Question

If an antiderivative of f(x) is $e^x$ and that of $g\left(x\right)$ is cos x, then $\int f\left(x\right)$ cos x dx + $\int g\left(x\right)$ $e^x$ dx is equal to

• A. $f\left(x\right).g\left(x\right)+C$
• B. $f\left(x\right)+g\left(x\right)+C$
• C. $e^x$ $cosx+C$
• D. $f\left(x\right)-g\left(x\right)+C$

Answer 1

The correct option is C $e^x$ $cosx+C$

$\int f\left(x\right)dx=e^x$

$\Rightarrow f\left(x\right)=e^x$

$\int g\left(x\right)dx=\cos x$

$\Rightarrow g\left(x\right)=-\sin x$

Now $\int f\left(x\right)\cos xdx+\int g\left(x\right)e^{x}dx$

$=\int e^x\cos xdx+\int e^x(-\sin x)dx$

$=\int e^x(\cos x-\sin x)dx$

$=e^x\cos x+c$

[using $\int e^x\left(f\right(x)+f^{\prime }(x\left)\right)dx=e^{x}f\left(x\right)+c$ property].