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[SOLVED] Trigonometric Substitutions

karush

Well-known member
Jan 31, 2012
2,716
$\displaystyle
\int {\frac{\sqrt{x^2-9}}{x}}\ dx
$
using
$\displaystyle
x=3\sec{\theta}\ \ \ dx=3\sin{\theta}\sec^2{\theta}\ d\theta
$
so then
$\displaystyle
\int {\frac{3\tan{\theta}}{3\sec{\theta}}}\ 3\sin{\theta}\sec^2{\theta}\ d\theta
\Rightarrow 3\int {\tan^2{\theta}}\ d\theta
$

the answer to this is
$\displaystyle
\sqrt{x^2-9}-3\sec^{-1}\left(x/3\right)+C
$

but after trying about 5 times can't seem to arrive at it..:confused:
 

ThePerfectHacker

Well-known member
Jan 26, 2012
236
Re: trig substitutions

so then
$\displaystyle
\int {\frac{3\tan{\theta}}{3\sec{\theta}}}\ 3\sin{\theta}\sec^2{\theta}\ d\theta
\Rightarrow 3\int {\tan^2{\theta}}\ d\theta
$

$$ \tan^2 \theta = \sec^2\theta - 1 $$
 

karush

Well-known member
Jan 31, 2012
2,716
Re: trig substitutions

$$ \tan^2 \theta = \sec^2\theta - 1 $$
so
$\displaystyle 3\int \sec^2(\theta)-1
\Rightarrow
3\left[\tan{\theta}-\theta\right]
\Rightarrow
3\left[\frac{\sqrt{x^2-9}}{3}-\sec^{-1}{\frac{x}{3}}\right]
\Rightarrow
\sqrt{x^2-9}-3\sec^{-1}\left(x/3\right)+C
$

however why couldn't we use $\tan^2{\theta} $
 

ThePerfectHacker

Well-known member
Jan 26, 2012
236
Re: trig substitutions

however why couldn't we use $\tan^2{\theta} $
You need to compute anti-derivative of $\tan^2 \theta$. The standard way to do this is to use identity involving $\sec^2\theta$.