Answer
$$\int x\cos^3xdx=\frac{3}{4}(x\sin x+\cos x)+\frac{1}{12}x\sin3x+\frac{1}{36}\cos3x+C$$
Work Step by Step
$$A=\int x\cos^3xdx=\int x\Big(\frac{3\cos x+\cos3x}{4}\Big)dx$$ $$A=\frac{3}{4}\int x\cos xdx+\frac{1}{4}\int x\cos3xdx$$ $$A=X+Y$$
1) For $X$:
Set $u= x$ and $dv=\cos xdx$
We would have $du=dx$ and $v=\sin x$
Applying Integration by Parts, we have
$$A=\frac{3}{4}(x\sin x-\int\sin xdx)$$ $$A=\frac{3}{4}(x\sin x+\cos x)+C$$
2) For $Y$:
Set $u= x$ and $dv=\cos3xdx$
We would have $du=dx$ and $v=\frac{1}{3}\sin3x$
Applying Integration by Parts, we have
$$A=\frac{1}{4}\Big(\frac{1}{3}x\sin3x-\frac{1}{3}\int\sin3xdx\Big)$$ $$A=\frac{1}{4}\Big(\frac{1}{3}x\sin3x+\frac{1}{9}\cos3x\Big)+C$$ $$A=\frac{1}{12}x\sin3x+\frac{1}{36}\cos3x+C$$
Therefore, $$A=\frac{3}{4}(x\sin x+\cos x)+\frac{1}{12}x\sin3x+\frac{1}{36}\cos3x+C$$
