int tan^3 2x sec 2x , dx = | vedclass

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(A) Let $I = \int 	an^3 2x \sec 2x \, dx$.We can rewrite $	an^3 2x$ as $	an^2 2x \cdot 	an 2x$.Since $	an^2 2x = \sec^2 2x - 1$,the integral beco

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$\frac{1}{6} \sec^3 2x - \frac{1}{2} \sec 2x + c$

Vedclass · Answer

(A) Let $I = \int 	an^3 2x \sec 2x \, dx$.We can rewrite $	an^3 2x$ as $	an^2 2x \cdot 	an 2x$.Since $	an^2 2x = \sec^2 2x - 1$,the integral becomes:$I = \int (\sec^2 2x - 1) \sec 2x 	an 2x \, dx$.Let $u = \sec 2x$. Then $du = 2 \sec 2x 	an 2x \, dx$,which implies $\sec 2x 	an 2x \, dx = \frac{1}{2} du$.Substituting these into the integral:$I = \int (u^2 - 1) \cdot \frac{1}{2} du$$I = \frac{1}{2} \int (u^2 - 1) \, du$$I = \frac{1}{2} \left( \frac{u^3}{3} - u ight) + c$$I = \frac{u^3}{6} - \frac{u}{2} + c$Substituting $u = \sec 2x$ back into the equation:$I = \frac{1}{6} \sec^3 2x - \frac{1}{2} \sec 2x + c$.

$\int \tan^3 2x \sec 2x \, dx = $

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