Help me with these integrals?

Question

How do I solve these integrals using substitution??

a. integral of ((cos (square root of x)) / (square root of x)) dx
b. integral of z (z+1)^1/3 dz

Answer 1

a. Integrate ∫{cos(√x)/√x}dx.

Let u = √x
Then
du = dx/(2√x)
2du = dx/√x

∫{cos(√x)/√x}dx = ∫(cos u)(2du)
= 2sin u + C = 2sin(√x) + C
_______________

b. Integrate ∫{z(z+1)^(1/3)}dz

Let u = z + 1
Then
du = dz
z = u - 1

∫{z(z+1)^(1/3)}dz = ∫{(u-1)u^(1/3)}du
= ∫{u^(4/3) - u^(1/3)}du = (3/7)u^(7/3) - (3/4)u^(4/3) + C
= (3/7)(z+1)^(7/3) - (3/4)(z+1)^(4/3) + C

Answer 2

1) int [cos(sqrt x) dx/ sqrt x] =

= 2 int [cos(sqrt x) (dx/2 sqrt x)] =

= 2 int cos u du = 2 sin u + C = 2 sin (sqrt x) + C

sqrt x = u

du = dx/2sqrt x

2) Multiply and divide by 3/4 and use part integral:

(3/4) int {z [(4/3)(z+1)^1/3] dz} =
= (3/4)z (z+1)^1/3 - (3/4)int (4/3)(z+1)^4/3 dz =
=(3/4) z (z+1)^4/3 - int (z+1)^4/3 =
= (3/4) z(z+1)^4/3 - (3/7)(z+1)^7/3 + C=
= (3/4) z(z+1)^4/3 - (3/7)(z+1)(z+1)^4/3 + C
= [(3/4) z - (3/7)(z+1)] (z+1)^4/3 + C


u = (z+1)^4/3

u' = (4/3)(z+1)^(1/3)

v = z

v' = 1

Ana

Answer 3

Go to http://integrals.wolfram.com/index.jsp

This will do symbolic integration. It will give you an answer, but doesn't show you how it is computed

a. 2 * sin( sqrt(x) )

b. (x*z*(1 + z))/3