Calculus 2 master needed; evaluate the integral PLEASE SHOW STEPS IF IM WRONG I split off the sin^3 so i can use the pythag identity and allows for u substitution u=cosx du=-sinx dx -du=sin dx I move the negative towards the outside of the integral. then i divide the terms by sqroot 2||| I eventually get to a=1/2 b =5/2 did I miss anything? Or is this the final answer?

Your process is indeed correct!

The full solution is:

Step-by-step explanation:

We want to evaluate the integral:

As you had done, we can rewrite our integral as:

Using the Pythagorean Identity, this is:

Now, we can make a substitution. Let u = cos(x). Then:

Substitute:

Simplify:

Rewrite:

By the Reverse Power Rule:

Simplify:

Back-substitute:

=  - 2 + 2 cos⁵/₂ (x)  + C

                          5

Step-by-step explanation:

∫ sin³ (x)    dx

= ∫ sin² (x) sin (x)    dx

= ∫ (1 - cos² (x) sin (x)  dx

= ∫ - 1 - u²   du

        √u

= ∫ -   1    +  u³/₂   du

        √u

= - ∫   1    du  +  ∫ u³/₂   du

        √u

substitute it back

=  - 2 √u + 2 cos⁵/₂ (x)

                 5

add constant, therefore

=  - 2 + 2 cos⁵/₂ (x)  + C

                         5

area = length x width

length = 2 x width

32 = 2 x width x width

32 = 2 x width^2

36 = width ^2  

width = 6

length = 12

area is 32

perimeter = 2 x 12 + 2 x 6 = 24 + 12 = 36

step-by-step explanation:

the function makes a strait line horizontally with no x intercept but a y intercept at 9.