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?
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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?
$\displaystyle \int \frac{1}{\sqrt{a^2 - x^2}}\, dx = \arcsin{\frac{x}{a}} + C$.Quote:
Originally Posted by Retromingent
In this case, $\displaystyle a = 4 $.
So $\displaystyle \int \frac{1}{\sqrt{16 - x^2}}\, dx = \int \frac{1}{\sqrt{4^2 - x^2}}\, dx = \arcsin {\frac{x}{4}} + C.$
