# Thread: Basic integration problem, general power formula

1. ## Basic integration problem, general power formula

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?

2. ## Re: Basic integration problem, general power formula

Originally Posted by togo

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 $x = \frac{sin( \theta )}{4}$ right? In that case the integral turns into
$\int \frac{sin^{-1} ( sin( \theta ))}{4}d \theta = \int \frac{ \theta }{4} d \theta$

where the extra 2 comes from the integration.

-Dan

3. ## Re: Basic integration problem, general power formula

I would prefer to make substitution sin^-1 ( 4x) = t that gives [1/sqrt(1-16 x^2) ] *4 dx =dt Thus we have [1/sqrt(1-16 x^2) ] dx =dt/4
now the integral becomes t/4 dt. Now you can take it further.