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

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(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 = $

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