# Thread: integration question no.5

1. ## integration question no.5

integration question no.5

2. You could then let $\displaystyle {\theta}=sin^{-1}(u), \;\ d{\theta}=\frac{1}{\sqrt{1-u^{2}}}du$

Then making the subs, it boils down to:

$\displaystyle 2\int\frac{u}{(u^{2}+3)^{\frac{1}{3}}}$

May as well finish it:

If you let $\displaystyle w=u^{2}+3, \;\ dw=2udu$

That turns into an easy integral: $\displaystyle \int\frac{1}{w^{\frac{1}{3}}}dw=\frac{3}{2}w^{\fra c{2}{3}}$

Resub: $\displaystyle \frac{3}{2}(u^{2}+3)^{\frac{2}{3}}$

Resub: $\displaystyle \frac{3}{2}(sin^{2}{\theta}+3)^{\frac{2}{3}}$

3. Use u = sinT?
[Let me write T as theta---for less typing.]

Okay.

u = sinT
So,
du = cosT dT
dT = du/cosT

INT.[sin(2T) / cbrt(sin^2(T) +3)]dT
= INT.[2sinTcosT *(sin^2(T) +3)^(-1/3)]dT
Substitutions,
= INT.[2u*cosT*(u^2 +3)^(-1/3)](du/cosT)
= INT.[2u*(u^2 +3)^(-1/3)]du
= INT.[(u^2 +3)^(-1/3)](2u du)

d(u^2 +3) = 2u du

= INT.[(u^2 +3)^(-1/3)]d(u^2 +3)
= 1/(-1/3 +1) * (u^2 +3)^(-1/3 +1) +C
= (3/2)(u^2 +3)^(2/3) +C

Going back to u = sinT

= (3/2)(sin^2(T) +3)^(2/3) +C -----------------answer.