If $∫xe^x\,cosx\,dx=f(x)+c$, then f(x) is equal to |
$\frac{e^x}{2}\{(1-x)sinx-x\,cosx\}$ $\frac{e^x}{2}\{(x-1)sinx+x\,cosx\}$ $\frac{e^x}{2}\{(1+x)sinx-x\,cosx\}$ none of these |
$\frac{e^x}{2}\{(x-1)sinx+x\,cosx\}$ |
Here I = Real part of $∫xe^{(1+i)x}dx$ and $∫xe^{(1+i)x}dx=\frac{xe^{(1+i)x}}{(1+i)}-\frac{e^{(1+i)x}}{(1+i)}dx=\frac{xe^{(1+i)x}}{(1+i)}-\frac{e^{(1+i)x}}{(1+i)^2}$ $=e^{(1+i)x}\begin{Bmatrix}\frac{x(1+i)-1}{(1+i)^2}\end{Bmatrix}$ = ex [cos x + i sin x] $[\frac{(x-1)ix}{1+2i-1}]=\frac{e^x}{-2}\{cosx-sinx\}\{(x-1)+ix\}$ Hence $I =\frac{e^x}{-2}$ [(1 - x) sin x – x cos x] =$\frac{e^x}{2}$ [x cos x + (x – 1) sin x] . Hence (B) is the correct answer. |