Why do I get two different indefinite integrals by using two distinct methods of integration?

Question

This is the integral:

METHOD 1:

$\int {\sin (x){{\cos }^3}(x)dx} = \int {{{\cos }^2}(x)\sin (x)\cos (x)dx = \int {(1 - \sin^2(x))\sin (x)\cos (x)dx} } $

Let $u = \sin (x)$, and $du = \cos(x) dx$. Continuing where we left:

$ = \int {(1 - {u^2})u\,du} = \int {u - {u^3}du = \frac{{{u^2}}}{2}} - \frac{{{u^4}}}{4} = \frac{{2{{\sin }^2}(x) - {{\sin }^4}(x)}}{4}$

METHOD 2:

Let $u = \cos (x)$, and $du = -\sin(x) dx$.

$ \int {\sin (x){{\cos }^3}(x)dx} = - \int {{u^3}} du = - \frac{{{u^4}}}{4} = - \frac{1}{4}{\cos ^4}(x) $

I added a constant to the primitive obtained by the second method so that both primitives "touched" each other on $\pi n$ for every integer $n$ but they are not the same function at all.

Answer 1

These functions actually differ by a constant, as we have the identity $$\sin^4(x)-\cos^4(x)=2\sin^2(x)-1$$

So, reminder, never forget your $+C$!

Answer 2

They are actually the same, within a constant... which is why you need to remember the $+C$ at the end. In reality, $$\frac{2\sin^2x-\sin^4 x}{4}=\frac{1}{4}-\frac{1}{4}\cos^4 x$$ This is why you need to take into account the constant!