,$\cos(n+m)\pi = -1$ and $\cos(n-m)\pi = -1$. $I = -\frac{1}{2} \left[ \left( \frac{-1}{n+m} - \frac{1}{n-m} \right) - \left( \frac{1}{n+m} + \frac{1}{n-m} \right) \right]$ $I = -\frac{1}{2} \left[ -2 \left( \frac{1}{n+m} + \frac{1}{n-m} \right) \right] = \frac{1}{n+m} + \frac{1}{n-m}$ $I = \frac{(n-m) + (n+m)}{n^2 - m^2} = \frac{2n}{n^2 - m^2}$.

If $(n - m)$ is odd and $|m| \ne |n|,$ then $\int_0^\pi {\cos mx \sin nx} \,dx$ is

Similar Questions

$\int_0^\pi \sqrt{1+4 \sin^2 \frac{x}{2}+4 \sin \frac{x}{2}} \, dx$ is equal to

$\int_0^1 \sin \left(2 \tan ^{-1} \sqrt{\frac{1+x}{1-x}}\right) d x$ is equal to :

$\int_0^4 |2x - 5| \, dx = $

$\int_0^\pi \frac{dx}{1 + \sin x} = $

Let $g(x) = \int_0^{|x|^{3/4}} t^{2/3} \sin \frac{1}{t} \, dt$,for all real $x$. Then,$\lim_{x \rightarrow 0} \frac{g(x)}{x}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module