Question

Please integrate 2xcos3x dx (Using integral by part)

Answer 1

Alright, so we need to pick a "u" and a "v" for our by parts. The "u" we choose is a part of the function thatwill be easy or convenient to differentiate and the "v" will be the part that is easy or convenient to integrate. In this case the choice is obvious, we will choose u to be 2x, since one derivative of it will make the x disappear. Thismeans ourv will be cos(3x). So the formula for by parts is:
\[u \int\limits_{}^{}v-(\int\limits_{}^{}[u'\int\limits_{}^{}v])\]
So u' will be 2 and integral v will be (1/3)sin3x. Now we just plug in these values:
\[ \frac{ 2x }{ 3 }\sin(3x)-\int\limits_{}^{}\frac{ 2 }{ 3 }\sin(3x)\]
So now I can just factor out the (2/3) constant and integrate the remaining sin(3x) to get:
\[\frac{ 2x }{ 3 }\sin(3x)-\frac{ 2 }{ 3 }(-\frac{ 1 }{ 3 }\cos(3x))\]
\[\frac{ 2x }{ 3 }\sin(3x)+\frac{ 2 }{ 9 }\cos(3x)+C\]

Answer 2

Wow, Thank you so much!

Answer 3

Np ^_^