Can someone help me with this integration by parts equation x^2sin(3x+1) Thank you.
You need to use integration by parts twice, applying once using Mathstud28 choice of u and v gives:
$\displaystyle I=\int x^2 \sin(3x+1) \ dx=-x^2 \left[ \frac{1}{3} \cos(3x+1)\right] + \int 2x \left[ \frac{1}{3} \cos(3x+1)\right] \ dx$
Use integration by parts again to do the integral on the right.
CB
Don't integrate - balloontegrate!
Straight lines integrate up, differentiate down (making it easier to check). The triangular networks satisfy the product rule - you may prefer to draw a zoomed-out version of these...
The challenge of integration by parts is simply to fill out a product rule shape starting with one of the bottom balloons instead of (as happens in differentiation) the top one.
See Balloon Calculus: worked examples from past papers
or http://www.mathhelpforum.com/math-he...tegration.html
to see the process.
Have you ever seen tabular integration. Here it is for fun.
Let $\displaystyle u=x^{2}, \;\ dv=v'dx=sin(3x+1)$
Take the derivatives of u until you get to 0. Take the antiderivatives of v'.
$\displaystyle \text{sign \;\ \;\ u and its derivatives \;\ \;\ \;\ v' and its antiderivatives }$
$\displaystyle \begin{array}{ccc}+& \;\ \;\ \;\ x^{2}\searrow& \;\ \;\ sin(3x+1)\\-&2x\searrow&\frac{-1}{3}cos(3x+1)\\-&2\searrow&\frac{-1}{9}sin(3x+1)\\-&0&\frac{1}{27}cos(3x+1)\end{array}$
Now, take the signed products and add up the diagonal entries:
$\displaystyle \frac{-1}{3}x^{2}cos(3x+1)+\frac{2}{9}xsin(3x+1)+\frac{2} {27}cos(3x+1)+C$