Mastering Definite Integrals for Solving Tricky Problems

sinClair
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Homework Statement


Integrate \int_{0}^{1}\sqrt{\frac{4x^2-4x+1}{x^2-x+3}}dx

Homework Equations

The Attempt at a Solution


U sub: let u=x^2-x+3 Then du=2x-1 and then have to evaluate \int_{3}^{3}\sqrt{\frac{du^2}{u}}dx But how with these limits of integration should this be 0? Not sure how to evaluate this...
 
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Complete the square in both numerator and denominator, then use a u-sub for the radican.
 
rocophysics said:
Complete the square in both numerator and denominator, then use a u-sub for the radican.
I see no reason to do that.

sinClair said:

Homework Statement


Integrate \int_{0}^{1}\sqrt{\frac{4x^2-4x+1}{x^2-x+3}}dx


Homework Equations




The Attempt at a Solution


U sub: let u=x^2-x+3 Then du=2x-1 and then have to evaluate \int_{3}^{3}\sqrt{\frac{du^2}{u}}dx But how with these limits of integration should this be 0? Not sure how to evaluate this...
\sqrt{x^2}= |x| and 2x- 1 changes sign at x= 1/2. This integral is the same as
\int_0^1 \frac{|2x-1|}{\sqrt{x^2- x+ 3}}dx= -\int_0^{1/2}\frac{2x-1}{\sqrt{x^2-x+ 3}}dx+ \int_{1/2}^1 \frac{2x-1}{\sqrt{x^2- x+ 3}}dx
 
Thank you so much. I was getting confused with changing the limits of integration back and forth but got it now, thanks.
 
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...

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