\magnification=\magstep1
\centerline{Mathematics 136 -- AP Calculus}
\centerline{Summary of Trigonometric Substitutions}
\centerline{Fall 2009}
\bigskip
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:
$$\eqalign{\sqrt{a^2 - (a\sin\theta)^2} &= a\cos\theta\cr
           \sqrt{(a\tan\theta)^2 + a^2} &= a\sec\theta\cr
           \sqrt{(a\sec\theta)^2 - a^2} &= a\tan\theta\cr}$$
Hence, 
\bigskip
\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 $a \tan\theta$.
\bigskip
\noindent
Then we substitute for the rest of the integral, integrate
the resulting trigonometric form, and convert
back to the original variable. 
\bigskip
\noindent
{\it Two Examples}
\bigskip
\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 reduction formula for powers of $\sin$ (this is \# 73 in Stewart's table
of integrals from the Reference Pages handout from last week):
$$a^2 \int \sin^2\theta\ d\theta = {-a^2\over 2}\sin\theta\cos\theta + {a^2\over 2}\int\ d\theta = 
{-a^2\over 2}\sin\theta\cos\theta + {a^2 \over 2}\theta + C.$$
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
$$\int {u^2\over \sqrt{a^2 - u^2}}\ du = {-1\over 2}u\sqrt{a^2 - u^2} + {a^2\over 2}\arcsin(u/a)
 + C.$$ 
Note that the final answer here matches \# 34 in the table of integrals.
\bigskip
\noindent
B)  Compute $\int{dx\over x\sqrt{x^2 + 16}}$.  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:
$$\int {4\sec^2\theta\ d\theta \over 4\tan\theta\cdot 4 \sec\theta}
={1\over 4}\int {\sec\theta\over\tan\theta}\ d\theta = {1\over 4}\int {1\over \sin\theta}\ d\theta$$
Since $1/\sin(\theta) = \csc(\theta)$, from \# 15 in Stewart's table of integrals,
$${1\over 4}\int \csc(\theta) \ d\theta = 
{1\over 4}\ln\left| \csc(\theta)- \cot(\theta)\right| + C$$
Then, from $x = 4\tan\theta$, we get $\cos\theta = {4\over\sqrt{x^2+16}}$ and 
$\sin\theta = {x\over\sqrt{x^2+16}}$.  Hence
$$\csc(\theta) = {\sqrt{x^2+16}\over x} \quad {\rm and} \quad \cot(\theta) = {4\over x}.$$
The integral equals:
$${1\over 4} \ln\left| {\sqrt{x^2+16}\over x} - {4\over x} \right| + C.$$

\bye
