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?
Follow Math Help Forum on Facebook and Google+
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? . In this case, . So
View Tag Cloud