1. ## Trigonometric Substitution Q2

Question:
$\displaystyle \int\frac{dx}{\sqrt{4x-x^2}}$

How to solve it?

2. Originally Posted by mj.alawami
Question:
$\displaystyle \int\frac{dx}{\sqrt{4x-x^2}}$

How to solve it?
Here, there is no solution with steps.
we give a "start".
Try to complete the square for x in $\displaystyle 4x-x^2=-(x^2-4x)$.

On the other hand, the substitution $\displaystyle x=\frac{1}{u}$ will make the original integral = -$\displaystyle \int \frac{du}{u\sqrt{4u-1}}$ which can be evaluated easily by the substitution $\displaystyle z=\sqrt{4u-1}$.

3. Originally Posted by mj.alawami
Question:
$\displaystyle \int\frac{dx}{\sqrt{4x-x^2}}$

How to solve it?
Here are some steps to get you started.

First complete the square on the denominator to get

$\displaystyle 4x-x^2=-(x^2-4x)=-(x^2-4x+4)+4=4-(x-2)^2$

From here make the trig sub

$\displaystyle (x-2)=2\sin(y) \implies dx=2\cos(y)dy$

Just sub into the integral and simplify you should end up with

$\displaystyle \int dy$