Trig Substitution

**Retromingent** · September 24th 2008, 09:16 PM

The problem:
integrate 1/sqr(16-x^2) dx

My answer: (@ = theta) (trig substitution)
x=sin @ a = 16 dx = cos @ sqr(16-x^2) = cos @
= integrate cos @ / cos @ = 1

The book's answer:
arcsin(x/4)+c

Where did I go wrong?

**Prove It** · September 24th 2008, 09:21 PM

Originally Posted by Retromingent

The problem:
integrate 1/sqr(16-x^2) dx

My answer: (@ = theta) (trig substitution)
x=sin @ a = 16 dx = cos @ sqr(16-x^2) = cos @
= integrate cos @ / cos @ = 1

The book's answer:
arcsin(x/4)+c

Where did I go wrong?

$\int \frac{1}{\sqrt{a^2 - x^2}}\, dx = \arcsin{\frac{x}{a}} + C$ .

In this case, $a = 4$ .

So $\int \frac{1}{\sqrt{16 - x^2}}\, dx = \int \frac{1}{\sqrt{4^2 - x^2}}\, dx = \arcsin {\frac{x}{4}} + C.$

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