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

$\int e^x(1-\cot x+\cot^2 x) dx =$

$\int e^{x} \frac{(x-1)}{x^{2}} d x$ is equal to

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

$\int {{e^x}(1 + \tan x)\sec x\,dx = }$

If $\int \left\{ \cos^{-1} x - (1-x^2)^{-\frac{1}{2}} \right\} k \, dx = k \cdot \cos^{-1} x + c$,then $k = $ . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module