# More trig substitution

• February 26th 2009, 08:32 PM
silencecloak
More trig substitution
$\int \frac{x^2}{(3+4x-4x^2)^{3/2}}$

I'm pretty sure I need to complete the square...but when I start that process the problem gets really nasty.

A push por favor?
• February 26th 2009, 08:48 PM
ThePerfectHacker
Quote:

A push
A push down a cliff? :)

Quote:

Originally Posted by silencecloak
$\int \frac{x^2}{(3+4x-4x^2)^{3/2}}$

I'm pretty sure I need to complete the square...but when I start that process the problem gets really nasty.

$3+4x-4x^2 = - (4x^2 -4x - 3) = - (4x^2 - 4x + 1 - 4)$.
Therefore, $-((2x-1)^2-4) = 4 - (2x-1)^2$.
• February 26th 2009, 09:03 PM
silencecloak
This problem is super confusing to me
$\int \frac{x^2}{(3+4x-4x^2)^{3/2}}$

=

$\int \frac{x^2}{(4-(2x-1)^2)^{3/2}}$

Would I set $u = 2sec\theta$

as my next step? And after that I have no clue
• February 26th 2009, 09:11 PM
ThePerfectHacker
Quote:

Originally Posted by silencecloak
This problem is super confusing to me
$\int \frac{x^2}{(3+4x-4x^2)^{3/2}}$

=

$\int \frac{x^2}{(4-(2x-1)^2)^{3/2}}$

Would I set $u = 2sec\theta$

as my next step? And after that I have no clue

First $t=2x-1$.
Second use a trignometric substitution.
• February 26th 2009, 09:16 PM
Soroban
Hello, silencecloak!

This is a nasty problem . . .

Quote:

$\int \frac{x^2\,dx}{(3+4x-4x^2)^{3/2}}$
Yes, we must complete the square . . .

We have: . $3 + 4x - 4x^2 \;=\;4 - 1 + 4x - 4x^2\;=\; 4 - (1 - 4x + 4x^2) \;=\;1 - (2x - 1)^2$

The integral becomes: . $\int\frac{x^2\,dx}{\left[1 - (2x-1)^2\right]^{\frac{3}{2}}}$

Let: $2x-1 \:=\:\sin\theta \quad\Rightarrow\quad 2\,dx \:=\:\cos\theta\,d\theta \quad\Rightarrow\quad dx \:=\:\tfrac{1}{2}\cos\theta\,d\theta$

. . and: . $x \:=\:\frac{\sin\theta + 1}{2}$

Substitute: . $\int\frac{\left(\frac{\sin\theta+1}{2}\right)^2} {(\cos\theta)^3}\left(\tfrac{1}{2}\cos\theta\,d\th eta\right) \;=\; \frac{1}{8}\int \frac{\sin^2\!\theta + 2\sin\theta + 1}{\cos^2\theta}\,d\theta$

. . . $= \;\frac{1}{8}\int\bigg[\frac{\sin^2\!\theta}{\cos^2\!\theta} + \frac{2\sin\theta}{\cos^2\!\theta} + \frac{1}{\cos^2\!\theta}\bigg]\,d\theta$

. . . $=\;\frac{1}{8}\int\bigg[\left(\frac{\sin\theta}{\cos\theta}\right)^2 + 2\left(\frac{1}{\cos\theta}\right)\left(\frac{\sin \theta}{\cos\theta}\right) + \left(\frac{1}{\cos\theta}\right)^2 \bigg]\,d\theta$

. . . $= \;\frac{1}{8} \int\bigg[\tan^2\!\theta + 2\sec\theta\tan\theta + \sec^2\!\theta\bigg]\,d\theta$

Can you finish it?

• February 26th 2009, 09:24 PM
silencecloak
Well now I'm just more confused

Two different answers from completing the square

@Soroban

you lost me at

$2x-1 \:=\:\sin\theta \quad\Rightarrow\quad 2\,dx