int sec^{2/3} x csc^{4/3} x , dx = | vedclass

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

Question

(B) We have $I = \int \sec^{2/3} x \csc^{4/3} x \, dx = \int \frac{1}{\cos^{2/3} x \sin^{4/3} x} \, dx$. Multiplying the numerator and denominator by

Vedclass · Accepted Answer

$-3(	an x)^{-1/3} + c$

Vedclass · Answer

(B) We have $I = \int \sec^{2/3} x \csc^{4/3} x \, dx = \int \frac{1}{\cos^{2/3} x \sin^{4/3} x} \, dx$. Multiplying the numerator and denominator by $\sec^2 x$,we get: $I = \int \frac{\sec^2 x}{\frac{\cos^{2/3} x \sin^{4/3} x}{\cos^2 x}} \, dx = \int \frac{\sec^2 x}{\frac{\sin^{4/3} x}{\cos^{4/3} x}} \, dx = \int \frac{\sec^2 x}{(	an x)^{4/3}} \, dx$. Let $	an x = t$,then $\sec^2 x \, dx = dt$. Substituting these into the integral,we get: $I = \int t^{-4/3} \, dt = \frac{t^{-4/3 + 1}}{-4/3 + 1} + c = \frac{t^{-1/3}}{-1/3} + c = -3t^{-1/3} + c$. Replacing $t$ with $	an x$,we get $I = -3(	an x)^{-1/3} + c$.

$\int \sec^{2/3} x \csc^{4/3} x \, dx = $

Similar Questions

If $\int \frac{3}{2 \cos ^3 x \sqrt{2 \sin 2 x}} d x = \frac{3}{2}(\tan x)^B + \frac{1}{10}(\tan x)^A + c$,then $A =$

Integrate the function $\frac{(\log x)^{2}}{x}$.

$\int \frac{dx}{5 + 4\cos x} = $

If $\int \frac{2 \sin 2x - 3 \cos x}{2 \sin^2 x - 3 \sin x + 4} dx = f(x) + c$ where $c$ is the constant of integration,then $f\left(\frac{\pi}{2}\right) - f(0) =$

If $\int \frac{x+1}{\sqrt{2x-1}} \, dx = f(x) \sqrt{2x-1} + C$,where $C$ is an arbitrary constant,then $f(x)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module