How do I find the integral int(x^2*sin(pix))dx ?
1 Answer
Using Integration by parts,
intx^2sinpixdx
=
(-1/pi)x^2cospix + ((2)/pi^2)xsinpix + (2/pi^3)cospix + C
Remember that Integration by parts uses the formula:
intu dv =uv - intv du
Which is based off of the product rule for derivatives:
uv = vdu + udv
To use this formula, we must decide which term will be
Inverse Trig
Logarithms
Algebra
Trig
Exponentials
This gives you an order of priority of which term is used for "
We now have:
u = x^2 ,dv = sinpix
The next items we need in the formula are "
The derivative is obtained using the power rule:
d/dxx^2 = 2x = du
For the integral, we can use substitution.
using
We now have:
du = 2x dx ,v = (-1/pi)cospix
Plugging into our original Integration by Parts formula, we have:
intu dv =uv - intv du
=
intx^2sinpixdx = (-1/pi)x^2cospix - (-1/pi)int2xcospixdx
We are now left with another integral which we must use Integration by Parts once more to resolve. By pulling the
intxcospixdx = (1/pi)xsinpix - (1/pi)intsinpixdx
This last integral we can solve with a final round of substitution, giving us:
(1/pi)intsinpixdx = (-1/pi^2)cospix
Placing everything we've found together, we now have:
(-1/pi)x^2cospix - (-2/pi)[(1/pi)xsinpix - (-1/pi^2)cospix]
Now we can simplify the negatives and parenthesis to get our final answer:
intx^2sinpixdx =
(-1/pi)x^2cospix + ((2)/pi^2)xsinpix + (2/pi^3)cospix + C
The key is to remember that you will end up with a chain of multiple terms being added or subtracted together. You are continuously splitting the integral into smaller, manageable parts that you must keep track of for the final answer.