Question

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

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

Answer 1

The correct option is C $e^x\cos x+c$

$\int \underset{I}{f\left(x\right)}\underset{II}{\cos x}dx+\int \underset{I}{g\left(x\right)}\underset{II}{e^x}d\left(x\right)$

Using by parts

$f\left(x\right)\int \cos xdx-\int f^{\prime }\left(x\right)\cdot \sin xdx+g\left(x\right)\int e^x-\int g^{\prime }\left(x\right)e^x$

$\int \underset{II}{f\left(x\right)}\underset{I}{\cos x}dx+\underset{II}{g\left(x\right)}\underset{I}{e^x}dx$

Using by parts

$\cos x\int f\left(x\right)+\int \sin x(\int f\left(x\right)dx)dx+e^x\int g\left(x\right)-\int e^x(\int g\left(x\right)dx)$

$=\cos x{e}^x+\int \sin x{e}^{x}dx+e^x\cos x-\int e^x\cos xdx$

$=2\cos x{e}^x+\int \sin x{e}^{x}dx+\int e^x\cos x-\int e^x\cos xdx$

$=2\cos x{e}^x+\int \sin x{e}^{x}dx+\int \underset{II}{e^x}\underset{I}{(-\cos x)}dx$

$=2\cos x{e}^x+\int \sin x{e}^{x}dx+(-\cos x)e^x-\int \sin x{e}^{x}dx$

$=\cos x{e}^x+c$.