Is $\frac{cos^4(x)}{4}-\frac{cos^2(x)}{2}+C$ a correct evalution of $\int sin^3(x)cos(x)\ dx$ The answer on Khan Academy states that the integral evaluates to $1 \over 4$$\sin^4(x)+C$
However, I performed a u-substitution that I cannot find a mistake in (maybe I am blind).
Here's the working:
$I=\int \sin^3(x)\cos(x)dx$
let $\ u=\cos(x)$
$du=-\sin(x)dx$
$dx= \frac{-1}{\sin(x)}\ du$
$I=\int \sin^3(x)u\cdot-\frac{1}{\sin(x)}dx$
$I=\int -\sin^2(x)\cdot u\ du$
$I=\int -(1-\cos^2(x))\ u\ du$
$I=\int (\cos^2(x)-1)\cdot u\ du$
$I=\int u(u^2-1)\ du$
$I=\int u^3-u\ du$
$I=\frac{u^4}{4}-\frac{u^2}{2}$
$I=\frac{\cos^4(x)}{4}-\frac{\cos^2(x)}{2}+C$

Both the answers are same but they differ by constants. If you want the same answer as stated then just simply substitute $u=\sin x$
Hence $du=\cos x$
Hence $$\int \sin^3 x\cos xdx=\int u^3 du=\frac {u^4}{4}+C=\frac {\sin^4 x}{4}+C$$
