Trigonometric Substitution for $\int\sqrt{9-x^2}\,\mathrm dx$ Question: Use the substitution $x=3\sin(t)$ to evaluate the integral of $\int\sqrt{9-x^2}\,\mathrm dx$.
I started by making a right triangle and solving for $\sin(t)$ and $\cos(t)$. 


*

*$\sin(t)=\frac{x}{3}$ and $\cos(t)=\frac{\sqrt{9-x^2}}{3}$
Then, I solved for the values $\mathrm dx$ and $\sqrt{9-x^2}$.

*$\sqrt{9-x^2}=3\cos(t)$

*$\mathrm dx=3\,\cos(t)\,\mathrm dt$
Then, I got the integral of $3\,\cos^2(t)\,\mathrm dt$. 

*$3\,(\frac{1}{2}\cos(t)\sin(t)+ \frac{1}{2}t)$

*$\frac{3}{2}\cos(t)\sin(t)+\frac{3}{2}t$
Then, I substituted in the values I found for $\sin(t)$, $\cos(t)$, etc.

*$\frac{3}{2}(\frac{\sqrt{9-x^2}}{3})\times\frac{x}{3}+\frac{3}{2}\arcsin(\frac{x}{3})$


That was the wrong answer and I do not know why. Where did I go wrong?
Thank you for your help!

$$
\int \sqrt{9-x^2} dx
$$
let $x=3\sin\phi$,$dx=3\cos\phi d\phi$, thus we have
$$
3\int d\phi\cos\phi\sqrt{9(1-sin^2\phi)}=9\int d\phi \cos^2\phi =9\int d\phi \frac{1}{2}(1+cos(2\phi))=\frac{9\phi}{2}+\frac{9\sin(2\phi)}{4}+C
$$
Thus we see that
$$
\int \sqrt{9-x^2} dx=\frac{9\phi}{2}+\frac{9\sin(2\phi)}{4}+C 
$$
where $\phi=\sin^{-1}(x/3)$.  Simplifying we obtain
$$
\int \sqrt{9-x^2} dx=\frac{1}{2}\big( \sqrt{9-x^2}+9\sin^{-1}\big(\frac{x}{3}\big)     \big)
$$
