Integration by Trigonometric Substitution

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Trignomometric Substitution

Expression

$\sqrt{a^2-x^2}$

 

Substitution Needed

$\begin{aligned}x&=a\sin\theta\\dx&=a\cos\theta d\theta\end{aligned}$

Identity Needed

$1-\sin^2\theta=\cos^2\theta$

Right Angle Triangle

$\sin\theta=\frac{opp}{hyp}=\frac xa$

$\sqrt{a^2+x^2}$

$\begin{aligned}x&=a\tan\theta\\dx&=a\sec^2\theta d\theta\end{aligned}$

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

$\tan\theta=\frac{opp}{adj}=\frac xa$

$\sqrt{x^2-a^2}$

$\begin{aligned}x&=a\sec\theta\\dx&=a\sec\theta\tan\theta d\theta\end{aligned}$

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

$\sec\theta=\frac{hyp}{adj}=\frac xa$

Example. Compute $\int\frac{\sqrt{x^2-9}}xdx.$ 

Solution. $\mathrm{Let~}x=3\sec\theta.\text{ Then, }dx=3\sec\theta\tan\theta d\theta$

The trigonometric identity needed is $\sec^2\theta-1=\tan^2\theta.$

The right angle triangle needed is:

 

 

$\begin{aligned} \int\frac{\sqrt{x^2-9}}xdx& =\int\frac{\sqrt{(3\sec\theta)^2-9}}{3\sec\theta}3\sec\theta\tan\theta d\theta   \\ &=\int\sqrt{9\sec^2\theta-9}\tan\theta d\theta  \\ &=\int\sqrt{9\tan^2\theta}\tan\theta d\theta  \\ &=\int3\tan\theta\tan\theta d\theta=3\int\tan^2\theta d\theta  \\ &=3\int(\sec^2\theta-1)d\theta=3(\tan\theta-\theta)+C \\ &=3{\left(\frac{\sqrt{x^2-9}}3-arcsec(x/3)\right)}+C \end{aligned}$

$\text{Note that }\theta=\mathrm{arcsec}(x/3)\text{ (from the substitution equation)}.$