[SOLVED] Solving This Indefinite Integral With A Trig Substitution

**fattydq** · April 5th 2009, 08:57 AM

The problem is the integral of x/(4-x^2), and I'm asked to solve it using a trig substitution despite there being no square root present in the problem, how can I go about doing this?

**Soroban** · April 5th 2009, 09:28 AM

Hello, fattydq!

$\int\frac{x}{4-x^2}\,dx$ . . . . using a trig substitution.

Let $x \:=\:2\sin\theta \quad\Rightarrow\quad dx \:=\:2\cos\theta\,d\theta$

. . and: . $4 - x^2 \:=\:4 - (2\sin\theta)^2 \:=\:4-4\sin^2\!x \:=\:4(1-\sin^2\!x) \:=\:4\cos^2\!x$

Substitute: . $\int\frac{2\sin\theta}{4\cos^2\!\theta}\,(2\cos\th eta\,d\theta) \;=\;\int\frac{\sin\theta}{\cos\theta}\,d\theta \;=\;\int\tan\theta\,d\theta \;=\;-\ln(\cos\theta) + C$ .[1]

Bacl-substitute: . $2\sin\theta \:=\:x \quad\Rightarrow\quad \sin\theta \:=\:\frac{x}{2} \:=\:\frac{opp}{hyp}$

We have: . $opp = x,\;hyp = 2\quad\Rightarrow\quad adj = \sqrt{4-x^2}$

. . Hence: . $\cos\theta \:=\:\frac{adj}{hyp} \:=\:\frac{\sqrt{4-x^2}}{2}$

Then [1] becomes: . $-\ln\left(\frac{\sqrt{4-x^2}}{2}\right) + C \;=\;-\bigg[\ln(\sqrt{4-x^2}) - \ln(2)\bigg] + C$

. . . . . . $= \;-\ln\left(\sqrt{4-x^2}\right) + \underbrace{\ln(2) + C}_{\text{This is a constant}} \;=\; -\ln\left(\sqrt{4-x^2}\right) + C$

. . . . . . $= \;-\ln\left(4-x^2\right)^{\frac{1}{2}} + C \;=\;-\tfrac{1}{2}\ln\left(4-x^2\right) + C$

**Jhevon** · April 5th 2009, 10:03 AM

Originally Posted by fattydq

The problem is the integral of x/(4-x^2), and I'm asked to solve it using a trig substitution despite there being no square root present in the problem, how can I go about doing this?

for the record, if the method of integration was not specified, a regular u-substitution of $u = 4 - x^2$ would take care of this guy with much less difficulty.

**fattydq** · April 5th 2009, 10:14 AM

Originally Posted by Jhevon

for the record, if the method of integration was not specified, a regular u-substitution of $u = 4 - x^2$ would take care of this guy with much less difficulty.

Yep, I actually had to solve it that way first, and then in part 2 I was asked to solve it with a trig substitution, which did indeed prove to be much more painful

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