int {frac{{sec x(1 + tan x)dx}}{{({e^{ - x}} | vedclass

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Question

(A) Given integral is $I = \int \frac{\sec x(1 + 	an x)dx}{e^{-x} + \sec x}$.Multiply numerator and denominator by $e^x$:$I = \int \frac{e^x \sec x(1

Vedclass · Accepted Answer

$\ln \left( {1 + {e^{\frac{\pi }{4}}}\sqrt 2 } \right)$

Vedclass · Answer

(A) Given integral is $I = \int \frac{\sec x(1 + 	an x)dx}{e^{-x} + \sec x}$.Multiply numerator and denominator by $e^x$:$I = \int \frac{e^x \sec x(1 + 	an x)dx}{1 + e^x \sec x}$.Let $t = 1 + e^x \sec x$. Then $dt = (e^x \sec x + e^x \sec x 	an x) dx = e^x \sec x(1 + 	an x) dx$.Thus,$I = \int \frac{dt}{t} = \ln |t| + C = \ln |1 + e^x \sec x| + C$.So,$f(x) = \ln(1 + e^x \sec x) + C$.Given $f(0) = \ln 2$,we have $\ln(1 + e^0 \sec 0) + C = \ln 2 \Rightarrow \ln(1 + 1) + C = \ln 2 \Rightarrow \ln 2 + C = \ln 2 \Rightarrow C = 0$.Therefore,$f(x) = \ln(1 + e^x \sec x)$.Now,$f\left( \frac{\pi}{4} ight) = \ln(1 + e^{\frac{\pi}{4}} \sec \frac{\pi}{4}) = \ln(1 + e^{\frac{\pi}{4}} \sqrt{2})$.The correct option is $A$.

$\int {\frac{{\sec x(1 + \tan x)dx}}{{({e^{ - x}} + \sec x)}}} = f(x) + C$ where $f(0) = \ln 2$,then $f\left( {\frac{\pi }{4}} \right)$ is -

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