# Easy integration problem.

• June 13th 2009, 07:19 AM
usagi_killer
Easy integration problem.
• June 13th 2009, 08:23 AM
TheEmptySet
Quote:

Originally Posted by usagi_killer

let $x=3\sin(t) \implies dx=3\cos(t)dt$

$\int \sqrt{9-x^2}dx=\int \sqrt{9-9\sin^2(t)}(3\cos(t))dt=$

$9\int \cos^2(t)dt =\frac{9}{2}\int 1+\cos(2t)dt=\frac{9}{2}t+\frac{9}{4}\sin(2t)+c=$

$\frac{9}{2}t+\frac{9}{2}\sin(t)\cos(t)+C$

Now we need to solve for t and cos(t)

$x=3\sin(t) \iff \frac{x}{3}=\sin(t) \iff \frac{x^2}{9}=\sin^2(t) \iff \frac{x^2}{9}=1-\cos^2(t)$

$\iff \cos^2(t)=1-\frac{x^2}{9} \iff \cos(t)=\frac{\sqrt{9-x^2}}{3}$

$x=3\sin(t) \iff \sin^{-1}\left( \frac{x}{3}\right)=t$

$\frac{9}{2}\sin^{-1}\left( \frac{x}{3}\right)+\frac{9}{2}\frac{x\sqrt{9-x^2}}{9}+C$

$\frac{9}{2}\sin^{-1}\left( \frac{x}{3}\right)+\frac{x\sqrt{9-x^2}}{2}+C$
• June 13th 2009, 08:25 AM
Twig
Let $\frac{x}{3} = sin(t)$

$\int \sqrt{9-x^{2}} dx = \int 3 \sqrt{1-\frac{x^{2}}{9}} = 3 \int \sqrt{1-sin^{2}(t)} \cdot 3 cos(t) dt \Rightarrow$

$9 \int cos^{2}(t) dt$

This can be re-written using the double angle formula $cos(2t)=2cos^{2}(t)-1$

$\frac{9}{2} \int (1+cos(2t)) dt = \frac{9}{2}(t+\frac{1}{2}sin(2t)) + C$

I gotten beaten to it here, the above solution is better! I just realized I didnt solve for x in the end.. =)
• June 13th 2009, 08:31 AM
Amer
Quote:

Originally Posted by usagi_killer

let $3sinu =x$ ...... $(3cosu) du = dx$

from assumptions $3sinu=x$ we can find cosu

$sin^2u + cos^2u =1$ but $sinu =\frac{x}{3}$

$\frac{x^2}{9} + cos^2u = 1$

$cosu = \frac{\sqrt{9-x^2}}{3}$

$\int \sqrt{9-x^2} dx$

$\int {\color{red}\sqrt{9-x^2}} \left(\frac{3}{{\color{red}3}}\right) (3cosu) du$

the red terms is cosu

$\int (3cos^2u) du$ you can solve it right .