Question

$\int x{e}^x\cos xdx=\frac{e^x}{2}[(x-1)\sin x+x\cos x]$. If this is true enter 1, else enter 0.

Answer 1

$\cos x$ is R.P. of $e^{ix}$ i.e. of $\cos x+i\sin x\therefore I=R.P.$ of $\int x{e}^x{e}^{ix}dx=\int x{e}^{(1+i)x}dx$
Intergrating by parts
$I=x.\frac{e^{1+i}x}{(1+i)}-\frac{1}{(1+i)}\int e^{(1+i)x}.1dx$
$I=x.\frac{e^{1+i}x}{(1+i)}-\frac{1}{(1+i)}{e}^{(1+i)x}$
$=\frac{e^{(1+i)x}}{{(1+i)}^2}[x(1+i)-1]=\frac{e^x.e^{ix}}{2i}[x+xi-1]$
$=\frac{e^x[(x-1)+ix](\cos x+i\sin x)}{2i}$
Real part of above is imaginary part of N'
$I=\frac{e^x}{2}[(x-1)\sin x+x\cos x]$