Evaluate the following definite integral: int | vedclass

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

Question

(C) Let $I = \int_0^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x} d x \quad \dots(1)$Using the property $\

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

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

Evaluate the following definite integral: $\int_0^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x} d x$

Similar Questions

The value of the integral $\int_0^{\pi / 2} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x$ is

$\int_{-1}^{1} x \tan^{-1} x \, dx$ equals

$\int_{-1}^3 \left(\tan^{-1}\left(\frac{x}{x^2+1}\right) + \tan^{-1}\left(\frac{x^2+1}{x}\right)\right) dx =$

Evaluate $\int_{-1}^{1} \sin^{5} x \cos^{4} x \, dx$

The value of $\int_{0}^{2\pi} [\sin 2x(1 + \cos 3x)] \,dx$,where $[t]$ denotes the greatest integer function,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module