# Thread: Am I on the right track?: Trigonometric Substitution

1. ## Am I on the right track?: Trigonometric Substitution

Use the indicated substitution to evaluate the integral

1
(integral) dx/(x^2)sqrt{(x^2)+4), x=2tan(θ)
1/2

first got the derivative of x=2tan(θ)-->dx=2sec^2 (θ) dθ

then I evaluated the denominator, which becomes 8tan^2(θ)secθ

1
(integral) dx/(x^2)sqrt{(x^2)+4) =
1/2

integral 1,1/2: 2sec^2(θ)dθ/8tan^2(θ)sec(θ) =

(1/4) integral 1,1/2: 1dθ/cos(θ)(sin(θ)/cos(θ))(tan(θ))

(1/4) integral 1,1/2: 1/sin(θ)tan(θ) dθ

(1/4 integral 1,1/2: csc(θ) dθ) x (1/4 integral 1,1/2: cot(θ) dθ)

1
(1/4)(ln|csc(θ)-cot(θ)|) x (ln|sin(θ)|) + C
1/2

Thanks!

2. Hello, wyhwang7!

You went off near the end . . .

Use the indicated substitution to evaluate the integral.
. . . $\displaystyle \int^1_{\frac{1}{2}} \frac{dx}{x^2\sqrt{x^2+4}} \quad x = 2\tan\theta$

First got the derivative of: $\displaystyle x\,=\,2\tan\theta \quad\Rightarrow\quad dx \,=\,2\sec^2\!\theta\,d\theta$

then I evaluated the denominator, which becomes: $\displaystyle 8\tan^2\!\theta\sec\theta$

$\displaystyle \int \frac{dx}{x^2\sqrt{x^2+4}} \;= \; \int \frac{2\sec^2\!\theta\,d\theta}{8\tan^2\!\theta\se c\theta} \;=\;\frac{1}{4}\int\frac{\sec\theta\,d\theta}{\ta n^2\!\theta}$

We have: .$\displaystyle \frac{1}{4}\int\frac{\left(\frac{1}{\cos\theta}\ri ght)}{\left(\frac{\sin^2\!\theta}{\cos^2\!\theta}\ right)} \,d\theta \;=\;\frac{1}{4}\int\frac{\cos\theta}{\sin^2\!\the ta}\,d\theta$

. . $\displaystyle =\;\frac{1}{4}\int\frac{1}{\sin\theta}\cdot\frac{\ cos\theta}{\sin\theta}\,d\theta \;=\;\frac{1}{4}\int\csc\theta\cot\theta\,d\theta \;=\;-\frac{1}{4}\csc\theta$

Back-substitute: .$\displaystyle 2\tan\theta \:=\:x\quad\Rightarrow\quad \tan\theta \:=\:\frac{x}{2} \:=\:\frac{opp}{adj}$

. . Then: .$\displaystyle hyp \,=\,\sqrt{x^2+4} \quad\Rightarrow\quad \csc\theta \:=\:\frac{\sqrt{x^2+4}}{x}$

Now we must evaluate: .$\displaystyle -\frac{1}{4}\cdot\frac{\sqrt{x^2+4}}{x}\,\bigg]^1_{\frac{1}{2}}$