I'm trying to calculate $$\int cos^5(x)dx$$ with the reduction formula.
$$\int cos^5(x)dx=\frac{1}{5}cos^4(x)sin(x)+\frac{4}{5}\int cos^3(x)dx$$
then
$$\int cos^3(x)dx=\frac{1}{3}cos^2(x)sin(x)+\frac{2}{3}\int cos(x)dx$$
so
$$\int cos^5(x)dx=\frac{1}{5}cos^4(x)sin(x)+\frac{4}{5}(\frac{1}{3}cos^2(x)sin(x)+\frac{2}{3}sin(x))$$
But this looks different from:
$$\frac{1}{5}sin^5(x)-\frac{2}{3}sin^3(x)+sin(x)$$
given by:
Best Answer
Converting the cosines in your answer to sines via the pythagorean identity, $\cos ^2x=1-\sin ^2x$ yields:
$\frac{1}{5}\cos^4(x)\sin(x)+\frac{4}{5}(\frac{1}{3}\cos^2(x)\sin(x)+\frac{2}{3}\sin(x))$ $=\frac{1}{5}(1-\sin^2(x))^2\sin(x)+\frac{4}{5}(\frac{1}{3}(1-\sin^2(x))\sin(x)+\frac{2}{3}\sin(x))$
Simplifying from there:
$=\frac{1}{5}(1-2\sin^2(x)+\sin^4(x))\sin(x)+\frac{4}{5}(\frac{1}{3}\sin(x)-\frac{1}{3}\sin^3(x))+\frac{2}{3}\sin(x))$
$=\frac{1}{5}\sin(x)-\frac{2}{5}\sin^3(x)+\frac{1}{5}\sin^5(x)+\frac{4}{15}\sin(x)-\frac{4}{15}\sin^3(x))+\frac{8}{15}\sin(x)$
$=\frac{1}{5}\sin^5(x)-\frac{2}{3}\sin^3(x)+\sin(x)$
Which agrees with the answer from the solver.