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