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?

Printable View

- Sep 24th 2008, 09:16 PMRetromingentTrig Substitution
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? - Sep 24th 2008, 09:21 PMProve It