Definite Integral of a Semi-Circle. Substitution.

[sepoto]

So there is a substitution and the definite integral becomes:

\(\displaystyle \frac{\pi}{2}=\int^a_{-a}\sqrt{1-\frac{t^2}{a^2}\;\frac{dt}{a}}\)

So then if both sides of the definite integral are multiplied by \(\displaystyle a^2\) I have:

\(\displaystyle \frac{\pi a^2}{2}=\int^a_{-a}\sqrt{a^2-t^2}\;dt\)

The solutions manual seems to say that:

\(\displaystyle \frac{\pi a^2}{2}=\frac{1}{a^2}\int^a_{-a}\sqrt{a^2-t^2}\;dt\)

So my question is how would there be a \(\displaystyle \frac{1}{a^2}\)? I don't really see where the \(\displaystyle \frac{1}{a^2}\) is coming from.

Thanks in advance for any help on this problem...

 

[sepoto]

\(\displaystyle \frac{\pi}{2}=\int^a_{-a}\sqrt{1-\frac{t^2}{a^2}\;\frac{dt}{a}}\)

is incorrect. The correct integral is

\(\displaystyle \frac{\pi}{2}=\int^a_{-a}\sqrt{1-\frac{t^2}{a^2}}\;\frac{dt}{a}\)

then

\(\displaystyle \frac{\pi}{2}=\frac{1}{a}\int^a_{-a}\sqrt{1-\frac{t^2}{a^2}}\;dt\)

\(\displaystyle \frac{\pi}{2}=\frac{1}{a}\int^a_{-a}\frac{a}{a}\sqrt{1-\frac{t^2}{a^2}}\;dt\)

\(\displaystyle \frac{\pi}{2}=\frac{1}{a^2}\int^a_{-a}\sqrt{a^2-t^2}\;dt\)

I'm still confused about how things go from:

\(\displaystyle \frac{\pi}{2}=\frac{1}{a}\int^a_{-a}\frac{a}{a}\sqrt{1-\frac{t^2}{a^2}}\;dt\)
to:
\(\displaystyle \frac{\pi}{2}=\frac{1}{a^2}\int^a_{-a}\sqrt{a^2-t^2}\;dt\)

The solutions manual seems to imply that the value of the definite integral then is changed to \(\displaystyle \frac{\pi a^2}{2}\) is that correct?