Hi. I'm wondering how come the answer has an 8 in the denominator. My answer has 2, although the nominator remains the same?
u = (sin(4x)^-1)
du = (1 - (16x^2))^(-1/2)
n = 1
n + 1 = 2
any hints, thanks?
I'm not completely sure you have the method. Just to verify, you are using the substitution $\displaystyle x = \frac{sin( \theta )}{4}$ right? In that case the integral turns into
$\displaystyle \int \frac{sin^{-1} ( sin( \theta ))}{4}d \theta = \int \frac{ \theta }{4} d \theta$
where the extra 2 comes from the integration.
-Dan