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    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.
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    (Original post by MartyO)
    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.
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    (Original post by Zacken)
    Yes, that's fine. You simply need your substitutions to be piecewise injective.
    Lol u waste
 
 
 
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Updated: February 10, 2017
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