# Trig substitution calc 2

• Oct 1st 2010, 07:15 PM
highc1157
Trig substitution calc 2
Dear MHF experts (btw, you guys rock!! lol),

$\displaystyle \displaystyle \int \frac{x^2}{\sqrt{25-x^2}}\,dx$

(someone else's calculations for this, last step is correct, i took the derivative and it equals the original integral)

let sin(θ) = x/5 ; 5 cos(θ) dθ = dx
= 25 ∫ sin²(θ) dθ
= 25/2 ∫dθ - 25/2 ∫cos(2θ) dθ ............. sin²(θ) = (1/2) [1 - cos(2θ)]
= 25/2 θ - 25/2 sin(θ)cos(θ) + c
= 25/2 arcsin(x/5) - x/2 √(25 - x²) + c

----------------------------------------------------------------------------
I got to the step in bold no problem. I got the first term in the last step right; however, I cannot seem get from the second term in bold to the second term in the last step. I only need help getting from the bold underlined to the last underlined term xD

In the last step that's underlined, I can substitute (x/5)=sinθ ... but cannot figure out what to substitute for cosθ
• Oct 1st 2010, 07:41 PM
Ithaka
$\displaystyle sin \theta = x/5; cos ^2 \theta + sin ^2 \theta =1; cos \theta = \sqrt{1-x^2/25}; cos \theta = 1/5 \sqrt{25-x ^2}$