Evaluate this definite integral please I need help bad!?

Question

Integral from Pi/6 to Pi/2 for cos(x)/sin^9(x)dx! the last time I posted this is was answered wrong!

Answer 1

Substitution: Let
u = sin^9 x du = cos x dx
So we get ∫ u^-9 du = u^-8/-8 = sin ^-8 x / -8
Now plug in your limits
You get -1/8[ 1 - (1/2)^-8] = 1/8[2^8-1] = 255/8

Answer 2

First, you have to antidifferentiate the equation. You should know that cosx is the derivative of sinx and, if you would look closely, you could see that you can change the equation to sin^-9(x)cosx. That should help you see that the antiderivative is (sin^-8(x))/(-8) because the antiderivative of (f(x)^n) times fprime(x) is (f(x)^n+1)/(n+1). Now that you have the antiderivative, you evaluate it from pi/6 to pi/2. You should know from the unit circle that sin(pi/2) is 1 and 1^-8 all over -8 is -1/8. sin(pi/6) is .5 and .5^-8 all over -8, us a calculater, is -32. So, you subtract the two numbers, -1/8-(-32) and get 31 and 7/8.

Answer 3

∫cos(x)/sin^9(x) dx
u = sin(x)

changing our limits as well so we don't have to convert back
sin(pi/6) = 1/2
sin(pi/2) = 1

du/dx = cos(x)
dx = du/cos(x)

∫cos(x)*du/[cos(x)*u^-9] so the cos(x) cancel leaving us with just u's and du. good.

∫du/u^9
∫u^-9 du
(-1/8) u^-8
-1/(8u^8) + C

-1/(8u^8) evaluated from 1/2 to 1

-1/[8(1/2)^8] - 1/[8(1)^8]
-256/8 + 1/8
-255/8

-31.875