int frac{sec x}{3(sec x+tan x)+2} d x= | vedclass

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(A) Let $I = \int \frac{\sec x}{3(\sec x+	an x)+2} dx$.Multiply the numerator and denominator by $(\sec x - 	an x)$:$I = \int \frac{\sec x(\sec x -

Vedclass · Accepted Answer

$\frac{1}{2} \log \left|\frac{	an \frac{x}{2}+1}{	an \frac{x}{2}+5}ight|+c$

Vedclass · Answer

(A) Let $I = \int \frac{\sec x}{3(\sec x+	an x)+2} dx$.Multiply the numerator and denominator by $(\sec x - 	an x)$:$I = \int \frac{\sec x(\sec x - 	an x)}{3(\sec^2 x - 	an^2 x) + 2(\sec x - 	an x)} dx$.Since $\sec^2 x - 	an^2 x = 1$,we have:$I = \int \frac{\sec^2 x - \sec x 	an x}{3 + 2(\sec x - 	an x)} dx$.Let $u = \sec x - 	an x$. Then $du = (\sec x 	an x - \sec^2 x) dx$,so $-du = (\sec^2 x - \sec x 	an x) dx$.$I = \int \frac{-du}{3 + 2u} = -\frac{1}{2} \ln|3 + 2u| + C = -\frac{1}{2} \ln|3 + 2(\sec x - 	an x)| + C$.Using half-angle identities $\sec x = \frac{1+	an^2(x/2)}{1-	an^2(x/2)}$ and $	an x = \frac{2	an(x/2)}{1-	an^2(x/2)}$:$I = -\frac{1}{2} \ln \left| \frac{3(1-	an^2(x/2)) + 2(1+	an^2(x/2)) - 4	an(x/2)}{1-	an^2(x/2)} ight| + C$.$I = -\frac{1}{2} \ln \left| \frac{5 - 4	an(x/2) - 	an^2(x/2)}{(1-	an(x/2))(1+	an(x/2))} ight| + C$.$I = -\frac{1}{2} \ln \left| \frac{(5+	an(x/2))(1-	an(x/2))}{(1-	an(x/2))(1+	an(x/2))} ight| + C$.$I = -\frac{1}{2} \ln \left| \frac{5+	an(x/2)}{1+	an(x/2)} ight| + C = \frac{1}{2} \ln \left| \frac{1+	an(x/2)}{5+	an(x/2)} ight| + C$.

$\int \frac{\sec x}{3(\sec x+\tan x)+2} d x=$

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