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

*print*Print*list*Cite

### 2 Answers

We have to find the definite integral of y= - cos 3x * sin x, in the limit x = 0 to x = pi.

Int [ - cos 3x * sin x]

=> Int [ cos 3x * sin dx]

=> Int [ (4(cos x)^3 - 3 cos x) sin x dx]

let cos x = u

=> du = -sin x dx

=> Int [ (2u^3 - 3u) du]

=> 2 u^4 / 4 - 3u^2 / 2

substitute u = cos x

=> 2* (cos x)^4 / 4 - 3* (cos x)^2 / 2

At x = pi

=> 2* 1/ 4 - 3 / 2 + C

At x = 0

=> 2* 1/ 4 - 3 / 2 + C

Subtracting the two we have

2* 1/ 4 - 3 / 2 + C - ( 2* 1/ 4 - 3 / 2 + C)

=>0

**Therefore the required value of the area is 0.**

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.**