Integrate [sqrt(sinx+1)]cosx dx from pi to 3pi/2

If you can solve this for me I will love you forever (I'm reviewing for a test)

FYI, the answer is -2/3 I just don't understand how to get there.

Start out by substituting:
u = sin(x) + 1 ==> du = cos(x) dx.

So, we have:
∫ cos(x)√[sin(x) + 1] dx
= ∫ √[sin(x) + 1] [cos(x) dx], by re-arranging
= ∫ √u du, by applying substitutions
= (2/3)u^(3/2) + C, by integrating
= (2/3)[sin(x) + 1]^(3/2) + C, by back-substituting u = sin(x) + 1.

Applying the limits:
∫ cos(x)√[sin(x) + 1] dx (from x=π to 3π/2)
= (2/3)[sin(x) + 1]^(3/2) (evaluated from x=π to 3π/2)
= (2/3){[sin(3π/2) + 1]^(3/2) - [sin(π) + 1]^(3/2)}
= (2/3)[(-1 + 1)^(3/2) - (0 + 1)^(3/2)]
= (2/3)(0 - 1)
= -2/3, as required.

I hope this helps!

substitute 1+ sin x = u

cosx dx = du

when x = pi u = 1+sin x = 1+sin pi = 1
when x = 3pi/2 u = 1+sin x = 1+ sin 3pi/2 = 0

we get

∫ [sqrt(sinx+1)]cosx dx becomes ∫ sqrt(u) du from 1 to 0


∫ u^(1/2) du = u^(3/2) / 3/2

0^(3/2) /3/2 - 1(3/2) /3/2

- 2/3

use substitution
u=sin(x)+1
rest is easy