QueenFisher said:
integrate with respect to x: (sinx)^3 * cosx
i have no idea where to start, can anyone help me? I've looked at differentials of other trig functions but i can't see any that would help
If your sine function is raised to an
odd power, it's commonly to let u = cos x, and work from there.
If your cosine function is raised to an
odd power, it's commonly to let u = sin x, and work from there.
If both are raised to an
odd power, then you can either let u = sin x, or u = cos x.
Note that, you should sometimes need to use the
Pythagorean identity: sin
2x + cos
2x = 1, to solve your problem.
I'll give you an example:
-----------
Example:
\int \cos x \sin ^ 2 x dx
cos x is raised to the power 1, hence it's an odd power, let u =
sin x.
u = sin x => du = cos x dx, right? Substitute that into your integral, we have:
\int u ^ 2 du = \frac{u ^ 3}{3} + C
Change u back to x, gives:
\int \cos x \sin ^ 2 x dx = \frac{\sin ^ 3 x}{3} + C
Can you go from here? :)