My Math Forum Integration by parts

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March 14th, 2014, 06:44 AM   #1
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Integration by parts

i don't know how to solve the red part
Thanks!
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 March 14th, 2014, 07:04 AM #2 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Integration by parts Your problem is to integrate $\int x sin(3x) dx$. What you give is NOT "integration by parts". You seem to have substituted u for sin(3x) but then left the other "x" without substituting. You can't do that- to integrate you have to have the integral in terms of a single variable. Integration by parts: IF you let u= sin(3x), dv= x dx then du= 3cos(x)dx and $v= x^2/2$ so $uv- \int v du= \frac{1}{2}x^2 cos(3x)- \frac{3}{2}\int x^2 cos(x) dx$ BUT I can see no good reason to do that, since that last integral is harder than the first! Instead, let u= x, dv= sin(3x)dx so that du= dx, v= -(1/3) cos(3x) and the integral becomes $uv- \int v du= -\frac{1}{3}x cos(3x)+ \frac{1}{3}\int cos(3x) dx$
 March 14th, 2014, 06:21 PM #3 Senior Member   Joined: Mar 2014 Posts: 112 Thanks: 8 I think the procedure by Zynoakib is not necessarily correct. du = 3 cos(3x) dx and dv = x dx i.e. v = x²/2. I got x²sin(3x)/2 - ? x²/2 3 cos(3x) dx = x²sin(3x)/2 - 3/2 ? x² cos(3x) dx. HallsofIvy correctly pointed out the correct procedure for the integration by parts. $\int \cos{(3x)}\,dx\,=\,\frac{1}{3}\int 3\cos{(3x)}\,dx\,=\,\frac{1}{3}\sin{(3x)}\,+\,C\te xt{.}$

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