If P = int_0^{3pi} f(cos^2 x) dx and Q = int_ | vedclass

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

Question

(C) Given $P = \int_0^{3\pi} f(\cos^2 x) dx$ and $Q = \int_0^{\pi} f(\cos^2 x) dx$. Using the property of definite integrals,$\int_0^{na} f(x) dx = n

Vedclass · Accepted Answer

$P - 3Q = 0$

Vedclass · Answer

(C) Given $P = \int_0^{3\pi} f(\cos^2 x) dx$ and $Q = \int_0^{\pi} f(\cos^2 x) dx$. Using the property of definite integrals,$\int_0^{na} f(x) dx = n \int_0^a f(x) dx$ if $f(x+a) = f(x)$. Here,$f(\cos^2(x+\pi)) = f(\cos^2 x)$ because $\cos^2(x+\pi) = (-\cos x)^2 = \cos^2 x$. Thus,the function $f(\cos^2 x)$ is periodic with period $\pi$. Therefore,$P = \int_0^{3\pi} f(\cos^2 x) dx = 3 \int_0^{\pi} f(\cos^2 x) dx = 3Q$. This implies $P = 3Q$,or $P - 3Q = 0$.

If $P = \int_0^{3\pi} f(\cos^2 x) dx$ and $Q = \int_0^{\pi} f(\cos^2 x) dx$,then:

Similar Questions

The value of the integral $\int_{0}^{1} x \cot^{-1}(1 - x^2 + x^4) dx$ is

$\int_0^{\frac{\pi}{4}} \frac{\cos ^2 x}{\cos ^2 x+4 \sin ^2 x} d x=$

By using the properties of definite integrals,evaluate the integral $\int_{0}^{1} x(1-x)^{n} d x$.

$\int_0^{\frac{\pi}{2}} \frac{\sin^2 x}{\sin x + \cos x} dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module