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**coolio** 1. The integral: (sinˆ2 (3x) cos(3x) ) dx

2. The integral: (x+2)(x-3)ˆ20 dx

3. The integral from 0 to pi/4: sqrt(tan(x)) secˆ2 (x) ) dx

4. The integral from 0 to 1: ( eˆ(cubed root of x) ) / 6xˆ(2/3) dx

5. Find the area that is enclosed by the curve y=2x-xˆ3 , the line x=1, the line x=5, and the x axis.

Hi,

to 1.) Use substitution:

u(x) = sin(3x) ===> du/dx = 3*cos(3x) ===> du = 3*cos(3x) * dx

Your integral becomes:

(1/3)∫(sin²(3x) * 3*cos(3x) dx = (1/3) ∫u²*du = 1/9*u³ + C

resubstitute:

∫(sin²(3x) * cos(3x) dx = 1/9*sin³(3x) + C

to 2.)

Use partial integration:

Code:

∫(x+2)(x-3)^20 dx = 1/21*(x+2)(x-3)^21 - ∫1*1/21*(x-3)^21 dx
u = x+2 v = 1/21*(x-3)^21
u' = 1 v' = (x-3)^20

Thus your integral becomes:

∫(x+2)(x-3)^20 dx = 1/21*(x+2)(x-3)^21 - 1/462*(x-3)^22 + C

EB