int {{e^x}[tan x - log (cos x)],dx = } | vedclass

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Question

(A) We know that the integral of the form $\int e^x [f(x) + f'(x)]\,dx = e^x f(x) + c$.Given the integral $\int e^x [	an x - \log (\cos x)]\,dx$.Let

Vedclass · Accepted Answer

${e^x}\log (\sec x) + c$

Vedclass · Answer

(A) We know that the integral of the form $\int e^x [f(x) + f'(x)]\,dx = e^x f(x) + c$.Given the integral $\int e^x [	an x - \log (\cos x)]\,dx$.Let $f(x) = -\log (\cos x) = \log (\cos x)^{-1} = \log (\sec x)$.Then,$f'(x) = \frac{d}{dx} [\log (\sec x)] = \frac{1}{\sec x} \cdot (\sec x 	an x) = 	an x$.Substituting these into the formula,we get $\int e^x [	an x + \log (\sec x)]\,dx = e^x \log (\sec x) + c$.Thus,the correct option is $A$.

$\int {{e^x}[\tan x - \log (\cos x)]\,dx = }$

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