anonymous
  • anonymous
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) \[\int\limits_{}^{}\sin ^{3}(x) \cos(x) dx\] \[\int\limits_{}^{}(1-\cos ^{2}(x))\cos(x)\sin(x) dx\] \[u= \cos(x); -du=\sin(x) dx\] \[-\int\limits_{}^{}(1-u ^{2})u du= -\int\limits_{}^{}u-u ^{3} du=-\frac{ u ^{2} }{ 2 }+\frac{ u ^{4} }{ 4} +C\] \[=-\frac{\cos ^{2}(x) }{ 2 }+\frac{\cos ^{4}(x) }{ 4}+C\]
Mathematics
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anonymous
  • anonymous
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) \[\int\limits_{}^{}\sin ^{3}(x) \cos(x) dx\] \[\int\limits_{}^{}(1-\cos ^{2}(x))\cos(x)\sin(x) dx\] \[u= \cos(x); -du=\sin(x) dx\] \[-\int\limits_{}^{}(1-u ^{2})u du= -\int\limits_{}^{}u-u ^{3} du=-\frac{ u ^{2} }{ 2 }+\frac{ u ^{4} }{ 4} +C\] \[=-\frac{\cos ^{2}(x) }{ 2 }+\frac{\cos ^{4}(x) }{ 4}+C\]
Mathematics
schrodinger
  • schrodinger
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hba
  • hba
@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
hba
  • hba
Take*
hba
  • hba
That is what wolf says :)

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anonymous
  • anonymous
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?
anonymous
  • anonymous
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?
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=+sin^4+x++%2F4%3D+-cos^2+x+%2F2%2Bcos^4+x+%2F4
hartnn
  • hartnn
both the expressions are equivalent, \[sin^4=[1-cos^2]^2=[1-2cos^2+cos^4]\]
hartnn
  • hartnn
the 1/4 which we get extra here, is combined with +c .
anonymous
  • anonymous
So both answers are equally right?
hartnn
  • hartnn
exactly.
anonymous
  • anonymous
Thanks a lot, both of you! :)
hartnn
  • hartnn
welcome ^_^

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