Calc Solutions: Day 3 Trig Substitution

Citation: "Calculus Early Transcendentals, 6th Edition", by James Stewart Section 7.3 Trigonometric Substitution Integrals Problem 4:

$\int \frac{dx}{\sqrt{16-x^2}}$
Solution: This problem will be a very basic and instructive example of integrating with trig substitutions. The key to doing these problems is the following identity:

$Identity: sin^2(\theta)+cos^2(\theta)=1$
Suppose we make a substitution for x:
$$x=4sin(\theta)$
Then,
$\int \frac{dx}{\sqrt{16-x^2}}=\int \frac{4cos(\theta)d\theta}{\sqrt16-16sin^2(\theta)}=\int \frac{cos(\theta)d\theta}{\sqrt{1-sin^2(\theta)}}=\int \frac{cos(\theta)}{cos(\theta)}= \theta$
Noting that the following comes from the Pythagorean Identity written above:
$$cos^2(\theta)=1-sin^2(\theta)$
Finally, we rewrite theta in terms of x.
$\theta=sin^{-1}(x)$

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Calc Solutions

Sunday, January 22, 2012

Day 3 Trig Substitution

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