# Trig Integration By Substitution

Mod note: Moved from technical math section
∫(2x+6)/sqrt(5-4x-x^2)

I have 2/3(ln|tan(theta)+sec(theta)|-3|cos(theta)|) where x=sin^-1((x+2)/3)

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SteamKing
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∫(2x+6)/sqrt(5-4x^2-x^2)

Are there supposed to be two x^2 expressions under the SQRT?

Apologies no it should be 5-4x-x^2

HallsofIvy
Complete the square in the square root: $$5- 4x- x^2= 5- (x^2+ 4x+ 4- 4)= 5- (x+ 2)^2+ 4= 9- (x+ 2)^2. Now make the substitution u= x+ 2, du= dx, x= u- 2 so 2x+ 6= 2u+ 2. The integral becomes [tex]\int \frac{2u+ 2}{\sqrt{9- u^2} du$$
Now let $u= 3 sin(\theta)$.