# 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 **ILATE** method.

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.