integral Trig substitution

[jwpaine]

I think all of my work is correct thus far. It is at the end that I am having difficulty when substituing back in for x
.
\(\displaystyle \int \frac{x^2dx}{\sqrt{9-x^2}}\)
\(\displaystyle a^2 - x^2 = 3^2 - x^2\)
\(\displaystyle x = 3sin(b)\)
\(\displaystyle dx = 3cos(b)db\)
\(\displaystyle a^2 - x^2 = 9 - sin^2(b)\)
\(\displaystyle = 9(1 - sin^2(b)\)
\(\displaystyle = 9cos^2(b)\)
\(\displaystyle a^2 - x^2 = 9cos^2(b)\)
\(\displaystyle \sqrt{9 - x^2} = 3cos(b)\)

==>\(\displaystyle \int \frac{(9sin^2(b)(3cos(b)db)}{3cos(b)} = 9\int sin^2(b)db\)

I know that \(\displaystyle \int sin^2(b)db = \frac{b}{2} - \frac{sin(2b)}{4} + C\)
I need help substituing back in, in terms of x.

Thanks!

 

[royhaas]

Solve for \(\displaystyle b\) in terms of \(\displaystyle x\) and use \(\displaystyle sin(2b)=2sin(b)cos(b)\).

 

[jwpaine]

Solve for \(\displaystyle b\) in terms of \(\displaystyle x\) and use \(\displaystyle sin(2b)=2sin(b)cos(b)\).

Well even if I say that \(\displaystyle 9 \int sin^2(b)db = 9[ \frac{b}{2} - \frac{1}{2}sin(b)cos(b)] + C\)
and \(\displaystyle b = sin^{-1}(\frac{x}{3})\)

I still can't just plugin for b, as I would be left with an inverse sin function in there

and the book gives an answer as \(\displaystyle \frac{1}{3}tan^3(x) + C\)

 

[galactus]

Your method looks sound, JW. Are you sure you're looking at the correct problem in the back of the book?.

You should get \(\displaystyle \frac{9sin^{-1}(x/3)}{2}-\frac{x\sqrt{9-x^{2}}}{2}\)

Just for kicks:

\(\displaystyle \int{sin^{2}(x)(sec^{4}(x))}dx=\frac{tan^{3}(x)}{3}\)

 

[jwpaine]

How did you handle the cos(b) in \(\displaystyle \frac{b}{2} - \frac{1}{2}sin(b)cos(b)\) ?

 

[stapel]

How did you handle the cos(b) in \(\displaystyle \frac{b}{2} - \frac{1}{2}sin(b)cos(b)\) ?

If x = 3sin(b), then sin(b) = x/3 = (opposite)/(adjacent). Draw the triangle and copy down the "value" of cos(b). :wink:

(The "b" remains an arc-function.)

Eliz.

 

[galactus]

After making our sub and whittling it down, we get \(\displaystyle 9\int{sin^{2}(b)}db=\frac{9b-9sin(b)cos(b)}{2}\)

Now, if we sub in \(\displaystyle b=sin^{-1}(x/3)\) to get it back in terms of x, we get our antiderivative.

We can write it as \(\displaystyle \frac{18b-9sin(2b)}{4}\)

\(\displaystyle \frac{18(sin^{-1}(x/3))-9sin(2(sin^{-1}(x/3)))}{4}\)

Which should give you your answer. Take note that \(\displaystyle 9sin(2sin^{-1}(x/3))=2x\sqrt{9-x^{2}}\)

 

[jwpaine]

Understood. Thank you all for taking the time to reply to this thread.

John.