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How do I solve this? $$\int _{0+0}^{1-0}\frac{dx}{\left(4-3x\right)\sqrt{x-x^2}}\:dx$$ I know it's a type 3 improper integral, and I'm having issues with these. I think that I need to write it as a sum of limits and then try to compute the values of those limits and the value of the sum would be my improper integral value. Do I need to try and work it around with a substitution? I was thinking I could use trigonometric substitution for this, but I don't think it applies here. Can anyone give me a hint or help me with this?

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Sub $x=u^2$; then the integral is

$$2 \int_0^1 \frac{du}{(4-3 u^2) \sqrt{1-u^2}} = \int_{-1}^1 \frac{du}{(4-3 u^2) \sqrt{1-u^2}}$$

We can evaluate this using a trig substitution and then another step which I will explain. Let $u = \sin{t}$; then the integral is

$$\int_{-\pi/2}^{\pi/2} \frac{dt}{4 - 3 \sin^2{t}} $$

which we can rewrite, using the double angle formula and subbing again, as

$$\int_{-\pi}^{\pi} \frac{dt}{5+3 \cos{t}} $$

We may convert this to an integral over the unit circle in the complex plane and use the residue theorem. Let $z=e^{i t}$ and the integral is

$$-i \oint_{|z|=1} \frac{dz}{z} \frac1{5+\frac32 (z+z^{-1})} = -i 2 \oint_{|z|=1} dz \, \frac1{3 z^2 + 10 z + 3}$$

The poles of the denominator are at $z=-3$ and $z=-1/3$; only $z=-1/3$ is in the unit circle. The residue theorem states that the integral is equal to $i 2 \pi$ times the residue at that pole, or

$$i 2 \pi (-i 2) \frac1{6(-1/3)+10} = \frac{\pi}{2}$$


Alternatively, if you do not like residues, you could simply use the substitution $v=\tan{(t/2)}$; then the integral becomes

$$\int_{-\infty}^{\infty} \frac{dv}{4+v^2} = \frac{\pi}{2} $$

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\begin{align} I&=\int_{0}^{1}\!{\frac {1}{ \left( 4-3\,x \right) \sqrt {-{x}^{2}+x}}} \,{\rm d}x\\ &=\int_{-1/2}^{1/2}\!-4\,{\frac {1}{\sqrt {-4\,{u}^{2}+1} \left( -5+6\,u \right) }}\,{\rm d}u,\,\,[\mbox{sub}, u = x-1/2]\\ &=-4\,\int_{-1/2}^{1/2}\!{\frac {1}{\sqrt {-4\,{u}^{2}+1} \left( -5+6\,u \right) }}\,{\rm d}u\\ &=-4\,\int_{\infty }^{0}\! \left( 8\,{{\it u_1}}^{2}+2 \right) ^{-1} \,{\rm d}{\it u_1},\,\,\,[\mbox{sub}, u_1 = ((1/2-u)/(1/2+u))^{(1/2)}]\\ &=-4\,\int_{\pi /2}^{0}\!\frac{1}{4}\,{\rm d}{\it u_2},\,\,\,[\mbox{sub}, u_1 = 1/2\tan(u_2)]\\ &=\frac{\pi}{2} \end{align}

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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\large A\ \Re\mbox{-}{\it Integration}}$: \begin{align} &\bbox[#ffd]{\int_{0}^{1}{\dd x \over \pars{4 - 3x}\root{x - x^{2}}}} = \int_{\infty}^{1}{-\dd x/x^{2} \over \pars{4 - 3/x} \root{1/x - 1/x^{2}}} \\[5mm] = &\ \int_{1}^{\infty}{\dd x \over \pars{4x - 3}\root{x - 1}} = \int_{0}^{\infty}{\dd x \over \pars{4x + 1}\root{x}} \\[5mm] = &\ \half\int_{0}^{\infty}{\dd x \over \pars{x + 1}\root{x}} \end{align}


\begin{align} &\mbox{With}\quad x = t^{2}: \\ &\bbox[#ffd]{\int_{0}^{1}{\dd x \over \pars{4 - 3x}\root{x - x^{2}}}} = \int_{0}^{\infty}{\dd t \over t^{2} + 1} = \bbox[10px,border:1px groove navy]{\pi \over 2} \\ & \end{align}
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