int sec^{frac{2}{3}} x cdot operatorname{cose | vedclass

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Question

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

Vedclass · Accepted Answer

$-3 	an^{-\frac{1}{3}} x + c$,where $c$ is the constant of integration

Vedclass · Answer

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

$\int \sec^{\frac{2}{3}} x \cdot \operatorname{cosec}^{\frac{4}{3}} x \, dx =$

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