Question

integration

X^3(x^2+4)^1/2
using subtitution or by part
which on is u

Answer 1

First you desegregate.

Answer 2

Do a trig sub. \(x = 2\tan(\theta)\)

Answer 3

what is desegregate???

Answer 4

It's over your head...
Do the substution \(x = 2\tan(\theta), dx=2\sec^2(\theta)d\theta \)

Answer 5

@arifadli8 Do you know how to do this trig sub?

Answer 6

i know a little about a trig sub

Answer 7

\[
\int x^3\sqrt{x^2+4} dx
=
\int [2\tan(\theta)^2]^3\sqrt{[2\tan(\theta)]^2+4]}\sec^2(\theta)d\theta
\]

Answer 8

Use \(\tan^2(\theta)+1 = \sec^2(\theta)\) in the radical.

Answer 9

Can you try it?

Answer 10

i will.. but right know i want to understand about desegregate.
if u have a link about desegregate. please give me

Answer 11

I'll tell you about it once you finish integration, as a reward. Ok?

Answer 12

why (2tan^2)^3...
its should (2 tan)^3
@Wio please explain

Answer 13

@Meepi can u explain

Answer 14

You don't need to use the substitution \(x = \tan \theta \) for this one:
\[\int x^3\sqrt{x^2 + 4} \, \mbox{d}x\]
use the substitution \(u = x^2\), \(\frac{1}{2} \mbox{d}u = x\, \text{d}x\):
\[\int u \sqrt{u + 4} \frac{1}{2}\, \text{d}u = \frac{1}{2} \int u \sqrt{u + 4}\, \text{d}u\]

I'm pretty sure you can solve this one, just use the substitution v = u + 4 :)

At the end substitute back u + 4 for v, then x^2 for u

Answer 15

\(x = 2 \tan \theta\), not \(x = \tan \theta\), that was a typo :p But you don't need to use it anyway so if you don't know it yet don't worry about it

Answer 16

You could probably do it in one substitution as well if you let u = x^2 + 4
um
\[\int x^2 \sqrt{x^2 + 4} x dx\]
u = x^2 + 4, du = 2x dx
\[\frac{1}{2}\int (u - 4)\sqrt{u}du\]
\[\frac{1}{2}\int u^{3/2} du - 2\int u^{1/2} du\]