The value of the integral int_{-pi}^{pi} (cos | vedclass

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

Question

(D) Let $I = \int_{-\pi}^{\pi} (\cos ax - \sin bx)^2 dx$.Expanding the integrand,we get $I = \int_{-\pi}^{\pi} (\cos^2 ax + \sin^2 bx - 2\cos ax \sin

Vedclass · Accepted Answer

$2\pi$

Vedclass · Answer

(D) Let $I = \int_{-\pi}^{\pi} (\cos ax - \sin bx)^2 dx$.Expanding the integrand,we get $I = \int_{-\pi}^{\pi} (\cos^2 ax + \sin^2 bx - 2\cos ax \sin bx) dx$.Using the linearity of the integral,$I = \int_{-\pi}^{\pi} (\cos^2 ax + \sin^2 bx) dx - 2\int_{-\pi}^{\pi} \cos ax \sin bx dx$.Since $\cos ax \sin bx$ is an odd function,the integral $\int_{-\pi}^{\pi} \cos ax \sin bx dx = 0$.Thus,$I = \int_{-\pi}^{\pi} (\cos^2 ax + \sin^2 bx) dx$.Since the integrand is an even function,$I = 2 \int_{0}^{\pi} (\cos^2 ax + \sin^2 bx) dx$.Using the identities $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$ and $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$,we have:$I = 2 \int_{0}^{\pi} (\frac{1 + \cos 2ax}{2} + \frac{1 - \cos 2bx}{2}) dx = \int_{0}^{\pi} (2 + \cos 2ax - \cos 2bx) dx$.Integrating term by term,$I = [2x + \frac{\sin 2ax}{2a} - \frac{\sin 2bx}{2b}]_{0}^{\pi}$.Since $a$ and $b$ are integers,$\sin(2a\pi) = 0$ and $\sin(2b\pi) = 0$.Therefore,$I = 2\pi + 0 - 0 = 2\pi$.

The value of the integral $\int_{-\pi}^{\pi} (\cos ax - \sin bx)^2 dx$,where $a$ and $b$ are integers,is

Similar Questions

Let $\phi (x) = \int_{0}^{1} e^{x} e^{t} \phi (t) dt + x$. If $\phi (\ln (e^{2} - 3))$ is equal to $A$,then find the value of $A$.

If $\int_0^a \sqrt{\frac{a-x}{x}} dx = \frac{k}{2}$,then $k = $

$\int_0^2 \frac{2x-2}{2x-x^2} dx$ is equal to

The integral $\int_{0}^{1} \frac{1}{7^{\left[\frac{1}{x}\right]}} dx$,where $[.]$ denotes the greatest integer function,is equal to:

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module