Calculus integration problem: $\int \sin^5 (x) \cos^2 (x)\,dx$ What's the integration of $$\int \sin^5 (x) \cos^2 (x)\,dx?$$
 A: Hint: Write 
$$
\sin^5(x)\cos^2(x)=(\sin^2(x))^2\cos^2(x)\sin(x).
$$
Now use $\cos^2(x)+\sin^2(x)=1$ and do the appropriate change of variable.
This is the general method to integrate functions of the type
$$
\cos^n(x)\sin^m(x)
$$
when one of the integers $n,m$ is odd.
A: $$ \int \sin^5 (x) \cos^2(x) dx $$
$$= \int(\sin^2(x))^2 \cos^2(x) \sin(x) dx$$ 
$$=-\int(1 - \cos^2(x))^2 cos^2(x) (-sin(x) dx) $$
Let $u = \cos(x)$ 
$\implies du = -\sin(x) dx$ 
$$= -\int(1 - u^2)² u² (du)$$ 
$$= -\int(1 - 2u^2 + u^4) u^2 du $$
$$= -\int(u^2 - 2u^4+ u^6) du$$ 
$$= -\left(\frac{u^3}{3} - \frac{2u^5}{5} + \frac{u^7}{7}\right) + C$$ 
$$= -u^3\left(\frac{1}{3} - \frac{2u^2}{5} +\frac{ u^4}{7}\right) + C $$
$$= -\cos^3(x) \left(\frac{1}{3} - \frac{2\cos^2(x)}{5} + \frac{\cos^4(x)}{7}\right) + C $$
$$= -\cos^3(x)\frac{15\cos^4(x) - 42\cos^2(x) + 35}{105} + C $$
A: Using trig identities, you can show that:
$$\sin ^5(x) \cos ^2(x)=\frac{5 \sin (x)}{64}+\frac{1}{64} \sin (3
   x)-\frac{3}{64} \sin (5 x)+\frac{1}{64} \sin (7 x)$$
To do this, first use the "Power-reduction formulas" to reduce to get:
$$\sin^5(x)=\frac{10 \sin x - 5 \sin 3 x+ \sin 5 x}{16}$$
$$\cos^2(x)=\frac{1 + \cos (2 x)}{2}$$
And then use:
$$\cos (2 x) \sin (nx) = {{\sin((n+2)x) - \sin((n-2)x)} \over 2}$$
