Answer
$\dfrac{\sin^6 x}{6}+\dfrac{\sin^{10} x}{10}-\dfrac{\sin^8 x}{4}+c$
Work Step by Step
Consider the integral $\int \cos^5 x \sin^5 x dx$
The given integral can be re-written as:
$\int \cos^4 x \sin^5 xdx (\cos x dx)=\int (1-sin^2 x)^2 \sin^5 x (\cos x dx)$
Plug in $a= \sin x \implies da= \cos x dx$
This implies that $\int (1-a^2)^2 a^5 da =\int (a^5+a^9-2a^7) da$
We use the formula: $\int x^n dx=\int\dfrac{x^{n+1}}{n+1} dx$
$\dfrac{a^6}{6}+\dfrac{a^{10}}{10}-\dfrac{a^{8}}{4}+c=\dfrac{\sin^6 x}{6}+\dfrac{\sin^{10} x}{10}-\dfrac{\sin^8 x}{4}+c$
