The value of the definite integral int_{pi / | vedclass

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

Question

(D) Let $I = \int_{\pi / 24}^{5 \pi / 24} \frac{1}{1 + (	an 2x)^{1/3}} dx = \int_{\pi / 24}^{5 \pi / 24} \frac{(\cos 2x)^{1/3}}{(\cos 2x)^{1/3} + (\s

Vedclass · Accepted Answer

$\frac{\pi}{12}$

Vedclass · Answer

(D) Let $I = \int_{\pi / 24}^{5 \pi / 24} \frac{1}{1 + (	an 2x)^{1/3}} dx = \int_{\pi / 24}^{5 \pi / 24} \frac{(\cos 2x)^{1/3}}{(\cos 2x)^{1/3} + (\sin 2x)^{1/3}} dx \dots (i)$Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,where $a+b = \frac{\pi}{24} + \frac{5\pi}{24} = \frac{6\pi}{24} = \frac{\pi}{4}$.Then $I = \int_{\pi / 24}^{5 \pi / 24} \frac{(\cos(2(\frac{\pi}{4} - x)))^{1/3}}{(\cos(2(\frac{\pi}{4} - x)))^{1/3} + (\sin(2(\frac{\pi}{4} - x)))^{1/3}} dx$$I = \int_{\pi / 24}^{5 \pi / 24} \frac{(\cos(\frac{\pi}{2} - 2x))^{1/3}}{(\cos(\frac{\pi}{2} - 2x))^{1/3} + (\sin(\frac{\pi}{2} - 2x))^{1/3}} dx$$I = \int_{\pi / 24}^{5 \pi / 24} \frac{(\sin 2x)^{1/3}}{(\sin 2x)^{1/3} + (\cos 2x)^{1/3}} dx \dots (ii)$Adding $(i)$ and $(ii)$:$2I = \int_{\pi / 24}^{5 \pi / 24} \frac{(\cos 2x)^{1/3} + (\sin 2x)^{1/3}}{(\cos 2x)^{1/3} + (\sin 2x)^{1/3}} dx = \int_{\pi / 24}^{5 \pi / 24} 1 dx$$2I = [x]_{\pi / 24}^{5 \pi / 24} = \frac{5\pi}{24} - \frac{\pi}{24} = \frac{4\pi}{24} = \frac{\pi}{6}$$I = \frac{\pi}{12}$

The value of the definite integral $\int_{\pi / 24}^{5 \pi / 24} \frac{d x}{1+\sqrt[3]{\tan 2 x}}$ is

Similar Questions

$\int_0^1 x(1-x)^n dx =$

$\int_{-1}^1 \frac{\log (1+x)}{1+x^2} d x = \int_0^1 \frac{\log (1+x)}{1+x^2} d x + \int_0^1 f(x) d x$, then $f(x) =$

The value of $\int_0^{n\pi + v} {|\sin x|\,dx} $ is

The greater of $\int_0^{\pi /2} \frac{\sin x}{x} \, dx$ and $\frac{\pi}{2}$ is:

The value of $\int_{0}^{4042} \frac{\sqrt{x} \, dx}{\sqrt{x}+\sqrt{4042-x}}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module