Integral of sin^2(x) using the Half Angle Formula

Later in this tutorial, we use an important trigonometric identity to help us solve the integral of the function sin²(3x). Your first thought might be to approach this problem with integration by parts, by writing...

∫sin²(3x)dx = ∫sin(3x)sin(3x)dx

Proceeding with I.B.P., we can let u = sin(3x) and dv = sin(3x)dx, and thus du = 3cos(3x)dx and v = -cos(3x)/3. So...

∫sin²(3x)dx = sin(3x)cos(3x)/3 - ∫-cos(3x)∙3cos(3x)/3dx = sin(3x)cos(3x)/3 + ∫cos²(3x)dx

Now, if we proceed in the same fashion with the integral of cos²(3x), we will end up with...

∫sin²(3x)dx = ∫sin²(3x)dx

Well, while this result is true, it not very useful. It is a trivial solution.

However, if we are smart enough to realise that cos²(3x) = 1 - sin²(3x), then the result becomes...

∫sin²(3x)dx = sin(3x)cos(3x)/3 + ∫1 - sin²(3x)dx = sin(3x)cos(3x)/3 + ∫dx - ∫sin²(3x)dx

So now we have an integral of sin²(3x) on the right hand side that we can add to the left hand side. Thus we have...

2∫sin²(3x)dx = sin(3x)cos(3x)/3 + ∫dx = sin(3x)cos(3x)/3 + x + C

Now dividing both sides by 2...

∫sin²(3x)dx = sin(3x)cos(3x)/6 + x/2 + D (where D = C/2)

Finally, using the sum difference formula, we can use the product-sum identity to abbreviate sin(3x)cos(3x) to...

sin(3x)cos(3x) = sin(6x)/2

Therefore, the final solution is....

∫sin²(3x)dx = x/2 + sin(6x)/12 + D

So the above was one method of finding the solution. As you can see, it involved quite a few steps, so is there a shortcut? Yes, there is... and the trick lies in realizing that

sin²θ = [1 - cos(2θ)]/2

This is what's called a power reducing or a half angle formula. So now our integral can be rewritten as...

∫sin²(3x)dx = ∫[1 - cos(2∙3x)]/2 dx = (1/2)∫1 - cos(6x)dx

Which evaluates to...

∫sin²(3x)dx = x/2 + sin(6x)/12 + C