Question

Evaluate
$\int \cos x.e^x.x^{2}dx$.

Answer 1

$\int \cos x.e^x.x^{2}dx$

Let $u=x^2\Rightarrow du=2xdx$ and $dv=e^x\cos x\Rightarrow v=\int e^x\cos xdx$

Using the above we have,

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

$=x^2\int e^x\cos xdx-e^x\cos x2\times \frac{x^2}{2}+c$

$=x^2\int e^x\cos xdx-e^x{x}^2\cos x+c$ where $c$ is the constant of integration. ..........$\left(1\right)$

Let $I=\int e^x\cos xdx$

Let $u=e^x\Rightarrow du=e^{x}dx$ and $dv=\cos xdx\Rightarrow v=\sin x$

$\Rightarrow I=e^x\sin x-\int e^x\sin xdx$ .......$\left(2\right)$

Consider $\int e^x\sin xdx$

Let $u=e^x\Rightarrow du=e^{x}dx$ and $dv=\sin xdx\Rightarrow v=-\cos x$

$\Rightarrow I=e^x\sin x-[-e^x\cos x-\int -e^x\cos xdx]$

$\Rightarrow I=e^x\sin x+e^x\cos x-\int e^x\cos xdx$

$\Rightarrow I=e^x(\sin x+\cos x)-I$ where $I=\int e^x\cos xdx$

$\Rightarrow 2I=e^x(\sin x+\cos x)$

$\Rightarrow I=\frac{e^x(\sin x+\cos x)}{2}$ ....$\left(3\right)$

Substituting $\left(3\right)$ in $\left(1\right)$ we get

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

$=x^{2}I-e^x{x}^2\cos x+c$

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

$=\frac{e^x{x}^2}{2}\sin x+e^x{x}^2\cos x(\frac{1}{2}-1)+c$

$=\frac{e^x{x}^2}{2}\sin x-\frac{e^x{x}^2}{2}\cos x+c$

$=\frac{e^x{x}^2}{2}(\sin x-\cos x)+c$ where $c$ is the constant of integration

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