integral help! can you solve this integral?

∫ 17 [ cos^5(x) dx / √(sin x)

= ∫ 17 [ cos^4(x) ( cos x dx) / √(sin x)

= 17 ∫ [1 - sin^2(x) ]^2 ( cos x dx) / √(sin x)

let sin x = u

cos x dx = du

now the integral becomes

17 ∫ [1 - u^2) ]^2 du / √u

= 17 ∫ [1 + u^4 - 2u^2 ] du / √u

= 17 ∫(u)^(- 1/2) du + 17 ∫ u^(7/2) du - 34 ∫ u^(3/2) du

= 17 * 2√u + 17(2/9) (u)^(9/2) - 34 (2/5) u^(5/2) + c

= 34√u + (34/9) (u)^(9/2) - (68 / 5) u^(5/2) + c

back substitute u = sin x

34√sin x + (34/9) (sin x)^(9/2) - (68 / 5) (sin x) ^(5/2) + C

You don't "solve" integrals. You solve equations. You evaluate integrals.

i) Try writing cos^5(x) as cos(x) (1-sin^2(x)]^2. and try u = sin(x)

ii) similar, but try u = sqrt(sin(x)) instead (if the first doesn't get you there)

lucky

you are missing a delimiter

17/180sin(x)^(1/2)(52cos(2x) + 5cos(4x) + 303)

chente