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frx
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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)
\[\int\limits_{}^{}\sin ^{3}(x) \cos(x) dx\]
\[\int\limits_{}^{}(1\cos ^{2}(x))\cos(x)\sin(x) dx\]
\[u= \cos(x); du=\sin(x) dx\]
\[\int\limits_{}^{}(1u ^{2})u du= \int\limits_{}^{}uu ^{3} du=\frac{ u ^{2} }{ 2 }+\frac{ u ^{4} }{ 4} +C\]
\[=\frac{\cos ^{2}(x) }{ 2 }+\frac{\cos ^{4}(x) }{ 4}+C\]
 one year ago
 one year ago
frx Group Title
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) \[\int\limits_{}^{}\sin ^{3}(x) \cos(x) dx\] \[\int\limits_{}^{}(1\cos ^{2}(x))\cos(x)\sin(x) dx\] \[u= \cos(x); du=\sin(x) dx\] \[\int\limits_{}^{}(1u ^{2})u du= \int\limits_{}^{}uu ^{3} du=\frac{ u ^{2} }{ 2 }+\frac{ u ^{4} }{ 4} +C\] \[=\frac{\cos ^{2}(x) }{ 2 }+\frac{\cos ^{4}(x) }{ 4}+C\]
 one year ago
 one year ago

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hba Group TitleBest ResponseYou've already chosen the best response.0
@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
 one year ago

hba Group TitleBest ResponseYou've already chosen the best response.0
That is what wolf says :)
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
Yepp, that's it but why doesn't the method I used work? I simply just used the fact that sin^2(x)=1cos^2(x), shouldn't that work, too?
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
http://www.wolframalpha.com/input/?i=%28sin^4%28x%29%29%2F4%3Dcos^2%28x%29%2F2%2Bcos^4%28x%29%2F4 They do look quite the same?
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
http://www.wolframalpha.com/input/?i=+sin^4+x++%2F4%3D+cos^2+x+%2F2%2Bcos^4+x+%2F4
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
both the expressions are equivalent, \[sin^4=[1cos^2]^2=[12cos^2+cos^4]\]
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
the 1/4 which we get extra here, is combined with +c .
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
So both answers are equally right?
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
Thanks a lot, both of you! :)
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
welcome ^_^
 one year ago
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