Question

Evaluate the definite integral: cos(x)/sin^9(x) on the interval from [pi/6,pi/2]

Answer 1

\[\text{Let } u = sin(x) \implies du = cos(x)\ dx\]

Should be pretty straight forward from there.

Answer 2

does sin^9(x) become sin^-9(x)?

Answer 3

If you like, sure. It actually becomes \(u^{-9}\)

Answer 4

so i will come to get -1/8(sin(x)^-8 and i will plug in pi/2 and pi/6 from there?

Answer 5

\[\text{Let } u = sin(x) \implies du = cos(x)\ dx\]\[\large\implies \int_{\pi/6}^{\pi/2}\frac{cos(x)}{sin^9(x)}\ dx\]\[\large=\int_{1/2}^1\frac{1}{u^9}\ du\]\[\large = \int_{1/2}^1 u^{-9} \]\[\large= -\left.\frac{1}{8u^8}\right|^1_{1/2}\]

Answer 6

Or yeah, plugging in the original limits is also fine if you convert back to the original function instead of using u.