Mastering Integrals: A Cute Factoring Trick

[ToxicBug]

Could anyone evaluate this integral for me? I got it in an exam and didn't know how to do it.

\int\sqrt{x-x^2}

 

[vincentchan]

let u = x-1/2 and do the substitude,...

 

[ToxicBug]

What do you mean? Can you evaluate it completely showing the steps?

 

[Jameson]

Ok, Mathematica was definitely wrong. There's going to be a trig substitution in this integral somewhere.

 

[ToxicBug]

Mathematica is **** and I'm sure that's not a right answer.

 

[vincentchan]

What do you mean? Can you evaluate it completely showing the steps?

what do I mean? I thought I make it very clear already!
\int\sqrt{x-x^2}dx=\int\sqrt{x(1-x)}dx

let u=x-\frac{1}{2}

It become:

\int\sqrt{(\frac{1}{2}-u)(\frac{1}{2}+u)}du

=\int\sqrt{\frac{1}{4}-u^2 }du

=\frac{u}{4}\sqrt{1-4u^2}+\frac{1}{8}sin^{-1}(2u)+C

=\frac{x-1/2}{4}\sqrt{1-4(x-1/2)^2}+\frac{1}{8}sin^{-1}(2x-1)+C

 

[ToxicBug]

Sick, thanks.

 

[dextercioby]

And one more thing:"Mathematica" is ALWAYS RIGHT...

Daniel.

 

[HallsofIvy]

Cute factoring.

What I would have done is "complete the square": x- x²= 1/4-1/4+ x-x²= 1/4-(x-1/2)² which would have led me to the u= x- 1/2 substitution. Once I had it in the form \sqrt{\frac{1}{4}- u^2} I would use a trig substitution.