int tan x sec^2 x sqrt{1 - tan^2 x} ; dx = | vedclass

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(A) Let $I = \int 	an x \sec^2 x \sqrt{1 - 	an^2 x} \; dx$. Substitute $	an x = t$,then $\sec^2 x \; dx = dt$. The integral becomes $\int t \sqrt{1

Vedclass · Accepted Answer

$-\frac{1}{3}(1 - 	an^2 x)^{3/2} + c$

Vedclass · Answer

(A) Let $I = \int 	an x \sec^2 x \sqrt{1 - 	an^2 x} \; dx$. Substitute $	an x = t$,then $\sec^2 x \; dx = dt$. The integral becomes $\int t \sqrt{1 - t^2} \; dt$. Now,substitute $1 - t^2 = u$,then $-2t \; dt = du$,or $t \; dt = -\frac{1}{2} du$. Substituting these into the integral: $I = \int \sqrt{u} \left(-\frac{1}{2}ight) du = -\frac{1}{2} \int u^{1/2} \; du$. Integrating with respect to $u$: $I = -\frac{1}{2} \left(\frac{u^{3/2}}{3/2}ight) + c = -\frac{1}{3} u^{3/2} + c$. Substituting $u = 1 - t^2 = 1 - 	an^2 x$ back: $I = -\frac{1}{3}(1 - 	an^2 x)^{3/2} + c$.

$\int \tan x \sec^2 x \sqrt{1 - \tan^2 x} \; dx = $

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