int_{-pi}^{pi} (cos px - sin qx)^2 dx is equa | vedclass

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

Question

(D) Let $I = \int_{-\pi}^{\pi} (\cos^2 px + \sin^2 qx - 2 \sin qx \cos px) dx$. Using the property $\int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx$ if

Vedclass · Accepted Answer

$2\pi$

Vedclass · Answer

(D) Let $I = \int_{-\pi}^{\pi} (\cos^2 px + \sin^2 qx - 2 \sin qx \cos px) dx$. Using the property $\int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx$ if $f(x)$ is even and $0$ if $f(x)$ is odd: $I = \int_{-\pi}^{\pi} \cos^2 px dx + \int_{-\pi}^{\pi} \sin^2 qx dx - 2 \int_{-\pi}^{\pi} \sin qx \cos px dx$. Since $\sin qx \cos px$ is an odd function,its integral over $[-\pi, \pi]$ is $0$. $I = \int_{-\pi}^{\pi} \frac{1 + \cos 2px}{2} dx + \int_{-\pi}^{\pi} \frac{1 - \cos 2qx}{2} dx$. $I = \frac{1}{2} [x + \frac{\sin 2px}{2p}]_{-\pi}^{\pi} + \frac{1}{2} [x - \frac{\sin 2qx}{2q}]_{-\pi}^{\pi}$. $I = \frac{1}{2} [(\pi - (-\pi)) + 0] + \frac{1}{2} [(\pi - (-\pi)) - 0] = \frac{1}{2} (2\pi) + \frac{1}{2} (2\pi) = \pi + \pi = 2\pi$.

$\int_{-\pi}^{\pi} (\cos px - \sin qx)^2 dx$ is equal to (where $p$ and $q$ are integers).

Similar Questions

If the value of the integral $\int_{0}^{5} \frac{x+[x]}{e^{x-[x]}} \,dx = \alpha e^{-1} + \beta$,where $\alpha, \beta \in R, 5\alpha + 6\beta = 0$,and $[x]$ denotes the greatest integer less than or equal to $x$; then the value of $(\alpha + \beta)^{2}$ is equal to:

The value of $\int_0^\pi \left| \sin x - \frac{2x}{\pi} \right| dx$ is

$\int_1^e \frac{1}{x} \, dx$ is equal to

Evaluate the definite integral $\int_{0}^{1} \left(x e^{x} + \sin \frac{\pi x}{4}\right) dx$.

$\int_{0}^{4/\pi} \left( 3x^2 \sin \frac{1}{x} - x \cos \frac{1}{x} \right) dx$ has the value:

Vedclass Test Series

Exam Paper Generator

Online Exam Module