- #1

vande060

- 186

- 0

## Homework Statement

this is the first problem like this ive ever tried so take it easy!

evaluate the integral

I = ∫ x^3/√(16-x^2) dx from 0 to 2√3

## The Attempt at a Solution

- π/2 < ϑ < π/2

x = 4sinϑ , dx = 4cosϑdϑ

*x = 2√3, x/4 = sinϑ , sinϑ = √3/2 , ϑ = π/3

x = 0 sinϑ = 0 , ϑ = 0

4cosϑ = √(16-x^2)

I = ∫ (4sinϑ)^3 * 4cosϑdϑ /4cosϑ

I = ∫ (4sinϑ)^3 dϑ

I = 64 ∫ sin^3ϑ dϑ

I = 64 ∫ (1-cos^2ϑ)*sinϑ dϑ

u = cosϑ

du = -sinϑ

I = -64 ∫ (1-u^2)du

I = -64( cosϑ - cos^3ϑ/3) + C

converting back to x

sinϑ = x/4

cosϑ = √(x^2 -16)/4

I = -64{[ √(x^2 -16)/4] - [√(x^2 -16)/4]^3/3]} from 0 to π/3

i think i made a mistake because i get imaginary numbers here

maybe i wasnt supposed to convert back to x, but change the solve in an integral set to theta bounds and expressed in theta like below

I = -64( cosϑ - cos^3ϑ/3) + C from 0 to pi/3

-64[( 1/2 - 1/24) - (1 - 1/3)] = 40/3

Last edited: