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anonymous

  • 5 years ago

Evaluate the integral with the aid of an appropriate u-sub: Integral sign { cos^19 (x)}^1/2 sinx dx

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  1. anonymous
    • 5 years ago
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    tailor made for u-sub. put \[u=cos(x)\] \[du=-sin(x)dx\] \[-\int u^{\frac{19}{2}}du\]

  2. anonymous
    • 5 years ago
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    I see what I did wrong, I messed up the power. The whole cos^19 (x) messed me up. Thanks for your help, I appreciate it. : )

  3. anonymous
    • 5 years ago
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    You should get \[-\int\limits_{}^{}\sin ^{1/2}{(u ^{19})} du\] From here you may have to jump into some trig identity im not sure

  4. anonymous
    • 5 years ago
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    actually there is no \[sin^{\frac{1}{2}}\] in it. just take the anti derivative of \[u^{\frac{9}{2}}\] which is \[\frac{2}{11} u^{\frac{11}{9}}\] and then replace u by cos(x) don't forget the "-" sign in front like i did.

  5. anonymous
    • 5 years ago
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    oops i meant the antiderivative is \[-\frac{2}{11}u^{\frac{11}{2}}\]

  6. anonymous
    • 5 years ago
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    the problem they wrote there is \[\int\limits_{}^{}\sin ^{1/2}[\cos ^{19}(x)]*\sin(x)*dx\]

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