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

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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)

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