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!