If an anti-derivate of $f(x)$ is $e^{x}$ and that of $g(x)$ is $\cos x$, then $\int_{}^{}{f(x)}\cos xdx + \int_{}^{}{g(x)}e^{x}dx$ is equal to

Question

Dr. Abubacker · Accepted Answer

(c)
Given, $\int_{}^{}{f(x)}dx = e^{x}$ and $\int_{}^{}{g(x)}dx = \cos x$
$\Rightarrow f(x) = e^{x}$ and $g(x) = - \sin x$
$	herefore\int_{}^{}{f(x)}\cos xdx + \int_{}^{}{g(x)}e^{x}dx$
$= \int_{}^{}e^{x}\cos xdx - \int_{}^{}{e^{x}\sin x}dx$
$= e^{x}\cos x + \int_{}^{}{\sin x}e^{x}dx - \int_{}^{}e^{x}\sin xdx = e^{x}\cos x + c$

Q. If an anti-derivate of $f(x)$ is $e^{x}$ and that of $g(x)$ is $\cos x$, then $\int_{}^{}{f(x)}\cos xdx + \int_{}^{}{g(x)}e^{x}dx$ is equal to

Explanation

Reviewed by Dr. Abubacker

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