int frac{sec^2 x}{(sec x + tan x)^2} dx = | vedclass

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(B) Let $I = \int \frac{\sec^2 x}{(\sec x + 	an x)^2} dx$.Let $t = \sec x + 	an x$. Then $\frac{1}{t} = \sec x - 	an x$.Adding these,$2 \sec x = t

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $I = \int \frac{\sec^2 x}{(\sec x + 	an x)^2} dx$.Let $t = \sec x + 	an x$. Then $\frac{1}{t} = \sec x - 	an x$.Adding these,$2 \sec x = t + \frac{1}{t} \Rightarrow \sec x = \frac{1}{2}(t + \frac{1}{t})$.Differentiating $t = \sec x + 	an x$,we get $dt = (\sec x 	an x + \sec^2 x) dx = \sec x(	an x + \sec x) dx = \sec x \cdot t \cdot dx$.Thus,$\sec x dx = \frac{dt}{t}$.Substituting these into the integral:$I = \int \frac{\sec x \cdot \sec x dx}{t^2} = \int \frac{\frac{1}{2}(t + \frac{1}{t}) \cdot \frac{dt}{t}}{t^2} = \frac{1}{2} \int \frac{t^2+1}{t^4} dt$.$I = \frac{1}{2} \int (t^{-2} + t^{-4}) dt = \frac{1}{2} [\frac{t^{-1}}{-1} + \frac{t^{-3}}{-3}] + C$.$I = -\frac{1}{2t} - \frac{1}{6t^3} + C = -\frac{3t^2 + 1}{6t^3} + C$.Substituting $t = \sec x + 	an x$,we get $I = -\frac{1+3(\sec x+	an x)^2}{6(\sec x+	an x)^3} + C$.

$\int \frac{\sec^2 x}{(\sec x + \tan x)^2} dx =$

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