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

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hba
 3 years ago
Best 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

frx
 3 years ago
Best ResponseYou've already chosen the best response.0Yepp, 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?

frx
 3 years ago
Best ResponseYou've already chosen the best response.0http://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?

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

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

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1the 1/4 which we get extra here, is combined with +c .

frx
 3 years ago
Best ResponseYou've already chosen the best response.0So both answers are equally right?

frx
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks a lot, both of you! :)
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