# Integrate$\int x \cos(x) \sin(x)\; dx$

Integrate: $$\int x \cos(x) \sin(x) \;dx$$

I've been trying to integrate by parts, but I can't! I know there's a trigonometric function about $$\cos (2x)$$ but I don't know how to integrate that function with $$\sin(x)$$.

• $\int \frac 12x \sin 2x dx$ as $\sin 2x=2\sin x\cos x$ – Mohammad Zuhair Khan Oct 25 '18 at 16:17
• Have you tried $u=x$, $dv=\sin x\cos xdx$? Note that $\sin x\cos x=\frac12\sin 2x$. – ajotatxe Oct 25 '18 at 16:18

Use integration by parts on $$\int x\cos (x)\sin(x)\ dx$$ with

$$u=x$$ and $$dv= \cos (x)\sin(x)\ dx$$

Then you will have $$du = dx$$ and $$v= (1/2)\sin ^2 x$$

You can finish the rest because you know how to integrate $$(1/2)\sin ^2 x$$

Given $$\int x\cos (x)\sin(x)\ dx$$

Use the identity $$\cos(x)\sin(x)=\dfrac{\sin(2x)}{2}$$

Now $$\int x\ \dfrac{\sin(2x)}{2}\ dx=\dfrac12\int x\sin(2x)\ dx$$

Now use Integration By Parts $$u=x$$ and $$v^{\prime}=\sin(2x)$$

Can you continue from here?

$$\int x \cos(x) \sin(x) dx = \int x \frac{1}{2} \sin(2x) dx = \frac{1}{2}\int x \sin(2x)dx \\= \frac{1}{2}x[-\frac{1}{2}\cos(2x)]-\frac{1}{2}\int [-\frac{1}{2}\cos(2x)]dx + C \\= -\frac{1}{4}x\cos(2x)+\frac{1}{8}\int \cos(2x)d(2x)+C\\ = -\frac{1}{4}x\cos(2x)+\frac{1}{8}\sin(2x)+C$$

• $\LaTeX \text{ Tip:}$ Use \sin x and \cos x to obtain $\sin x$ and $\cos x$ – Mohammad Zuhair Khan Oct 25 '18 at 16:31
• @Raptor Thanks for the tip. It has been fixed. – Banghua Zhao Oct 25 '18 at 16:48

There are several ways. Let us try by parts on $$\cos x$$:

$$I:=\int x\cos x\sin x\,dx=x\sin^2x-\int\sin x(x\cos x+\sin x)\,dx =x\sin^2x-I-\int\sin^2x\,dx.$$

Now,

$$J:=\int\sin^2x\,dx=-\cos x\sin x+\int\cos^2x\,dx =-\cos x\sin x-J+\int dx.$$

Grouping the results,

$$I=\frac{2x\sin^2x+\cos x\sin x-x}4.$$

$$I=\int x\cos(x)\sin(x)dx$$ Integration by parts time! $$dv=\cos(x)\sin(x)dx$$ $$v=\frac1{2}\sin^2(x)$$ and $$u=x\\du=dx$$ Giving $$I=\int udv=uv-\int vdu$$ $$I=\frac{x\sin^2(x)}{2}-\int\frac1{2}\sin^2(x)dx$$ Now we solve $$A=\int\frac1{2}\sin^2(x)dx$$ because $$I=\frac{x}{2}\sin^2(x)-A$$.

$$A=\int\frac1{2}\sin^2(x)dx$$ $$A=\frac1{2}\int\sin^2(x)dx$$ The sine reduction formula for positive whole powers $$n$$ gives $$\int\sin^n(x)dx=\frac{-\cos(x)\sin^{n-1}(x)}{n}+\frac{n-1}{n}\int\sin^{n-1}(x)dx$$ Applying it ($$n=2$$) gives $$A=\frac1{2}\int\sin^2(x)dx=\frac{1}{2}\bigg(\frac{-\cos(x)\sin^{2-1}(x)}{2}+\frac{2-1}{2}\int\sin^{2-1}(x)dx\bigg)$$ $$A=\frac{1}{2}\bigg(\frac{-\cos(x)\sin(x)}{2}+\frac{1}{2}\int\sin(x)dx\bigg)$$ And from $$\int\sin xdx=-\cos x$$ we have $$A=\frac{-\cos(x)\sin(x)}{4}-\frac{1}{4}\cos(x)$$ $$A=\frac{-1}{4}(\cos(x)\sin(x)+\cos(x))$$ $$A=\frac{-\cos(x)}{4}(\sin(x)+1)$$ And from $$I=\frac{x}{2}\sin^2(x)-A$$, we have $$I=\frac{x}{2}\sin^2(x)+\frac{\cos(x)}{4}(\sin(x)+1)$$ And since we're done integrating, we add the constant of integration $$I=\frac{x}{2}\sin^2(x)+\frac{\cos(x)}{4}(\sin(x)+1)+C$$

For future reference, this site is great for these sorts of integrals: it gives you detailed step-by-step solutions.