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 $\sin(x)$.
$\endgroup$ 25 Answers
$\begingroup$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?
$\endgroup$ $\begingroup$$\int x \cos(x) \sin(x) dx = \int x \frac{1}{2} \sin(2x) dx = \frac{1}{2}\int x \sin(2x)dx \\= \frac{1}{2}x[-\frac{1}{2}\cos(2x)]-\frac{1}{2}\int [-\frac{1}{2}\cos(2x)]dx + C \\= -\frac{1}{4}x\cos(2x)+\frac{1}{8}\int \cos(2x)d(2x)+C\\ = -\frac{1}{4}x\cos(2x)+\frac{1}{8}\sin(2x)+C$
$\endgroup$ 2 $\begingroup$Use integration by parts on $$ \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 $$
You can finish the rest because you know how to integrate $(1/2)\sin ^2 x$
$\endgroup$ $\begingroup$There are several ways. Let us try by parts on $\cos x$:
$$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.$$
$\endgroup$ $\begingroup$$$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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