Integrate by substitution:

Question

anonymous · Answer

$$\sqrt{(a-x)(x-b)}$$

anonymous · Answer

@experimentX @amistre64 @TuringTest @sauravshakya 
How do I do this using substitution and not by converting it to a quadratic.

amistre64 · Answer

u = (a-x)(x-b)

derive to get:

du = (a)(x-b) + (a-x)(-b) dx

perhaps?

anonymous · Answer

Use 
x= acos^2 t + bsin^2t.
I'll leave it to you from here.

experimentx · Answer

clever

amistre64 · Answer

u = (a-x)(x-b)

forgot how to derive constants (im assuming a and b are constant)

derive to get:

du = -(x-b) - (a-x) dx

i cant verify sidd way tho

experimentx · Answer

he eliminated the both variable out of roots with single subs ... two kill with single stone.

anonymous · Answer

Exactly.

anonymous · Answer

nice :)

anonymous · Answer

Thanks :)

turingtest · Answer

I can't make it work :(
I will keep trying

turingtest · Answer

$$x=a\cos^2t+b\sin^2t$$$$dx=-2a\cos t\sin t+2b\sin t\cos t=-a\sin(2t)+b\sin(2t)$$

turingtest · Answer

am I supposed to make some clever simplification for a-x and x-b now or what?

experimentx · Answer

$$  (a-x)= (a-b) \sin ^2 t \
(x-b) = (a-b)\cos ^2 t$$

turingtest · Answer

my trig simplification is shot today, I can't seem to get that...

callisto · Answer

$$a-x $$$$= a-(acos^2t + bsin^2t) $$$$= a (1-cos^2t) - bsin^2t$$$$ = asin^2t-bsin^2t $$$$= (a-b)sin^2t$$

turingtest · Answer

ah, I think today is not going to be one of my good days :/
thanks for tolerating my slowness guys

callisto · Answer

$$x-b$$$$=(acos^2t+bsin^2t)-b $$$$= acos^2t + b(sin^2t-1)$$$$= acos^2t - bcos^2t$$$$=(a-b)cos^2t$$

experimentx · Answer

hold on .. lemme post a similar Q.

turingtest · Answer

yeah @Callisto I totally overlooked that we get 1-cos^2=sin^2 and vice versa, it just escaped me...

anonymous · Answer

Aha. Thankyou guys.