# Trig substitution integral(need a check on the solution)

## 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

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Dick
Homework Helper
i) You actually don't need to use a trig substitution here. u=16-x^2 will actually do the job. ii) If you do, then if sin(x)=x/4 then cos(x)=sqrt(16-x^2)/4. Not the x^2-16 thing. That's where your imaginaries are coming from. And iii) If you've changed the variable back to x then the limits are 0 to 2*sqrt(3), not the theta limit of pi/3. Actually a lot of the rest of it is correct, which is good. But I didn't check every line.

i) You actually don't need to use a trig substitution here. u=16-x^2 will actually do the job. ii) If you do, then if sin(x)=x/4 then cos(x)=sqrt(16-x^2)/4. Not the x^2-16 thing. That's where your imaginaries are coming from. And iii) If you've changed the variable back to x then the limits are 0 to 2*sqrt(3), not the theta limit of pi/3. Actually a lot of the rest of it is correct, which is good. But I didn't check every line.
i actually just realized that before i read your post, so is that last line (40/3) right

Dick