Trigonometric Substitions

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We practice executing trigonometric substitutions.

(Video) Calculus: Single Variable

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Examples

(c.f. APEX Calculus Example 6.4.2) Compute the indefinite integral

$\int \frac {dx}{\sqrt {5 + x^2}}.$

The structure of this integrand is adapted to a trigonometric substitution of
type, so we let $x = \sqrt {5} \answer {\tan \theta }$ (type out the word theta). This gives $dx = \sqrt {5} \answer {\sec ^2 \theta }~d \theta$ .
We use the identity $\sec ^2 \theta = \tan ^2 \theta + 1$ , the quantity $5 + x^2$ can be rewritten using our substitution to equal $\answer {5 \sec ^2 \theta }$ . Therefore $\int \frac {dx}{\sqrt {5+x^2}} dx = \int \answer {\sec \theta } \, d \theta = \answer {\ln | \sec \theta + \tan \theta |} + C.$ (When writing out the indefinite integral, don’t forget absolute values.)
We construct a reference triangle compatible with our substitution. In this case, that means we need a right triangle with an angle $\theta$ which satisfies $x/\sqrt {5} = \tan \theta$ :
Using this reference triangle, we get that $\tan \theta = \frac {x}{\sqrt {5}} \qquad \sec \theta = \answer { \frac {\sqrt {x^2+5}}{\sqrt {5}}}.$ Using these formulas, we can write $\ln |\sec \theta + \tan \theta | = \ln \left | \answer {\frac {x}{\sqrt {5}} + \frac {\sqrt {x^2+5}}{\sqrt {5}}} \right | .$
We conclude that $\int \frac {dx}{\sqrt {x^2+5}} = \ln | x + \sqrt {x^2 + 5} | + C.$ (Note that the factor of $\sqrt {5}$ in the denominator of the logarithm is not necessary when we already have an arbitrary constant $C$ in our answer.)

Compute the indefinite integral

$\int \frac {dx}{x^2 \sqrt {x^2 - 3}}.$

The structure of this integrand is adapted to a trigonometric substitution of
type, so we let $x = \sqrt {3} \answer {\tan \theta }$ . This gives $dx = \sqrt {3} \answer {\sec \theta \tan \theta }~d \theta$ .
We use the identity $\sec ^2 \theta = \tan ^2 \theta + 1$ , the quantity $x^2 - 3$ can be rewritten using our substitution to equal $\answer {3 \tan ^2 \theta }$ . Therefore $\int \frac {dx}{x^2 \sqrt {x^2 - 3}} = \int \answer { \frac {\sqrt {3} \sec \theta \tan \theta }{3 \sec ^2 \theta \sqrt {3} \tan \theta }} d \theta = \answer {\frac {\sin \theta }{3}} + C.$
We construct a reference triangle compatible with our substitution. In this case, that means we need a right triangle with an angle $\theta$ which satisfies $x/\sqrt {3} = \sec \theta$ :
Using this reference triangle, we get that $\sin \theta = \answer {\frac {\sqrt {x^2-3}}{x}}.$
We conclude that $\int \frac {dx}{x^2 \sqrt {x^2-3}} = \answer { \frac {\sqrt {x^2-3}}{3x}} + C.$

(c.f. APEX Calculus Example 6.4.1) Compute the indefinite integral

$\int \sqrt {9 - x^2} dx.$

The structure of this integrand is adapted to a trigonometric substitution of
type, so we let $x = 3 \answer {\sin \theta }$ . This gives $dx = 3 \answer {\cos \theta }~d \theta$ .
Using the identity $\cos ^2 \theta + \sin ^2 \theta = 1$ , the expression $9 - x^2$ can be rewritten using the formula above as $\answer {(3 \cos \theta )^2}$ . Therefore $\int \sqrt {9-x^2} dx = \int \answer {9 (\cos \theta )^2} d \theta .$
We use the power reduction formula $\cos ^2 t = (1 + \cos 2t)/2$ and conclude $\int \sqrt {9-x^2} dx = \frac {9}{2} \int \left ( 1 + \cos 2 \theta \right ) d \theta = \frac {9}{2} \left [ \theta + \frac {1}{2} \sin 2 \theta \right ] + C.$
Now we construct a reference triangle which is compatible with the substitution we made. In this case, this means we need a right triangle for which $x/3 = \sin \theta$ :
Using the reference triangle above, we have the identities $\sin \theta = \frac {x}{3} \qquad \cos \theta = \answer { \frac {\sqrt {9-x^2}}{3}} \qquad \theta = \arcsin \frac {x}{3}.$
The quantity $\sin 2 \theta$ is difficult to write as a function of $x$ as it stands, so we first use the double-angle formula $\sin 2 \theta = \cos \theta \sin \theta$ . We therefore conclude that $\int \sqrt {9-x^2} dx = \answer { \frac {9}{2} \left ( \arcsin \frac {x}{3} + \frac {x \sqrt {9-x^2}}{9} \right )} + C.$ (Note: Ximera will interpret $\sin ^{-1} x$ as $1/(\sin x)$ , so write $\arcsin$ when you mean inverse sine.)

Press...	...to do
left/right arrows	Move cursor
shift+left/right arrows	Select region
ctrl+a	Select all
ctrl+x/c/v	Cut/copy/paste
ctrl+z/y	Undo/redo
ctrl+left/right	Add entry to list or column to matrix
shift+ctrl+left/right	Add copy of current entry/column to to list/matrix
ctrl+up/down	Add row to matrix
shift+ctrl+up/down	Add copy of current row to matrix
ctrl+backspace	Delete current entry in list or column in matrix
ctrl+shift+backspace	Delete current row in matrix

Type...	...to get
norm	$\|\|\blue{[?]}\|\|$
text	$\text{\blue{[?]}}$
sym_name	$\backslash\texttt{\blue{[?]}}$
abs	$\left\|\blue{[?]}\right\|$
sqrt	$\sqrt{\blue{[?]}}$
paren	$\left(\blue{[?]}\right)$
floor	$\lfloor \blue{[?]} \rfloor$
factorial	$\blue{[?]}!$
exp	${\blue{[?]}}^{\blue{[?]}}$
sub	${\blue{[?]}}_{\blue{[?]}}$
frac	$\dfrac{\blue{[?]}}{\blue{[?]}}$
int	$\displaystyle\int{\blue{[?]}}d\blue{[?]}$
defi	$\displaystyle\int_{\blue{[?]}}^{\blue{[?]}}\blue{[?]}d\blue{[?]}$
deriv	$\displaystyle\frac{d}{d\blue{[?]}}\blue{[?]}$
sum	$\displaystyle\sum_{\blue{[?]}}^{\blue{[?]}}\blue{[?]}$
prod	$\displaystyle\prod_{\blue{[?]}}^{\blue{[?]}}\blue{[?]}$
root	$\sqrt[\blue{[?]}]{\blue{[?]}}$
vec	$\left\langle \blue{[?]} \right\rangle$
mat	$\left(\begin{matrix} \blue{[?]} \end{matrix}\right)$
*	$\cdot$
infinity	$\infty$
arcsin	$\arcsin\left(\blue{[?]}\right)$
arccos	$\arccos\left(\blue{[?]}\right)$
arctan	$\arctan\left(\blue{[?]}\right)$
sin	$\sin\left(\blue{[?]}\right)$
cos	$\cos\left(\blue{[?]}\right)$
tan	$\tan\left(\blue{[?]}\right)$
sec	$\sec\left(\blue{[?]}\right)$
csc	$\csc\left(\blue{[?]}\right)$
cot	$\cot\left(\blue{[?]}\right)$
log	$\log\left(\blue{[?]}\right)$
ln	$\ln\left(\blue{[?]}\right)$
alpha	$\alpha$
beta	$\beta$
gamma	$\gamma$
delta	$\delta$
epsilon	$\epsilon$
zeta	$\zeta$
eta	$\eta$
theta	$\theta$
iota	$\iota$
kappa	$\kappa$
lambda	$\lambda$
mu	$\mu$
nu	$\nu$
xi	$\xi$
omicron	$\omicron$
pi	$\pi$
rho	$\rho$
sigma	$\sigma$
tau	$\tau$
upsilon	$\upsilon$
phi	$\phi$
chi	$\chi$
psi	$\psi$
omega	$\omega$
Gamma	$\Gamma$
Delta	$\Delta$
Theta	$\Theta$
Lambda	$\Lambda$
Xi	$\Xi$
Pi	$\Pi$
Sigma	$\Sigma$
Phi	$\Phi$
Psi	$\Psi$
Omega	$\Omega$

(Video) Calculus: Single Variable

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