Question

$\int \left(e^{\log x}+\sin x\right)\cos xdx$

Answer 1

$$\begin{gathered} \int \left(e^{\log x}+\sin x\right)\cos xdx \\ =\int \left(x+\sin x\right)\cos xdx\left(\because e^{\log x}=x\right) \\ =\int \left(x\cos x+\sin x\cos x\right)dx \\ =\int x\cos xdx+\frac{1}{2}\int 2\sin x\cos xdx \\ =\int \underset{\mathrm{I}}{x}\underset{\mathrm{II}}{\cos x}dx+\frac{1}{2}\int \sin 2xdx \\ =\left[x\int \cos xdx-\int \left\{\frac{d}{dx}\left(x\right)\int \cos xdx\right\}dx\right]+\frac{1}{2}\int \sin 2xdx \\ =x\sin x-\int 1.\sin xdx+\frac{1}{2}\left[\frac{-\cos 2x}{2}\right]+C \\ =x\sin x-\left(-\cos x\right)-\frac{1}{4}\cos 2x+C \\ =x\sin x+\cos x-\frac{1}{4}\left(1-2{\sin }^{2}x\right)+C \\ =x\sin x+\cos x+\frac{{\sin }^{2}x}{2}-\frac{1}{4}+C \\ =x\sin x+\cos x+\frac{{\sin }^{2}x}{2}+C\text{'}\mathrm{where}C\text{'}=C-\frac{1}{4} \end{gathered}$$