The integral 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

(D) Let $I = \int \sec^{2/3} x \csc^{4/3} x \, dx$. We can rewrite the integrand as: $I = \int \frac{1}{(\cos x)^{2/3} (\sin x)^{4/3}} \, dx$. Divide

Vedclass · Accepted Answer

$-3 	an^{-1/3} x + C$

Vedclass · Answer

(D) Let $I = \int \sec^{2/3} x \csc^{4/3} x \, dx$. We can rewrite the integrand as: $I = \int \frac{1}{(\cos x)^{2/3} (\sin x)^{4/3}} \, dx$. Divide the numerator and denominator by $\cos^{4/3} x$: $I = \int \frac{1}{\frac{(\sin x)^{4/3}}{(\cos x)^{4/3}} \cdot (\cos x)^{2/3} \cdot (\cos x)^{4/3}} \, dx$. $I = \int \frac{1}{(	an x)^{4/3} \cdot \cos^2 x} \, dx$. $I = \int \frac{\sec^2 x}{(	an x)^{4/3}} \, dx$. Let $t = 	an x$,then $dt = \sec^2 x \, dx$. Substituting these into the integral: $I = \int t^{-4/3} \, dt$. Integrating with respect to $t$: $I = \frac{t^{-4/3 + 1}}{-4/3 + 1} + C = \frac{t^{-1/3}}{-1/3} + C = -3 t^{-1/3} + C$. Substituting back $t = 	an x$: $I = -3 	an^{-1/3} x + C$.

The integral $\int \sec^{2/3} x \csc^{4/3} x \, dx$ is equal to: (Here $C$ is a constant of integration)

Similar Questions

Integrate the function $\frac{1}{x(\log x)^{m}}$,where $x > 0$ and $m \neq 1$.

$\int \frac{\sin(2x)}{\sin^2(x) + 2\cos^2(x)} dx = $

If $f(x) = \int \frac{5x^8 + 7x^6}{(x^2 + 2x^7 + 1)^2} dx$ $(x \geq 0)$ and $f(0) = 0$,then the value of $f(1) =$ ?

The value of $\int \frac{e^{x}(1+x) dx}{\cos^{2}(x e^{x})}$ is equal to

The integral $\int \frac{\sin^2 x \cos^2 x}{(\sin^3 x + \cos^3 x)^2} dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module