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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 $\sin(x)$.

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    $\begingroup$ $\int \frac 12x \sin 2x dx$ as $\sin 2x=2\sin x\cos x$ $\endgroup$ – Mohammad Zuhair Khan Oct 25 '18 at 16:17
  • $\begingroup$ Have you tried $u=x$, $dv=\sin x\cos xdx$? Note that $\sin x\cos x=\frac12\sin 2x$. $\endgroup$ – ajotatxe Oct 25 '18 at 16:18
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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 $(1/2)\sin ^2 x$

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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?

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

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

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

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$$I=\int x\cos(x)\sin(x)dx$$ 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.

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