int_{pi / 6}^{pi / 3} frac{sin ^{3} x}{sin ^{ | vedclass

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

Question

(C) Let $I = \int_{\pi / 6}^{\pi / 3} \frac{\sin ^{3} x}{\sin ^{3} x+\cos ^{3} x} d x$  $(i)$ Using the property $\int_{a}^{b} f(x) d x = \int_{a}^{b}

Vedclass · Accepted Answer

$\frac{\pi}{12}$

Vedclass · Answer

(C) Let $I = \int_{\pi / 6}^{\pi / 3} \frac{\sin ^{3} x}{\sin ^{3} x+\cos ^{3} x} d x$  $(i)$ Using the property $\int_{a}^{b} f(x) d x = \int_{a}^{b} f(a+b-x) d x$,we have: $I = \int_{\pi / 6}^{\pi / 3} \frac{\sin ^{3}(\frac{\pi}{3} + \frac{\pi}{6} - x)}{\sin ^{3}(\frac{\pi}{3} + \frac{\pi}{6} - x) + \cos ^{3}(\frac{\pi}{3} + \frac{\pi}{6} - x)} d x$ $I = \int_{\pi / 6}^{\pi / 3} \frac{\sin ^{3}(\frac{\pi}{2} - x)}{\sin ^{3}(\frac{\pi}{2} - x) + \cos ^{3}(\frac{\pi}{2} - x)} d x$ Since $\sin(\frac{\pi}{2} - x) = \cos x$ and $\cos(\frac{\pi}{2} - x) = \sin x$,we get: $I = \int_{\pi / 6}^{\pi / 3} \frac{\cos ^{3} x}{\cos ^{3} x + \sin ^{3} x} d x$  (ii) Adding $(i)$ and (ii): $2I = \int_{\pi / 6}^{\pi / 3} \frac{\sin ^{3} x + \cos ^{3} x}{\sin ^{3} x + \cos ^{3} x} d x$ $2I = \int_{\pi / 6}^{\pi / 3} 1 d x$ $2I = [x]_{\pi / 6}^{\pi / 3} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$ $I = \frac{\pi}{12}$

$\int_{\pi / 6}^{\pi / 3} \frac{\sin ^{3} x}{\sin ^{3} x+\cos ^{3} x} d x$ is equal to

Similar Questions

Value of the definite integral,$\int\limits_{ - \frac{1}{2}}^{\frac{1}{2}} {\,(\,\,{{\sin }^{ - 1}}(3x - 4{x^3})\,\, - \,\,{{\cos }^{ - 1}}(4{x^3} - 3x)\,\,)\,dx} \,\,$

The value of $\int_{-\pi / 2}^{\pi / 2} \frac{\cos ^{2} x}{1+3^{x}} d x$ is

$\int_0^1 f(1 - x) \, dx$ has the same value as which of the following integrals?

The value of the definite integral $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{d x}{\left(1+e^{x \cos x}\right)\left(\sin ^{4} x+\cos ^{4} x\right)}$ is equal to:

$\int_0^\pi x \sin x \cos^4 x \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module