Question

Question #30055

Answer 1

Partial integration twice and the integral appears on the right hand side. The integral can then be rearranged for. Knowledge of some trigonometric formulae can also simplify this easily.

`\int sin(theta)sin(5theta) "d"theta = 1/24 ( 3sin(4theta) - 2sin(6theta)) + C`.

Explanation:

Knowledge of the cyclic derivatives of sine and cosine and the partial integration technique gave me an indication that the integral may appear again on the right hand side.

The partial integration technique states that,

`int "f"'(theta)"g"(theta) "d"theta = "f"(theta)"g"(theta) - int "f"(theta)"g'"(theta) "d"theta`

Pick `"f"'(theta)=sin(theta)` and `"g"(theta)=sin(5theta)`.

Then,

`int sin(5theta)sin(theta) "d"theta = -cos(theta)sin(5theta) - int (-cos(theta))(5cos(5theta)) "d"theta`,
`int sin(5theta)sin(theta) "d"theta = -cos(theta)sin(5theta) + 5int cos(theta)cos(5theta) "d"theta`.

Then consider `int cos(theta)cos(5theta) "d"theta`. Pick `"f"'(theta)=cos(theta)` and `"g"(theta)=cos(5theta)`.

Then.

`int cos(theta)cos(5theta) "d"theta = sin(theta)cos(5theta) - int sin(theta)(-5sin(5theta))"d"theta`
`int cos(theta)cos(5theta) "d"theta = sin(theta)cos(5theta) + 5int sin(theta)sin(5theta)"d"theta`

Notice the integral we want to solve has appeared on the right hand side here. If we substitute this into our earlier expression,

`int sin(theta)sin(5theta) "d"theta = -cos(theta)sin(5theta) +5sin(theta)cos(5theta)+25\int sin(theta)sin(5theta) "d"theta`,

`24\int sin(theta)sin(5theta) "d"theta - cos(theta)sin(5theta) + 5sin(theta)cos(5theta)=0`.

Solving for the integral,

`\int sin(theta)sin(5theta) "d"theta = 1/24 (cos(theta)sin(5theta)-5sin(theta)cos(5theta))+C`.

This technique of looking for the integral to appear on the other side is pretty common, so even if this isn't the easiest way to solve this integral it's a good technique to be aware of. Try it yourself with `\int e^(theta)sin(theta) "d" theta` :)

.... just realised, as may be expected, knowledge of some trigonometric formulae can help a good deal!

`sin(a)sin(b) = 1/2 (cos(a-b)-cos(a+b))`

(This can be verified by applying the cosine addition formulae to the right hand side).

Then,

`sin(theta)sin(5theta) = 1/2 (cos(4theta)-cos(6theta))`,
`\int sin(theta)sin(5theta) "d"theta = 1/2( sin(4theta)/4 - sin(6theta)/6)`,
`\int sin(theta)sin(5theta) "d"theta = 1/24 ( 3sin(4theta) - 2sin(6theta)) + C`.