Integration with a constant "a": $ \int_0^a \frac1{\sqrt {a^2-x^2}} dx $ Find the exact value of

$$ \int_{0}^{a} \frac{1}{\sqrt {a^2-x^2}} \mathrm {dx} $$

Where, $a$ is a positive constant
Hi, guys can give me tips to solve this ? Should we use like u substitution?
 A: Here is a tip, 
for $a>0$ 
Factor a out:
$$\frac {1} {a}\int_{0}^{a}\frac{1}{\sqrt{1-\dfrac{x^2}{a}}}\, \mathrm dx$$
And then set $t=x/a$ threfore $\mathrm dt=\frac 1 a\,\mathrm dx$
$$\int_{0}^{1}\frac{1}{\sqrt{1-t^2}}\,\mathrm dt\\
=\arcsin\left(\frac x a\right)\bigg|_0 ^a\\
$$

for $a<0$
switch the order of the integration bounds, 
factor $|a|$ out:
$$-\frac {1} {|a|}\int_{-a}^{0}\frac{1}{\sqrt{1-\dfrac{x^2}{a}}}\, \mathrm dx$$
And then set $t=x/a$ threfore $\mathrm dt=\frac 1 a\,\mathrm dx$
$$=-\int_{-1}^{0}\frac{1}{\sqrt{1-t^2}}\,\mathrm dt\\
=(-\arcsin\left(\frac x a\right))\bigg|_{-a} ^{0}\\
$$ 

For $a=0,$ $I=0$
A: Notice, $$\int_0^a\frac{1}{\sqrt{a^2-x^2}}\:\mathrm dx$$
Let $x=a\sin\theta\implies \mathrm dx=a\cos\theta\:\mathrm d\theta$, 
\begin{align}&=\int_{0}^{\pi/2}\frac{1}{a\cos \theta}(a\cos\theta\ \, \mathrm d\theta)\\ &=\int_{0}^{\pi/2}\,\mathrm d\theta \\& =\left(\theta\right)_{0}^{\pi/2}\\&=\frac \pi 2\end{align}
A: Different method here.
Consider $$\dfrac{d}{da} \int_0^a \sqrt{a^2-x^2} dx = a \int_0^a \frac{1}{\sqrt{a^2-x^2}} dx + \underbrace{\sqrt{a^2-a^2}}_0$$
by the Leibnitz rule of differentating under the integral.
Now, the integral on the left-hand side is just a quarter of the area of the circle of radius $a$  (because the integrand is the $y$-coordinate of the upper-right quadrant of the circle at $x$-coordinate $x$), so the LHS is $$\dfrac{d}{da} \frac{\pi a^2}{4} = \frac{\pi a}{2}$$
Therefore the required integral is $$\frac{\pi}{2}$$
