How to use this applet:

Notes on the maths used in the applet:

A question of this type usually occurs annually on the Higher Level Paper. There are generally 2 different variations of this question. Variation 1 occurs when one of the values is a function of x (e.g. ln(x) or tan(x) or e^x) and the other is simply Cx, with no multiplying power, where C is a constant. i.e. you would see a question similar to this: “Find ̣x.ln(x).dx or ̣3x.Sin(x).dx.” Variation 2 occurs when both the values are functions of x. Such a type would take the form “Find ̣e^5x.Sin(3x).dx.”
     The second type is certainly the more complicated and so will be explained by example. The first type is relatively self explanatory and most possible questions can be determined using the integrator below.
      The second variation is a special case of integration by parts, as already explained. In the Leaving Cert., once you have integrated by parts as normal you should eventually find a function the same as the original question itself. One example should be sufficient as the method for any question is practically identical.
     
Example for the special case:

Find ̣e^3x.Sin(2x).dx.

let u = e^3x and dv = 3e^3x.dx.

du/dx = 3e^3x du = 3e^3x.dx

dv = Sin(2x).dx and v = -(1/2).Cos(2x).

By using the formula ̣u.dv = uv - ̣v.du, we arrive at:

e^3x.-(1/2).Cos(2x) - ̣-(1/2).Cos(2x).3e3x.dx.

Multiplying constants may be brought outside the integral. Thus:

̣e^3x.Sin(2x).dx = e^3x.-(1/2).Cos(2x) + (3/2) ̣e3x.Cos(2x).dx.

we now must find ̣e^3x.Cos(2x).dx.

let u = e^3x and dv = Cos(2x).dx.

du/dx = 3e^3x du = 3e^3x.dx

v = (1/2).Sin(2x).

using the formula again:

̣e^3x.Cos(2x).dx. = e^3x.(1/2)Sin(2x) – (3/2) ̣Sin(2x).e^3x.dx (Equation 2).

NB Now note that the right hand side of the term is actually the question we began with, i.e. ̣Sin(2x).e^3x.dx.

We now go back to equation 1 above and substitute equation 2 into it (both equation 1 and 2 are underlined).

̣e^3x.Sin(2x).dx = e^3x.-(1/2).Cos(2x) + (3/2) ̣e^3x.Cos(2x).dx. is now equal to:

̣e^3x.Sin(2x).dx = e^3x.-(1/2).Cos(2x) + (3/2)( e^3x.(1/2)Sin(2x) – (3/2) ̣Sin(2x).e^3x.dx)

̣e^3x.Sin(2x).dx = e^3x.-(1/2).Cos(2x) + (3/4) ( e^3x. Sin(2x)) – (9/4) (̣Sin(2x).e^3x.dx)

We now have ̣Sin(2x).e^3x.dx on both sides…

Bringing – (9/4) (̣Sin(2x).e^3x.dx) to the right hand side gives:

(13/4) ̣e^3x.Sin(2x).dx = e^3x.-(1/2).Cos(2x) + (3/4) ( e^3x. Sin(2x))

= ̣e3x.Sin(2x).dx = (4/13)(e^3x.-(1/2).Cos(2x) + (3/4) ( e^3x. Sin(2x))) -(2/13)(e^3x.Cos(2x)) + (3/13)(e^3x.Sin(2x)

which gives:

(e^3x/13) (-2Cos(2x) + 3Sin(2x) + C

Other useful information: