What's the integral of (x)^1/2 * sin^2(x^3/2 - 1)dx ?

Question

What's the integral of (x)^1/2 * sin^2(x^3/2 - 1)dx ? I know this is solved by substitution given u = x^3/2 - 1. I can't seem to get past that part.

Answer 1

∫ x^(1/2) sin^2[x^(3/2) - 1] dx =

as you correctly suggest, the proper substitution is:

x^(3/2) - 1 = u

differentiate both sides:

d[x^(3/2) - 1] = du  ÃƒÂ¢Ã‚†Â’

(3/2) x^[(3/2) -1] dx = du  ÃƒÂ¢Ã‚†Â’

(3/2) x^(1/2) dx = du  ÃƒÂ¢Ã‚†Â’

x^(1/2) dx (what actually appears in your integral) = (2/3)du

then, substituting, you get:

∫ sin^2[x^(3/2) - 1] x^(1/2) dx = ∫ sin^2u (2/3)du =

(2/3) ∫ sin^2u du =

according to half-angle identities, replace sin^2u with (1/2)[1 - cos(2u)]:

(2/3) ∫ (1/2)[1 - cos(2u)] du =

(2/3)(1/2) ∫ [1 - cos(2u)] du =

(1/3) ∫ [1 - cos(2u)] du =

(1/3) ∫ du - (1/3) ∫ cos(2u) du =

(1/3)u - (1/3) (1/2)sin(2u) + C =

(1/3)u - (1/6) sin(2u) + C

then, substituting back u = x^(3/2) - 1, you get:

∫ x^(1/2) sin^2[x^(3/2) - 1] dx = (1/3)[x^(3/2) - 1] - (1/6) sin{2[x^(3/2) - 1]} + C =

(1/3)x^(3/2) - (1/3) - (1/6) sin[2x^(3/2) - 2] + C

being - (1/3) a mere constant, you can include it into C, yielding:

∫ x^(1/2) sin^2[x^(3/2) - 1] dx = (1/3)x^(3/2) - (1/6) sin[2x^(3/2) - 2] + C

I hope it helps...

Bye!

Answer 2

u = x^(3/2) - 1

du = (3/2)x^(1/2) dx

x^(1/2)dx = (2/3) du

substitute:

∫(2/3) sin²u du

trig identity:

sin²u = 1/2 - (1/2)cos(2u)

∫(2/3) [1/2 - (1/2)cos(2u)] du

distribute:

∫1/3 - (1/3)cos(2u) du

(1/3)u - (1/6)sin(2u) + C

(1/3)[x^(3/2) - 1] - (1/6) sin[2(x^(3/2) - 1)] + C

(1/3)[x^(3/2) - 1] - (1/6) sin[2x^(3/2) - 2] + C

hope it helps!