Calculating improper intergal $\displaystyle \int^a_0 \frac{1}{\sqrt{a^2-x^2}} dx$ I tried to solve the improper integral below, but the answer was $\displaystyle \frac{\pi}{2}$. What is wrong with the solution?
$\displaystyle \int^a_0 \frac{1}{\sqrt{a^2-x^2}} dx 
= \displaystyle \lim_{p \to a-0} \int^p_0 \frac{1}{\sqrt{a^2-x^2}} dx \\
= \displaystyle \frac{1}{a} \lim_{p \to a-0} \int^p_0 \frac{1}{\sqrt{1-(\frac{x}{a})^2}} dx \\
= \displaystyle \frac{1}{a} \lim_{p \to a-0} \left[ \arcsin{\frac{x}{a}} \right]^p_0 \\
= \displaystyle \frac{\pi}{2a}
$

Be careful about the change of variable. You need to complete the diferential and change the integration limits. Here $u=\frac{x}{a}$ and $du=\frac{dx}{a}$, then
$\displaystyle \int^a_0 \frac{1}{\sqrt{a^2-x^2}} dx 
= \displaystyle   \int^a_0 \frac{1}{\sqrt{1-(\frac{x}{a})^2}} \frac{dx}{a} \\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle   \int^1_0 \frac{1}{\sqrt{1-(u)^2}} du \\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{\pi}{2}.
$
Any doubts?
