Solving an Integral Problem: Finding the Minus Sign

  • Thread starter Thread starter Jalo
  • Start date Start date
  • Tags Tags
    Integral
Jalo
Messages
117
Reaction score
0

Homework Statement



2e2gumw.png

(ignore what's written, it isn't important for the problem)

I'm studying integrals and I came across this solved example. However I can't understand where the minus of the integral came from came from.

Homework Equations





The Attempt at a Solution



The primitive of a f^p * f ' function is f^(p+1) / (p+1) Therefore it should be ln^2(x-1)/2 and not -ln^2(x-1)/2

Thanks
 
Physics news on Phys.org
You could dispel any doubts by making a u-substitution. Try u = 1 - x and see where that takes you.
 
Jalo said:

Homework Statement



2e2gumw.png

(ignore what's written, it isn't important for the problem)

I'm studying integrals and I came across this solved example. However I can't understand where the minus of the integral came from came from.

Homework Equations



The Attempt at a Solution



The primitive of a f^p * f ' function is f^(p+1) / (p+1) Therefore it should be ln^2(x-1)/2 and not -ln^2(x-1)/2

Thanks

By the chain rule: \displaystyle \frac{d}{dx}\ln(1-x)=\frac{1}{1-x}\frac{d}{dx}(1-x)=-\frac{1}{1-x}\,.

So you have the anti-derivative of f^p * (- f ' ) .
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

Similar threads

Back
Top