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?
Hello, fattydq!
$\displaystyle \int\frac{x}{4-x^2}\,dx$ . . . . using a trig substitution.
Let $\displaystyle x \:=\:2\sin\theta \quad\Rightarrow\quad dx \:=\:2\cos\theta\,d\theta$
. . and: .$\displaystyle 4 - x^2 \:=\:4 - (2\sin\theta)^2 \:=\:4-4\sin^2\!x \:=\:4(1-\sin^2\!x) \:=\:4\cos^2\!x$
Substitute: .$\displaystyle \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: .$\displaystyle 2\sin\theta \:=\:x \quad\Rightarrow\quad \sin\theta \:=\:\frac{x}{2} \:=\:\frac{opp}{hyp}$
We have: .$\displaystyle opp = x,\;hyp = 2\quad\Rightarrow\quad adj = \sqrt{4-x^2}$
. . Hence: .$\displaystyle \cos\theta \:=\:\frac{adj}{hyp} \:=\:\frac{\sqrt{4-x^2}}{2}$
Then [1] becomes: .$\displaystyle -\ln\left(\frac{\sqrt{4-x^2}}{2}\right) + C \;=\;-\bigg[\ln(\sqrt{4-x^2}) - \ln(2)\bigg] + C$
. . . . . . $\displaystyle = \;-\ln\left(\sqrt{4-x^2}\right) + \underbrace{\ln(2) + C}_{\text{This is a constant}} \;=\; -\ln\left(\sqrt{4-x^2}\right) + C$
. . . . . . $\displaystyle = \;-\ln\left(4-x^2\right)^{\frac{1}{2}} + C \;=\;-\tfrac{1}{2}\ln\left(4-x^2\right) + C$