Question:
$\displaystyle \int\frac{dx}{\sqrt{4x-x^2}}$
How to solve it?
Please answer in steps , thank you
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 $\displaystyle z=\sqrt{4u-1}$.
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
$\displaystyle \int dy$