# Solve using trig sub: \int(\sec^2\theta\tan^2\theta)/\sqrt(9-\tan^2\theta)d\theta?

## Symbolab has it with u-sub, so... (Unless you have to do u-sub to get the trig-sub?)

Apr 30, 2018

$\frac{9}{2} {\sin}^{-} 1 \left(\frac{1}{3} \tan \theta\right) - \frac{1}{2} \tan \theta \sqrt{9 - {\tan}^{2} \theta} + C$

#### Explanation:

Substitute $\tan \theta = 3 \sin \phi$, to get ${\sec}^{2} \theta \setminus d \theta = 3 \cos \phi \setminus d \phi$ so that

$\setminus \int \frac{\setminus {\sec}^{2} \setminus \theta \setminus {\tan}^{2} \setminus \theta}{\setminus} \sqrt{9 - \setminus {\tan}^{2} \setminus \theta} d \setminus \theta = \int \frac{{\left(3 \sin \phi\right)}^{2} 3 \cos \phi \setminus d \phi}{3 \cos \phi}$
$q \quad = 9 \int \setminus {\sin}^{2} \phi \setminus d \phi = \frac{9}{2} \int \setminus \left(1 - \cos \left(2 \phi\right)\right) \setminus d \phi$
$q \quad = \frac{9}{2} \phi - \frac{9}{4} \sin \left(2 \phi\right) + C$
$q \quad = \frac{9}{2} \phi - \frac{9}{2} \sin \phi \cos \phi = C$
$q \quad = \frac{9}{2} {\sin}^{-} 1 \left(\frac{1}{3} \tan \theta\right) - \frac{9}{2} \frac{1}{3} \tan \theta \sqrt{1 - {\left(\frac{1}{3} \tan \theta\right)}^{2}} + C$
$q \quad = \frac{9}{2} {\sin}^{-} 1 \left(\frac{1}{3} \tan \theta\right) - \frac{1}{2} \tan \theta \sqrt{9 - {\tan}^{2} \theta} + C$

Apr 30, 2018

$- \frac{1}{2} \tan \theta \sqrt{9 - {\tan}^{2} \theta} + \frac{9}{2} a r c \sin \left(\frac{1}{3} \cdot \tan \theta\right) + C$.

#### Explanation:

As Second Method, let us solve this Problem without using

any Trigonometric Substitution.

If we take $\tan \theta = t , t h e n , {\sec}^{2} \theta d \theta = \mathrm{dt}$.

$\therefore I = \int \frac{{\tan}^{2} \theta {\sec}^{2} \theta}{\sqrt{9 - {\tan}^{2} \theta}} d \theta$,

$= \int {t}^{2} / \sqrt{9 - {t}^{2}} \mathrm{dt}$,

$= - \int \frac{\left(9 - {t}^{2}\right) - 9}{\sqrt{9 - {t}^{2}}} \mathrm{dt}$,

$= - \int \left\{\frac{9 - {t}^{2}}{\sqrt{9 - {t}^{2}}} - \frac{9}{\sqrt{9 - {t}^{2}}}\right\} \mathrm{dt}$,

$= - \int \sqrt{{3}^{2} - {t}^{2}} \mathrm{dt} + 9 \int \frac{1}{\sqrt{9 - {t}^{2}}} \mathrm{dt}$,

$= - \left\{\frac{t}{2} \sqrt{9 - {t}^{2}} + \frac{9}{2} a r c \sin \left(\frac{t}{3}\right)\right\} + 9 \arcsin \left(\frac{t}{3}\right)$,

$= - \frac{t}{2} \sqrt{9 - {t}^{2}} + \frac{9}{2} a r c \sin \left(\frac{t}{3}\right)$.

$\because t = \tan \theta ,$

$I = - \frac{1}{2} \tan \theta \sqrt{9 - {\tan}^{2} \theta} + \frac{9}{2} a r c \sin \left(\frac{1}{3} \cdot \tan \theta\right) + C$.

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