I have no issues with regular integrals, but when trig integral's like the one below pop up, I can't seem to grasp them. I've never taken a trig class before so maybe that is part of the reason. Would anyone be kind enough to explain the "easiest" way to deal with trig functions with powers in beginner laymen's terms? Is it possible to rewrite them in a form completely without the powers so I can solve with a u-sub or by parts? Any tips or tricks to dealing with integrals like these in an efficient manner? Thank you in advance.
Problem in question:
$\int \cos^3(x)\sin^4(x) dx$
Best Answer
Using the trig identity $\sin^2x + \cos^2x = 1$,we get:
$$\int \cos^3(x) \sin^4(x) dx$$ $$= \int \cos(x) \sin^4(x) (1 - sin^2(x)) dx$$
Using substitution $u = \sin(x) \Rightarrow \frac{du}{dx} = \cos(x) \Rightarrow dx = \frac{du}{\cos(x)}$, the integral is eqivalent with: $$\int u^4 (1 - u^2) du$$ $$= \int u^4 - u^6 du$$
Apply the power rule, we get:
$$\int u^4 - u^6 du = \frac{u^5}{5} - \frac{u^7}{7} + C$$
Undo substitution and get the final answer:
$$\int \cos^3(x) \sin^4(x) = \frac{\sin^5(x)}{5} - \frac{\sin^7(x)}{7} + C$$