If I_1 = int_0^{3 pi} f(cos^2 x) dx and I_2 = | vedclass

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

Question

(C) Let $g(x) = f(\cos^2 x)$. Since $\cos^2(x + \pi) = (-\cos x)^2 = \cos^2 x$, the function $g(x)$ is periodic with period $\pi$. Using the property

Vedclass · Accepted Answer

$I_1 = 3 I_2$

Vedclass · Answer

(C) Let $g(x) = f(\cos^2 x)$. Since $\cos^2(x + \pi) = (-\cos x)^2 = \cos^2 x$, the function $g(x)$ is periodic with period $\pi$. Using the property of definite integrals for periodic functions, $\int_0^{n T} g(x) dx = n \int_0^T g(x) dx$, where $T$ is the period. Here, $T = \pi$ and $n = 3$. Therefore, $I_1 = \int_0^{3 \pi} f(\cos^2 x) dx = 3 \int_0^\pi f(\cos^2 x) dx$. Since $I_2 = \int_0^\pi f(\cos^2 x) dx$, we have $I_1 = 3 I_2$.

If $I_1 = \int_0^{3 \pi} f(\cos^2 x) dx$ and $I_2 = \int_0^\pi f(\cos^2 x) dx$, then

Similar Questions

If $I_1 = \int\limits_0^1 {{e^{ - x}}} {\cos ^2}x\,dx$,$I_2 = \int\limits_0^1 {{e^{ - {x^2}}}} {\cos ^2}x\,dx$ and $I_3 = \int\limits_0^1 {{e^{ - {x^3}}}} dx$; then

If $\int_0^1 {{e^{{x^2}}}(x - \alpha )\,dx = 0,} $ then

$\frac{3}{25} \int_0^{25 \pi} \sqrt{|\cos x - \cos^3 x|} \, dx =$

If $\int_0^{\frac{\pi}{4}} \frac{\sin^2 x}{1+\sin x \cos x} dx = \frac{1}{a} \log_e\left(\frac{a}{3}\right) + \frac{\pi}{b \sqrt{3}}$,where $a, b \in N$,then $a+b$ is equal to ....................

$\left|\frac{120}{\pi^3} \int_0^\pi \frac{x^2 \sin x \cos x}{\sin^4 x + \cos^4 x} dx\right|$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module