Question

If $\int e^x\cos \left(x\right)dx=F\left(x\right)+c,$ where $c$ is an arbitrary constant, then $F\left(x\right)=$

$\left(a\right)e^x(\sin x-\cos x)$
$\left(b\right)e^x(\sin x+\cos x)$
$\left(c\right)\frac{e^x}{2}(\sin x-\cos x)$
$\left(d\right)\frac{e^x}{2}(\sin x+\cos x)$

Answer 1

$I=\int e^x\cos \left(x\right)dx$
By using integration by parts
$\Rightarrow I=e^x\cos \left(x\right)+\int e^x\sin \left(x\right)dx$
$\Rightarrow I=e^x\cos \left(x\right)+e^x\sin \left(x\right)-\int e^x\cos \left(x\right)dx$
$\Rightarrow 2I=e^x\cos \left(x\right)+e^x\sin \left(x\right)$
$\Rightarrow I=\frac{e^x\cos \left(x\right)+e^x\sin \left(x\right)}{2}$
$\Rightarrow I=\frac{e^x}{2}(\sin x+\cos x)+c$
$F\left(x\right)=\frac{e^x}{2}(\sin x+\cos x)$

Option $\left(d\right)$ is correct.