Question

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)

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

Answer 1

@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

Answer 2

Take*

Answer 3

That is what wolf says :)

Answer 4

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?

Answer 5

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?

Answer 6

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

Answer 7

both the expressions are equivalent,
\[sin^4=[1-cos^2]^2=[1-2cos^2+cos^4]\]

Answer 8

the 1/4 which we get extra here, is combined with +c .

Answer 9

So both answers are equally right?

Answer 10

exactly.

Answer 11

Thanks a lot, both of you! :)

Answer 12

welcome ^_^