The integral int sec^{frac{2}{3}} x cdot oper | vedclass

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let $I = \int \sec^{\frac{2}{3}} x \operatorname{cosec}^{\frac{4}{3}} x \, dx$. We can rewrite the integrand as: $I = \int \frac{1}{\cos^{\frac{2}{3}} x \sin^{\frac{4}{3}} x} \, dx$. Divide the numerator and denominator by $\cos^{\frac{4}{3}} x$: $I = \int \frac{1}{\left(\frac{\sin^{\frac{4}{3}} x}{\cos^{\frac{4}{3}} x}ight) \cdot \cos^{\frac{2}{3}} x \cdot \cos^{\frac{4}{3}} x} \, dx = \int \frac{1}{(	an x)^{\frac{4}{3}} \cdot \cos^2 x} \, dx$. Since $\frac{1}{\cos^2 x} = \sec^2 x$,we have: $I = \int \frac{\sec^2 x}{(	an x)^{\frac{4}{3}}} \, dx$. Let $t = 	an x$,then $dt = \sec^2 x \, dx$. Substituting these into the integral: $I = \int t^{-\frac{4}{3}} \, dt = \frac{t^{-\frac{4}{3} + 1}}{-\frac{4}{3} + 1} + c = \frac{t^{-\frac{1}{3}}}{-\frac{1}{3}} + c = -3t^{-\frac{1}{3}} + c$. Substituting $t = 	an x$ back: $I = -3(	an x)^{-\frac{1}{3}} + c$.

The integral $\int \sec^{\frac{2}{3}} x \cdot \operatorname{cosec}^{\frac{4}{3}} x \, dx$ is equal to

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