Do you HAVE always to use u-substitution?

[Superposed_Cat]

Do you always have to use u-substitution? i.e. sometimes will you get a different answer to if you didn't? I am asking because I had the integral ∫sinxcosx dx between 0 and π/2 and the answer was 1/2 but I got 0.

∫sinxcosx dx between 0 and π/2=-cos(π/2)sin(π/2)=0 according to my calculations.

What am I doing wrong?

I don't thin this counts as homework style as I am hoping just to find out whether you always have to use u-substitution in certain case using the above question as proof.

 

[arildno]

Do you always have to use u-substitution? i.e. sometimes will you get a different answer to if you didn't? I am asking because I had the integral ∫sinxcosx dx between 0 and π/2 and the answer was 1/2 but I got 0.

∫sinxcosx dx between 0 and π/2=-cos(π/2)sin(π/2)=0 according to my calculations.

What am I doing wrong?

Antidifferentiating incorrectly.

You are using a fallacious integration rule here:
\int{u}'(x)v'(x)dx=u(x)v(x)+C

 

[Superposed_Cat]

why is that not valid?
what should I have done?

 

[arildno]

why is that not valid?

You might try to differentiate u(x)*v(x).
Do you then get your integrand u'(x)*v'(x)?

If you don't get that result, then you've proven that u(x)*v(x) is not an antiderivative of u'(x)*v'(x)

 

[SteamKing]

Do you always have to use u-substitution? i.e. sometimes will you get a different answer to if you didn't? I am asking because I had the integral ∫sinxcosx dx between 0 and π/2 and the answer was 1/2 but I got 0.

∫sinxcosx dx between 0 and π/2=-cos(π/2)sin(π/2)=0 according to my calculations.

What am I doing wrong?

If you look at the original integral carefully, you would see that it has the form:

\int u\ du

and it should be pretty obvious how to handle it.

 

[jtbell]

For further advice on this exercise, please use the homework forums.