Integration with trigonometric substitution

[lil_hawk]

Please help with this problem. I have the answer but I can't get past the first part.
It's the integral from 0 to 3 (radical 3)/2 of (x^3)/(4x^2 + 9)^(3/2) dx. It says to let u = 2x which I understand because you can square that to get 4x^2 but then it says x = (3/2) tan theta. How do you get that because I would have said 3 tan theta since 3^2 = 9. Any help is appreciated. Thanks.

 

[soroban]

Hello, lil_hawk!

I'd say you are correct. . They are making two substitutions.
. . We can manage with just one.

\(\displaystyle \L\int^{\;\;\;\;\frac{3\sqrt{3}}{2}}_0 \frac{x^3}{(4x^2\,+\,9)^{3/2}}\, dx\)

Let \(\displaystyle 4x^2\:=\:9\tan^2\theta\;\;\Rightarrow\;\;x\:=\:\frac{3}{2}\tan\theta\;\;\Rightarrow\;\;dx\:=\:\frac{3}{2}\sec^2\theta\,d\theta\)

. . and \(\displaystyle \sqrt{4x^2\,+\,9}\:=\:\sqrt{9\tan^2\theta\,+\,9} \:=\:\sqrt{9(\tan^2\,+\,1)} \:=\:3\sec\theta\)


Substitute: \(\displaystyle \L\:\int\frac{\frac{27}{8}\tan^3\theta}{\left(9\sec^2\theta\right)^{\frac{3}{2}}}\left(\frac{3}{2}\sec^2\theta\,d\theta\right) \:=\:\frac{3}{16}\int\frac{\tan^3\theta}{\sec\theta}\,d\theta\)

. . \(\displaystyle \L=\;\frac{3}{16}\int\frac{\tan\theta(\tan^2\theta)}{\sec\theta}\,d\theta \;=\;\frac{3}{16}\int\frac{\tan\theta(\sec^2\theta\,-\,1)}{\sec\theta}\,d\theta\)

. . \(\displaystyle \L=\;\frac{3}{16}\int\left(\sec\theta\tan\theta\,-\,\frac{\tan\theta}{\sec\theta}\right)\,d\theta \;=\;\frac{3}{16}\int\left(\sec\theta\tan\theta\,-\,\sin\theta)\,d\theta\)

. . \(\displaystyle \L=\;\frac{3}{16}(\sec\theta\,+\,\cos\theta)\,+\,C\)


Back-substitute

From \(\displaystyle \tan\theta\:=\:\frac{2x}{3}\), we have: \(\displaystyle \,opp\:=\:2x,\;adj\,=\,3,\;hyp\,=\,\sqrt{4x^2\,+\,9}\)
. . Hence: \(\displaystyle \:\cos\theta\:=\:\frac{3}{\sqrt{4x^2+9}},\;\sec\theta\:=\:\frac{\sqrt{4x^2+9}}{3}\)

We have: \(\displaystyle \L\:\frac{3}{16}\left[\frac{\sqrt{4x^2+9}}{3}\,+\,\frac{3}{\sqrt{4x^2+9}}\right]^{\frac{3\sqrt{3}}{2}}_0\)

. . \(\displaystyle \L=\;\frac{3}{16}\left[\left(\frac{\sqrt{4\cdot\frac{27}{4}\,+\,9}}{3}\,+\,\frac{3}{\sqrt{4\cdot\frac{27}{4}+9}}\right)\:-\:\left(\frac{\sqrt{9}}{3}\,+\,\frac{3}{\sqrt{9}}\right)\right]\)

. . \(\displaystyle \L=\;\frac{3}{16}\left[\left(\frac{6}{3}\,+\,\frac{3}{6}\right)\:-\:\(1 + 1)\right] \;=\;\frac{3}{16}\left[2\,+\,\frac{1}{2}\,-\,2\right] \;=\;\frac{3}{16}\left(\frac{1}{2}\right)\;=\;\fbox{\frac{3}{32}}\)