I wanted to integrate $\int\frac{\sin^4(x)}{\cos(x)}dx$, using “Mathway” and it yielded $\frac{1}{5}\sin^5(x)+C$ by substituting $u=\sin(x)$, My problem is when I do it
$u = \sin(x)dx$
$du = \cos(x)dx$
$\frac{du}{\cos(x)} = dx$
I get,
$\int\frac{u^4}{\cos(x)*\cos(x)}du=\int\frac{u^4}{\cos^2(x)}du$
Instead of
$\int{u^4}du$ which was Mathway's output, Did I do something wrong?
I don't know what "Mathway" is, but whatever it's doing, it's hallucinating, because you are absolutely correct. If $u = \sin x$, then $du = \cos x \, dx$, which does not work: $$\int \frac{\sin^4 x}{\cos x} \, dx = \int \frac{u^4}{\cos^2 x} \, du.$$ The $\cos x$ in the denominator does not cancel out.
Another way to see that the claimed solution cannot be correct is to simply differentiate the answer:
$$\frac{d}{dx}\left[\frac{1}{5} \sin^5 x \right] = \sin^4 x \cos x,$$
and this is not equivalent to the given integrand.
A correct method of solution would be to write $$\frac{\sin^4 x}{\cos x} = \frac{\sin^4 x \cos x}{\cos^2 x} = \frac{\sin^4 x \cos x}{1 - \sin^2 x},$$ and now let $$u = \sin x, \quad du = \cos x \, dx,$$ to obtain $$I = \int \frac{\sin^4 x}{\cos x} \, dx = \int \frac{u^4}{1-u^2} \, du.$$ Now perform long division and partial fraction decomposition: $$\frac{u^4}{1-u^2} = -u^2 - 1 + \frac{1}{1-u^2} = -u^2 - 1 + \frac{1}{2(1-u)} + \frac{1}{2(1+u)},$$ so $$\begin{align} I &= \int -u^2 - 1 \, du + \frac{1}{2}\int \frac{1}{1-u} + \frac{1}{1+u} \, du \\ &= -\frac{u^3}{3} - u + \frac{1}{2} \left( \log|1+u| - \log|1-u| \right) + C \\ &= -\frac{\sin^3 x}{3} - \sin x + \frac{1}{2} \log \left|\frac{1+\sin x}{1-\sin x}\right| + C. \end{align}$$ Differentiation of this expression will confirm its derivative is equal to the given integrand.
This seems to be fully wrong
$$I=\int\frac{\sin^4(x)}{\cos(x)}dx$$
$$u=\sin(x) \implies x=\sin^{-1}(u) \implies dx=\frac{du}{\sqrt{1-u^2}}$$
$$I=\int \frac{u^4}{1-u^2}\,du=\int\left(-u^2-\frac{1}{2 (u-1)}+\frac{1}{2 (u+1)}-1\right)\,du$$
