How do I find the integral int(x^2*sin(pix))dx ?

1 Answer
Jul 29, 2014

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 u, and which will be dv. A useful way to figure out which term goes where is the ILATE method.

Inverse Trig
Logarithms
Algebra
Trig
Exponentials

This gives you an order of priority of which term is used for "u", so whatever is left over becomes our dv. Our function contains an x^2 and a sinpix, so the ILATE method tells us that x^2 should be used as our u, since it is algebraic and higher on the list than sinpix, which is trig.

We now have:

u = x^2, dv = sinpix

The next items we need in the formula are "du" and "v", which we get by finding the derivative of "u" and the integral of "dv".

The derivative is obtained using the power rule:

d/dxx^2 = 2x = du

For the integral, we can use substitution.

using w = pix, we end up with (-1/pi)cosw

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 2 out of the integral, we are left with u = x, dv = cospix. By going through the same process of before, we get:

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.