The integral int frac{sec^2 x}{(sec x+tan x)^ | vedclass

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Question

(A) Let $t = \sec x + 	an x$. Then $dt = (\sec x 	an x + \sec^2 x) dx = \sec x(	an x + \sec x) dx = \sec x \cdot t \cdot dx$. Thus,$\sec x dx = \fr

Vedclass · Accepted Answer

$\frac{-1}{(\sec x+	an x)^{11/2}} \left\{ \frac{1}{11} + \frac{1}{7}(\sec x+	an x)^2 ight\} + K$

Vedclass · Answer

(A) Let $t = \sec x + 	an x$. Then $dt = (\sec x 	an x + \sec^2 x) dx = \sec x(	an x + \sec x) dx = \sec x \cdot t \cdot dx$. Thus,$\sec x dx = \frac{dt}{t}$. Also,$\sec x - 	an x = \frac{1}{t}$. Adding the two equations: $2 \sec x = t + \frac{1}{t} \implies \sec x = \frac{1}{2} \left( t + \frac{1}{t} ight)$. Substituting into the integral: $I = \int \frac{\sec x \cdot (\sec x dx)}{t^{9/2}} = \int \frac{\frac{1}{2}(t + \frac{1}{t}) \cdot \frac{dt}{t}}{t^{9/2}} = \frac{1}{2} \int \frac{t + t^{-1}}{t^{11/2}} dt = \frac{1}{2} \int (t^{-9/2} + t^{-13/2}) dt$. Integrating: $I = \frac{1}{2} \left[ \frac{t^{-7/2}}{-7/2} + \frac{t^{-11/2}}{-11/2} ight] + K = -\left[ \frac{1}{7 t^{7/2}} + \frac{1}{11 t^{11/2}} ight] + K$. Factoring out $-\frac{1}{t^{11/2}}$: $I = -\frac{1}{t^{11/2}} \left[ \frac{t^2}{7} + \frac{1}{11} ight] + K$. Substituting $t = \sec x + 	an x$ back,we get option $A$.

The integral $\int \frac{\sec^2 x}{(\sec x+\tan x)^{9/2}} dx$ equals (for some arbitrary constant $K$):

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