Another Definite Integral?

Question

Integral from 0 to 9/2= sqroot(81-x^2)dx

Answer 1

1/2(sqrt81-x^2 x+81 sin-1(x/9))

Answer 2

Integral ( 0 to 9/2, sqrt(81 - x^2) dx )

To solve this, we need to use trigonometric substitution.
Let x = 9sin(t). Then
dx = 9cos(t) dt.

Now, we need to change the bounds of integration.

When x = 0:
0 = 9sin(t)
0 = sin(t)
t = 0

When x = 9/2, then
9/2 = 9sin(t)
1/2 = sin(t), so
t = pi/6

New bounds for integration: 0 to pi/6.
Remember not only to substitute x, but also dx with 9cos(t)dt.

Integral (0 to pi/6, sqrt(81 - 81sin^2(t)) 9cos(t) dt )

Factor out 81 within the square root,

Integral (0 to pi/6, sqrt( 81[1 - sin^2(t)] ) 9cos(t) dt )

Use the identity 1 - sin^2(t) = cos^2(t).

Integral (0 to pi/6, sqrt( 81[cos^2(t)] ) 9 cos(t) dt )

Apply the square root.

Integral (0 to pi/6, 9cos(t) 9cos(t) dt )
Integral (0 to pi/6, 81cos^2(t) dt )

Factor out 81.

81 * Integral (0 to pi/6, cos^2(t) dt )

We have to use the half angle identity to solve this. A reminder that cos^2(t) = (1/2)(1 + cos(2t))

81 * Integral (0 to pi/6, (1/2)(1 + cos(2t)) dt )

Factor out 1/2.

(81/2) * Integral (0 to pi/6, (1 + cos(2t)) dt )

Which we can now integrate directly. A reminder that the integral of cos(2t) is (1/2)sin(2t).

(81/2) * [t + (1/2)sin(2t)] {from 0 to pi/6}

(81/2) * ( ( pi/6 + (1/2)sin(2*pi/6) ) - (0 + (1/2)sin(0)) )

(81/2) * ( pi/6 + (1/2)sin(pi/3) - (0 + 0) )

(81/2) * ( pi/6 + (1/2)(sqrt(3)/2) )

(81/2) * ( pi/6 + sqrt(3)/4 )

(81/2) * ( 2pi/12 + 3sqrt(3)/12 )

(81/2) * (2pi + 3sqrt(3))/12

[162pi + 243sqrt(3)]/24