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.