The integral int_{pi /6}^{pi /3} {sec ^{2/3} | vedclass

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

Question

(C) Let $I = \int_{\pi/6}^{\pi/3} \sec^{2/3} x \csc^{4/3} x \, dx$.We can rewrite the integrand as:$I = \int_{\pi/6}^{\pi/3} \frac{1}{\cos^{2/3} x \si

Vedclass · Accepted Answer

$3^{7/6} - 3^{5/6}$

Vedclass · Answer

(C) Let $I = \int_{\pi/6}^{\pi/3} \sec^{2/3} x \csc^{4/3} x \, dx$.We can rewrite the integrand as:$I = \int_{\pi/6}^{\pi/3} \frac{1}{\cos^{2/3} x \sin^{4/3} x} \, dx = \int_{\pi/6}^{\pi/3} \frac{1}{\cos^{2/3} x \sin^{2/3} x \cdot \sin^{2/3} x} \, dx$$I = \int_{\pi/6}^{\pi/3} \frac{1}{(\sin x \cos x)^{2/3} \cdot \sin^{2/3} x} \, dx = \int_{\pi/6}^{\pi/3} \frac{1}{	an^{2/3} x \cdot \sin^2 x} \, dx$$I = \int_{\pi/6}^{\pi/3} \frac{\sec^2 x}{	an^{2/3} x} \, dx$.Let $	an x = t$,then $\sec^2 x \, dx = dt$.When $x = \pi/6$,$t = 	an(\pi/6) = 1/\sqrt{3} = 3^{-1/2}$.When $x = \pi/3$,$t = 	an(\pi/3) = \sqrt{3} = 3^{1/2}$.$I = \int_{3^{-1/2}}^{3^{1/2}} t^{-2/3} \, dt = \left[ \frac{t^{1/3}}{1/3} ight]_{3^{-1/2}}^{3^{1/2}} = 3 \left[ t^{1/3} ight]_{3^{-1/2}}^{3^{1/2}}$$I = 3 \left( (3^{1/2})^{1/3} - (3^{-1/2})^{1/3} ight) = 3 \left( 3^{1/6} - 3^{-1/6} ight)$$I = 3^{1 + 1/6} - 3^{1 - 1/6} = 3^{7/6} - 3^{5/6}$.

The integral $\int_{\pi /6}^{\pi /3} {\sec ^{2/3} x \, \csc ^{4/3} x \, dx}$ is equal to

Similar Questions

The value of $\int_1^{e^2} \frac{dx}{x(1 + \ln x)^2}$ is

$\int_0^{\pi /4} \tan^6 x \sec^2 x \, dx = $

$\int_0^{\frac{\pi}{2}} \sin^4 \theta \cos^3 \theta \, d\theta =$

$\int_1^e \frac{1 + \log x}{x} \, dx = $

$\int_{8}^{18} \frac{1}{(x+2) \sqrt{x-3}} \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module