# Thread: Trigonometric Substitutions (fixed)

1. ## Trigonometric Substitutions (fixed)

(a) If we use the substitution t = 6tan(θ), then which of the following integrals is equivalent to ?

1

(b) Use the integral in part (a) to evaluate

Please show all steps and tank you!

2. Sorry.
Can't believe I did that.

3. Hello, choshiwara!

(a) If we use the substitution $t = 6\tan\theta$,
then which of the following integrals is equivalent to: . $\int\frac{dt}{t^2\sqrt{t^2+36}}$

$(1)\;\int\frac{\sec\theta\,d\theta}{36\tan^2\!\the ta} \qquad (2)\;\int\frac{d\theta}{216\tan\theta\sec\theta} \qquad(3)\;\int\frac{d\theta}{36\tan^2\!\theta} \qquad(4)\;\int\frac{\sec^2\!\theta\,d\theta}{36\t an^2\!\theta}$

Let: $t \:=\:6\tan\theta \quad\Rightarrow\quad dt \:=\:6\sec^2\!\theta\,d\theta \quad\Rightarrow\quad \sqrt{t^2+36} \:=\:6\sec\theta$

Substitute: . $\int\frac{\overbrace{6\sec^2\!\theta\,d\theta}^{dt }}{\underbrace{36\tan^2\!\theta}_{t^2}\cdot\underb race{6\sec\theta}_{\sqrt{t^2+36}}} \;\;=\;\; \int\frac{\sec\theta\,d\theta}{36\tan^2\!\theta}$ . . . answer choice (1)

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We have: . $\frac{1}{36}\int\frac{\sec\theta}{\tan^2\!\theta}\ ,d\theta \;=\;\frac{1}{36}\int\frac{\frac{1}{\cos\theta}}{\ frac{\sin^2\!\theta}{\cos^2\!\theta}}d\theta \;=\;\frac{1}{36}\int\frac{\cos\theta}{\sin^2\!\th eta}\,d\theta$

. . $= \;\frac{1}{36}\int\frac{1}{\sin\theta}\!\cdot\frac {\cos\theta}{\sin\theta}\,d\theta \;=\;\frac{1}{36}\int\csc\theta\cot\theta\,d\theta \;=\;-\frac{1}{36}\csc\theta + C$ .[1]

Back-substitute: . $t \:= \:6\tan\theta \quad\Rightarrow\quad \tan\theta \:=\:\frac{t}{6} \:=\:\frac{opp}{adj}$

So $\theta$ is in a right triangle with: . $opp = t,\;adj = 6$
Using Pythagorus, we find that: . $hyp = \sqrt{t^2+36}$
. . Hence: . $\csc\theta \:=\:\frac{\sqrt{t^2+36}}{t}$

Therefore, [1] becomes: . $-\frac{\sqrt{t^2+36}}{36t} + C$