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\centerline{Mathematics 136 -- Calculus 2}
\centerline{Summary of Trigonometric Substitutions}
\centerline{October 17 and 18, 2016}
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The trigonometric substitution method handles many integrals 
containing expressions like
$$\sqrt{a^2 - x^2}, \sqrt{x^2 + a^2}, \sqrt{x^2 - a^2}$$
(possibly including expressions without the square roots!)
The basis for this approach is the trigonometric identities
$$\eqalign{1 &= \sin^2 \theta + \cos^2 \theta\cr
        \Rightarrow \sec^2\theta &= \tan^2\theta + 1.\cr}$$
from which we derive other related identities:
$$\eqalign{\sqrt{a^2 - (a\sin\theta)^2} &= \sqrt{a^2(1 - \sin^2\theta)} = \sqrt{a^2\cos^2\theta}= a\cos\theta\cr
           \sqrt{(a\tan\theta)^2 + a^2} &= \sqrt{a^2(\tan^2\theta + 1)} = \sqrt{a^2\sec^2\theta}= a\sec\theta\cr
           \sqrt{(a\sec\theta)^2 - a^2} &= \sqrt{a^2(\sec^2\theta - 1)} = \sqrt{a^2\tan^2\theta}= a\tan\theta\cr}$$
(Technical note:  We usually assume that $a > 0$ and $0 < \theta < \pi/2$ here so that all the trig
functions take positive values.)  Hence, 
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\item{1.}  If our integral contains $\sqrt{a^2 - x^2}$,
the substitution $x=a \sin\theta$ will convert this
radical to the simpler form $a \cos\theta$.
\item{2.}  If our integral contains $\sqrt{x^2 + a^2}$,
the substitution $x=a \tan\theta$ will convert this
radical to the simpler form $a \sec\theta$.
\item{3.}  If our integral contains $\sqrt{x^2 - a^2}$,
the substitution $x=a \sec\theta$ will convert this
radical to the simpler form $a \tan\theta$.
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\noindent
We substitute for the rest of the integral {\it including the $dx$}.
For instance if $x = a\sin\theta$, then the $dx = a\cos\theta\ d\theta$.
If $x = a\tan\theta$, then $dx = a\sec^2\theta\ d\theta$.  If 
$x = a \sec\theta$, then $dx = a\sec\theta\tan\theta\ d\theta$.
All these substitutions should produce an integral with no $x$ terms
left -- everything expressed in terms of $\theta$.   If desired, limits
of integration on definite integrals can also be converted to equivalent
$\theta$-values.

We then integrate the resulting trigonometric form using the trig methods discussed
last time, and convert back to the original variable (or substitute limits of 
integration in terms of $\theta$). 
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\noindent
{\it Two Worked Examples}
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\noindent
A)  Compute $\int {u^2\over \sqrt{a^2 - u^2}}\ du$.
Solution:  The $\sqrt{a^2 - u^2}$ tells us that we want the 
sine substitution:  $u = a\sin\theta$.  Then $du = a\cos\theta\ d\theta$,
and the integral becomes:
$$\int {a^3 \sin^2\theta \cos\theta\ d\theta\over a\cos\theta}
=a^2 \int \sin^2\theta\ d\theta$$
We apply the half-angle formula $\sin^2\theta = {1\over 2}(1 - \cos(2\theta))$ to this
and integrate
$$a^2 \int {1\over 2}(1 - \cos(2\theta))\, d\theta$$
to yield 
$$= {a^2\over 2}\theta - {a^2\over 4} \sin(2\theta) 
=  {a^2\theta\over 2}  - {a^2\over 2} \sin\theta\cos\theta + C.$$
Then, we convert back to functions of $u$ using the substitution
equation $u = a\sin\theta$.  From this, 
$$\theta = \arcsin(u/a), \qquad \cos\theta = \sqrt{a^2 - u^2}/a,
\qquad \sin\theta = u/a$$
so the integral is
$$\int {u^2\over \sqrt{a^2 - u^2}}\ du = {a^2\over 2}\arcsin(u/a) - {1\over 2}u\sqrt{a^2 - u^2} + 
 + C.$$ 
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\noindent
B)  Compute $\int{dx\over \sqrt{x^2 + 16}}$.  Solution:
The $\sqrt{x^2 + 16}$ indicates that we want the tangent
substitution $x = 4\tan\theta$.  Then $dx = 4\sec^2\theta\ d\theta$
and the integral becomes:
$$\int {4\sec^2\theta\ d\theta \over 4 \sec\theta}
= \int \sec\theta\ d\theta.$$
As we saw last time, this form can be integrated as follows:
$$ \ln\left| \sec\theta + \tan\theta\right| + C$$
Then, from $x = 4\tan\theta$, we get $\sec\theta = {\sqrt{x^2+16}\over 4}$ and 
$\tan\theta = {x\over 4}$.  Hence the integral equals:
$$\ln\left| {\sqrt{x^2+16}\over x} + {4\over x} \right| + C.$$

Note:  If this was a definite  integral, say
$$\int_0^4 {dx\over \sqrt{x^2 + 16}},$$
then the equivalent limits of integration in terms of $\theta$ would be
found from the formula $\theta = \tan^{-1}(x/4)$, so
$x = 0 \Leftrightarrow \theta = 0$ and $x = 4 \Leftrightarrow \theta = \pi/4$.
If we used these values, then we could apply the Evaluation Theorem to the 
trig integral, and it would not be necessary to convert back to the antiderivative
in terms of $x$.
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\noindent
{\it Practice Problems}
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\noindent
A.  $$\int x^2 \sqrt{25 - x^2}\ dx$$
(use the half-angle formulas to do the trigonometric integral you get).
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\noindent
B.  $$\int {1\over (x^2 + 1)^{3/2}}\ dx$$
(if you don't see how to integrate the trigonometric form, try writing it using
sines and cosines).  
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\noindent
C.  $$\int {1\over (x^2 - 4)^{3/2}}\ dx$$
using the substitution $x = 2\sec(\theta)$.  You will also want to write this
one using sines and cosines(!)
\bye
