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frx

  • 3 years ago

I'm a bit confused. I thought I'd solved the problem but when running the integral through Wolframalpha the answer is totally different, am I wrong or is it just the wolframalpha algorithm that doesn't process things like me? http://www.wolframalpha.com/input/?i=integrate+sin^3(x)cos(x) \[\int\limits_{}^{}\sin ^{3}(x) \cos(x) dx\] \[\int\limits_{}^{}(1-\cos ^{2}(x))\cos(x)\sin(x) dx\] \[u= \cos(x); -du=\sin(x) dx\] \[-\int\limits_{}^{}(1-u ^{2})u du= -\int\limits_{}^{}u-u ^{3} du=-\frac{ u ^{2} }{ 2 }+\frac{ u ^{4} }{ 4} +C\] \[=-\frac{\cos ^{2}(x) }{ 2 }+\frac{\cos ^{4}(x) }{ 4}+C\]

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  1. hba
    • 3 years ago
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    @frx ake the integral: integral sin^3(x) cos(x) dx For the integrand sin^3(x) cos(x), substitute u = sin(x) and du = cos(x) dx: = integral u^3 du The integral of u^3 is u^4/4: = u^4/4+constant Substitute back for u = sin(x): Answer: | | = (sin^4(x))/4+constant

  2. hba
    • 3 years ago
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    Take*

  3. hba
    • 3 years ago
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    That is what wolf says :)

  4. frx
    • 3 years ago
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    Yepp, that's it but why doesn't the method I used work? I simply just used the fact that sin^2(x)=1-cos^2(x), shouldn't that work, too?

  5. frx
    • 3 years ago
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    http://www.wolframalpha.com/input/?i=%28sin^4%28x%29%29%2F4%3D-cos^2%28x%29%2F2%2Bcos^4%28x%29%2F4 They do look quite the same?

  6. frx
    • 3 years ago
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    http://www.wolframalpha.com/input/?i=+sin^4+x++%2F4%3D+-cos^2+x+%2F2%2Bcos^4+x+%2F4

  7. hartnn
    • 3 years ago
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    both the expressions are equivalent, \[sin^4=[1-cos^2]^2=[1-2cos^2+cos^4]\]

  8. hartnn
    • 3 years ago
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    the 1/4 which we get extra here, is combined with +c .

  9. frx
    • 3 years ago
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    So both answers are equally right?

  10. hartnn
    • 3 years ago
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    exactly.

  11. frx
    • 3 years ago
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    Thanks a lot, both of you! :)

  12. hartnn
    • 3 years ago
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    welcome ^_^

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