int frac{sec^2 x}{(sec x + tan x)^{5/2}} dx = | vedclass

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(NONE) Let $I = \int \frac{\sec^2 x}{(\sec x + 	an x)^{5/2}} dx$.
We know that $\sec x + 	an x = t$.
Differentiating with respect to $x$,we get $(

Vedclass · Accepted Answer

$ -(\sec x + 	an x)^{-1/2} - \frac{1}{5}(\sec x + 	an x)^{-5/2} + c$

Vedclass · Answer

(NONE) Let $I = \int \frac{\sec^2 x}{(\sec x + 	an x)^{5/2}} dx$.
We know that $\sec x + 	an x = t$.
Differentiating with respect to $x$,we get $(\sec x 	an x + \sec^2 x) dx = dt$,which implies $\sec x(	an x + \sec x) dx = dt$.
Thus,$\sec x dx = \frac{dt}{t}$.
Also,$\sec x - 	an x = \frac{1}{t}$.
Adding the two equations: $2 \sec x = t + \frac{1}{t} = \frac{t^2+1}{t}$,so $\sec x = \frac{t^2+1}{2t}$.
Substituting these into the integral:
$I = \int \frac{\sec x \cdot \sec x dx}{t^{5/2}} = \int \frac{(\frac{t^2+1}{2t}) \cdot \frac{dt}{t}}{t^{5/2}} = \frac{1}{2} \int \frac{t^2+1}{t^{7/2}} dt$.
$I = \frac{1}{2} \int (t^{-3/2} + t^{-7/2}) dt$.
Integrating,we get $I = \frac{1}{2} [\frac{t^{-1/2}}{-1/2} + \frac{t^{-5/2}}{-5/2}] + c = -t^{-1/2} - \frac{1}{5} t^{-5/2} + c$.
Substituting $t = \sec x + 	an x$,we get $I = -(\sec x + 	an x)^{-1/2} - \frac{1}{5}(\sec x + 	an x)^{-5/2} + c$.

$\int \frac{\sec^2 x}{(\sec x + \tan x)^{5/2}} dx =$

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