# Thread: trigonometric substitution

1. ## trigonometric substitution

Hi

the integral is Integral sign ("S") dx/ (sqrt(21-4x-x^2))

So first I completed the square so 21 - (x+2)^2 +4 so
Let u = x+2 so du = dx

so the original integral becomes S du/ sqrt( 25 - u^2)

Then I let u = 5 sin t which makes sqrt (25 - u^2) = 5cos t
du = 5 cos t dt

so then S du/ sqrt(25 - u^2) = S 5 cos t dt / 5 cos t

is this right???
= S dt = t + C
= sin ^-1 (u/5) +C = sin^-1 ((x+2)/5) + C

I find the 5 sin t = u where t = sin^-1 u/5 so if this is wrong let me know.

Thanks
Calculus beginner

2. We may factor the denominator of the integrand:

$\frac{1}{21-4x-x^2}=\frac{1}{(x+7)(3-x)}.$

Now the method of Partial Fractions may be used to split the integrand into the sum of two fractions:

$\frac{1}{(x+7)(3-x)}=\frac{A}{x+7}+\frac{B}{3-x}.$