Integrating int(16+x^2)^-3/2 w/trig substitution

[riverjib]

Can somebody help me with this?

int(16+x^2)^-3/2

I'm pretty sure the best way to start is with the substitution x=atan@ where x=4, but I'm not sure what to do with the fact that (16+x^2) is cubed underneath the radical...we haven't had any problems yet that weren't straightforward.

Thanks!

 

[soroban]

Hello, riverjib!

Don't let the cube intimidate you . . .

\(\displaystyle \L\int \frac{dx}{(16\,+\,x^2)^{\frac{3}{2}}\)

Start off as you suggested . . .

Let \(\displaystyle x\:=\:4\cdot\tan\theta\;\;\Rightarrow\;\;dx\:=\:4\cdot\sec^2\theta\cdot d\theta\)

Watch . . . \(\displaystyle \left(16\,+\,x^2\right)^{\frac{3}{2}} \;=\;\left(16\,+\,16\tan^2\theta\right)^{\frac{3}{2}}\;=\;\left(16[1\,+\,\tan^2\theta]\right)^{\frac{3}{2}}\)

. . \(\displaystyle = \;\left(16\cdot\sec^2\theta)^{\frac{3}{2}} \;=\;16^{\frac{3}{2}}\cdot(\tan^2\theta)^{\frac{3}{2}}\;=\;64\cdot\tan^3\theta\)


Substitute: \(\displaystyle \L\:\int\frac{4\cdot\sec^2\theta\cdot d\theta}{64\cdot\tan^3\theta} \;=\;\frac{1}{16}\int\frac{\sec^2\theta}{\tan^3\theta}\,d\theta \;=\;\frac{1}{16}\int\frac{\frac{1}{\cos^2\theta}}{\frac{\sin^3\theta}{\cos^3\theta}}\,d\theta\)

. . \(\displaystyle \L=\;\frac{1}{16}\int\frac{\cos\theta}{\sin^3\theta}\,d\theta\;=\;\int(\sin\theta)^{-3}(\cos\theta\cdot d\theta)\)


Then let: \(\displaystyle \,u\:=\;\sin\theta\) . . . etc.