1. ## Evaluate this Integral

This was a test problem:
$\int^{1}_0\sqrt{\frac{1-x}{x}}dx$
I tried to stare it down until I figured out a solution. I thought about trig substitution, inverse trig, but in the end went with spliting the inside by
$\sqrt{\frac{1}{x}-\frac{x}{x}}}dx$
ran out of time and left it like that....

2. Hello, ugkwan!

This is a very messy problem . . . for a test question.

$\displaystyle \int^{1}_0\sqrt{\frac{1-x}{x}}dx$

$\displaystyle \text{Let: }\:u \:=\:\sqrt{\frac{1-x}{x}} \quad\Rightarrow\quad u^2 \:=\:\frac{1-x}{x} \quad\Righgtarrow\quad u^2x \:=\:1-x$

. . $\displaystyle u^2x + x \:=\:1 \quad\Rightarrow\quad (u^2+1)x \:=\:1 \quad\Rightarrow\quad x \:=\:\frac{1}{u^2+1}
$

. . $\displaystyle x \:=\:(u^2+1)^{\text{-}1} \quad\Rightarrow\quad dx \:=\:-(u^2+1)^{\text{-}2}\,2u\,du \;=\;\frac{-2u\,du}{(u^2+1)^2}$

$\displaystyle\text{Substitute: }\;\int u\cdot\frac{-2u\,du}{(u^2+1)^2} \;=\;-2\int\frac{u^2}{(u^2+1)^2}\,du$ .[1]

$\displaystyle \text{Partial Fractions: }\;\frac{u^2}{(u^2+1)^2} \;=\;\frac{Ax+B}{u^2+1} + \frac{Cx+D}{(u^2+1)^2}$

. . $\text{and we get: }\;A = 0,\;B = 1,\;C = 0,\;D = -1$

$\displaystyle \text{And we have: }\;-2\int\left(\frac{1}{u^2+1} - \frac{1}{(u^2+1)^2}\right)\,du$

Can you finish it?

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

I just noticed that [1] can be done "by parts" . . .

. . $\begin{array}{cccccccc}
U &=& u && dV &=& u(u^2+1)^{-2}du \\ \\[-3MM]
dU &=& du && V &=& -\frac{1}{2}{(u^2+1)^{-1}} \end{array}$

$\displaystyle \text{And we have: }\;-2\left(-\tfrac{1}{2}\,\frac{u}{u^2+1} + \tfrac{1}{2}\int\frac{du}{u^2+1}\right)$

3. We can use the substitution $x = \sin^2(y)$

$dx = 2\sin(y) \cos(y) dy$

The integral becomes :

$\int_0^{\frac{\pi}{2}} \sqrt{\frac{\cos^2(y)}{\sin^2(y)} } ~2\sin(y) \cos(y)~ dy$

$\int_0^{\frac{\pi}{2}} ( 1 + \cos(2y) )~dy$

$= \left[ y + \frac{\sin(2y) } {2} \right]_0^{\frac{\pi}{2} }$

$= \frac{\pi}{2}$