int sec x log(sec x + tan x) , dx = | vedclass

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(B) Let $t = \log(\sec x + 	an x)$.Then,differentiating with respect to $x$,we get $\frac{dt}{dx} = \frac{1}{\sec x + 	an x} \cdot (\sec x 	an x +

Vedclass · Accepted Answer

$\frac{1}{2}[\log(\sec x + 	an x)]^2 + c$

Vedclass · Answer

(B) Let $t = \log(\sec x + 	an x)$.Then,differentiating with respect to $x$,we get $\frac{dt}{dx} = \frac{1}{\sec x + 	an x} \cdot (\sec x 	an x + \sec^2 x) = \frac{\sec x(	an x + \sec x)}{\sec x + 	an x} = \sec x$.Thus,$dt = \sec x \, dx$.Substituting these into the integral,we get $\int t \, dt$.Integrating with respect to $t$,we have $\frac{t^2}{2} + c$.Substituting back the value of $t$,we get $\frac{1}{2}[\log(\sec x + 	an x)]^2 + c$.

$\int \sec x \log(\sec x + \tan x) \, dx = $

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