TITLE: The antiderivative of $\cos^5(x)\sin^5(x)$ - is this incorrect?
QUESTION [7 upvotes]: I like to check my answers with wolframalpha, and this one's stubbornly coming up as false when set equal to its answer for the antiderivative, but I can't figure out where I'm going wrong.
Using the identity $\cos^2x = (1-\sin^2x)$, I rewrote the integral as:
$$\int \cos(x)(1-\sin^2x)^2(\sin^5x)\,dx$$
$$u = \sin x\\
du = \cos x\,dx\\
dx = du/\cos x$$
Canceling out the stray cosine, that turns into $\int(1-u^2)^2u^5 \, du$
Expanding, it's $\int (1-2u^2+u^4)u^5\,du = \int (u^5-2u^7+u^9)\,du$
$$= \frac{\sin^6 x}{6}-\frac{\sin^8 x}{4}+\frac{\sin^{10} x}{10}+C$$
Basically, does anybody see an error here? Is there an error?

REPLY [3 votes]: Another approach: $\frac{\sin 2x}2 =\sin x \cos x$ yielding:
$$\frac{1}{2^5}\int \sin^5 2x\,dx$$
Letting $u=2x$ then $du=2dx$ and we get:
$$\frac{1}{2^6}\int \sin^5 u\,du$$
Writing $\sin^5 u = (1-\cos^2 u)^2 \sin u$, and letting $v=\cos u$, you get:
$$-\frac{1}{2^6}\int (1-v^2)^2 \, dv$$
This yields:
$$-\frac{1}{2^6}\left(v-\frac{2}{3}v^3+\frac{1}{5}v^5+C\right)$$
Substitute back in $v=\cos 2x$.