What is the area between the curve y=cos3x*sin-x, x axis and lines x=o and x=pi?

We'll have to calcuate the definite integral of the
function cos3x*sin(-x)

Int
cos3x*sin(-x)dx

We'll re-write cos 3x = 4(cos x)^3 - 3cos
x

We'll substitute cos x =
t

We'll differentiate both
sides:

- sin x dx = dt

We'll
re-write the integra in t:

Int (4t^3 -
3t)dtl

We'll apply the property of integral to be
additive:

Int (4t^3 - 3t)dt  = 4Int t^3dt - 3Int
tdt

4Int t^3dt - 3Int tdt = 4*t^4/4 -
3t^2/2

We'll simplify and we'll
get:

Int (4t^3 - 3t)dt = t^4 -
3t^2/2

We'll apply Leibniz-Newton
formula:

Int cos3x*sin(-x)dx = (cos pi)^4 - (cos 0)^4 -
(3/2)[(cos pi)^2 - (cos 0)^2]

Int cos3x*sin(-x)dx = (-1)^4
- 1 - (3/2)(1 - 1)

Int cos3x*sin(-x)dx =
0

The area between the given curve, x axis
and the given lines is 0 square units: Int cos3x*sin(-x)dx =
0.