## 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) $\int\limits_{}^{}\sin ^{3}(x) \cos(x) dx$ $\int\limits_{}^{}(1-\cos ^{2}(x))\cos(x)\sin(x) dx$ $u= \cos(x); -du=\sin(x) dx$ $-\int\limits_{}^{}(1-u ^{2})u du= -\int\limits_{}^{}u-u ^{3} du=-\frac{ u ^{2} }{ 2 }+\frac{ u ^{4} }{ 4} +C$ $=-\frac{\cos ^{2}(x) }{ 2 }+\frac{\cos ^{4}(x) }{ 4}+C$ one year ago one year ago

1. hba Group Title

@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 Group Title

Take*

3. hba Group Title

That is what wolf says :)

4. frx Group Title

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 Group Title

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 Group Title
7. hartnn Group Title

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

8. hartnn Group Title

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

9. frx Group Title

So both answers are equally right?

10. hartnn Group Title

exactly.

11. frx Group Title

Thanks a lot, both of you! :)

12. hartnn Group Title

welcome ^_^