Is $\frac{cos^4(x)}{4}-\frac{cos^2(x)}{2}+C$ a correct evalution of $\int sin^3(x)cos(x)\ dx$

Question

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 $\ u=\cos(x)$

$du=-\sin(x)dx$

$dx= \frac{-1}{\sin(x)}\ du$

$I=\int \sin^3(x)u\cdot-\frac{1}{\sin(x)}dx$

$I=\int -\sin^2(x)\cdot u\ du$

$I=\int -(1-\cos^2(x))\ u\ du$

$I=\int (\cos^2(x)-1)\cdot u\ du$

$I=\int u(u^2-1)\ du$

$I=\int u^3-u\ du$

$I=\frac{u^4}{4}-\frac{u^2}{2}$

$I=\frac{\cos^4(x)}{4}-\frac{\cos^2(x)}{2}+C$

Answer 1

$$\frac{\sin^4 x}4=\frac{(1-\cos^2x)^2}4=\frac{\cos^4x-2\cos^2x+1}4 =\frac{\cos^4x}4-\frac{\cos^2x}2+\frac14$$ so both these answers differ by a constant, and so if one is a valid indefinite integral, then so is the other.

Answer 2

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 $$\int \sin^3 x\cos xdx=\int u^3 du=\frac {u^4}{4}+C=\frac {\sin^4 x}{4}+C$$

Answer 3

Your solution is correct (with minor error at the end) and it is equivalent to the one given in Khan Academy: $$\begin{align}I&=\frac{\cos^4(x)}{4}\overbrace{\require{cancel}\cancel{+}}^{-}\frac{\cos^2(x)}{2}+C=I= \\ &=\frac{(1-\sin^2 x)^2}{4}-\frac{\cos^2(x)}{2}+C=\\ &=\frac14-\frac{\sin^2 x}{2}+\frac{\sin^4 x}{4}-\frac{\cos^2 x}{2}+C= \\ &=\frac{\sin^4 x}{4}+C+\frac14-\frac12=\frac{\sin^4 x}{4}+A.\end{align}$$