int_0^{3 pi / 2} frac{cos ^3 x}{cos ^3 x+sin | vedclass

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

Question

(C) Let $I = \int_0^{3 \pi / 2} \frac{\cos ^3 x}{\cos ^3 x+\sin ^3 x} d x$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get: $I = \i

Vedclass · Accepted Answer

$\frac{3 \pi}{4}$

Vedclass · Answer

(C) Let $I = \int_0^{3 \pi / 2} \frac{\cos ^3 x}{\cos ^3 x+\sin ^3 x} d x$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get: $I = \int_0^{3 \pi / 2} \frac{\cos ^3 (3 \pi / 2 - x)}{\cos ^3 (3 \pi / 2 - x) + \sin ^3 (3 \pi / 2 - x)} d x$. Since $\cos(3 \pi / 2 - x) = -\sin x$ and $\sin(3 \pi / 2 - x) = -\cos x$,we have: $I = \int_0^{3 \pi / 2} \frac{(-\sin x)^3}{(-\sin x)^3 + (-\cos x)^3} d x = \int_0^{3 \pi / 2} \frac{-\sin ^3 x}{-\sin ^3 x - \cos ^3 x} d x = \int_0^{3 \pi / 2} \frac{\sin ^3 x}{\sin ^3 x + \cos ^3 x} d x$. Adding the two expressions for $I$: $2I = \int_0^{3 \pi / 2} \frac{\cos ^3 x + \sin ^3 x}{\cos ^3 x + \sin ^3 x} d x = \int_0^{3 \pi / 2} 1 d x$. $2I = [x]_0^{3 \pi / 2} = \frac{3 \pi}{2}$. Therefore,$I = \frac{3 \pi}{4}$.

$\int_0^{3 \pi / 2} \frac{\cos ^3 x}{\cos ^3 x+\sin ^3 x} d x=$

Similar Questions

If $I_n = \int_{-\pi}^{\pi} \frac{\sin(nx)}{(1+\pi^x) \sin x} dx$,$n=0, 1, 2, \ldots$,then
$(A)$ $I_n = I_{n+2}$
$(B)$ $\sum_{m=1}^{10} I_{2m+1} = 10\pi$
$(C)$ $\sum_{m=1}^{10} I_{2m} = 0$
$(D)$ $I_n = I_{n+1}$

The value of $\int_{0}^{1} \frac{dx}{x + \sqrt{1 - x^2}}$ is

$\int_{0}^{\frac{\pi}{2}} \frac{\sin^{\frac{2}{3}} x}{\sin^{\frac{2}{3}} x + \cos^{\frac{2}{3}} x} dx =$

$\int_{-\pi / 2}^{\pi / 2} \sin |x| \, dx$ is equal to

The value of the integral $I = \int_{1/2014}^{2014} \frac{\tan^{-1} x}{x} dx$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module