3
$\begingroup$

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!

$\endgroup$
1
  • $\begingroup$ your third line, "then i got the integral of 3 cos(t)^2 dt" is not correct. You should have two factors of 3 (you only have 1), you get two because: one from the change of variables, and the other from the $\sqrt{9-x^2}$ term. see solution below $\endgroup$ – Jeff Faraci Feb 28 '14 at 2:39
1
$\begingroup$

$$ \int \sqrt{9-x^2} dx $$ 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 $$ \int \sqrt{9-x^2} dx=\frac{1}{2}\big( \sqrt{9-x^2}+9\sin^{-1}\big(\frac{x}{3}\big) \big) $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.