Integration by substitution manipulation error?

We've just discussed u-substitution in our class, and I was able to solve

\int { \sin (x)\cos ^{ 4 } (x) } dx

with

u=\cos ^{ 4 }(x)

, hence

\displaystyle dx=\frac{du}{-4\cos^3(x)\sin(x)}

(I know I could/should use

u=\cos(x)

instead):

\displaystyle \int { \sin (x)\cos ^{ 4 } (x)dx } =\int { \sin (x)u^{ \frac { 1 }{ 4 } }\cos ^{ 3 } (x)\frac { du }{ -4\cos ^{ 3 } (x)\sin (x) } } \\ =\frac { -1 }{ 4 } \int { u^{ \frac { 1 }{ 4 } }du } \\ =\frac { -1 }{ 4 } \frac { u^{ \frac { 5 }{ 4 } } }{ \frac { 5 }{ 4 } } +C\\ =\frac { -\cos ^{ 5 } (x) }{ 5 } +C

Though this gives me the right answer, on the second step:

\displaystyle \int { \sin (x)u^{ \frac { 1 }{ 4 } }\cos ^{ 3 } (x)\frac { du }{ -4\cos ^{ 3 } (x)\sin (x) } } \Rightarrow \cos ^{ 4 } (x)=|\cos (x)|\cos ^{ 3 } (x)

.
But at the end, when I substitute back

u=\cos^4(x)

, the inconsistency sort of cancels out:

(\cos ^{ 4 } (x))^{ \frac { 5 }{ 4 } }

.

Would you say my steps work? Halfway through the problem the equality does not hold up. Like I said,

u=\cos(x)

would have been the better choice for substitution.

Would you say my steps work? Halfway through the problem the equality does not hold up. Like I said,

u=\cos(x)

would have been the better choice for substitution.

Yes, that's fine. You simply need your substitutions to be piecewise injective.

Yes, that's fine. You simply need your substitutions to be piecewise injective.

Lol u waste

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