Integrate sin^33x
To integrate sin^33x, also written as ∫sin33x dx, sin cubed 3x, sin^3(3x), and (sin 3x)^3, we start by using standard trig identities to simplify the integral.
We can change the expression to another one that is identical, but now has a squared term.
We recall the Pythagorean trig identity and multiply the angles throughout by 3. We then rearrange it for the sin squared term so that we can substitute for it in our new integration expression.
We substitute the rearranged trig identity.
We let u = cos3x.
Then, du/dx = -3sin3x. If you wish to know how to differentiate this, I have shown the steps in the aside section at the bottom of this page.
We rearrange the previous expression for du in terms of dx. As you can see, the sin3x dx terms look like the terms outside the brackets in our integration problem. The only issue we have is that there is a -3 constant that will prevent us from making a straight substitution. To solve this issue, we play a trick as shown below.
If we multiply -⅓ with -1 then the result is +1 and we are not changing anything, however; we now have -3sin3x dx that we can replace by du.
We also replace cos3x with u so that the integration problem is now in terms of u alone.
On the LHS, we multiply out the constant resulting in a change in signs. On the RHS we integrate each term separately, which means the same thing as the LHS.
Hence, this is the final integration, and all we now need to do is to replace u with cos3x.
This is the final solution.
Differentiate cos3x
