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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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 $A$.

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

Similar Questions

If $\int e^x(x^3+x^2-x+4) dx = e^x f(x) + c$, then $f(1) =$

$\int e^x \left( \frac{1-x}{1+x^2} \right)^2 dx = $ . . . . . . + $C$

$\int \frac{x-3}{(x-1)^3} e^x \, dx =$

$\int \log x \cdot [\log (ex)]^{-2} dx = . . . . . .$

If $\int {{e^{{x^2}}}\left( {2 - \frac{1}{{{x^2}}}} \right)dx = {e^{{x^2}}}f(x) + C} $ and $f\left( {\frac{1}{2}} \right) = 2$,then $f(1)$ is equal to (where $C$ is an arbitrary constant).

Vedclass Test Series

Exam Paper Generator

Online Exam Module