# Thread: Integrals w/ trig substitutions

1. ## Integrals w/ trig substitutions

I'm having some trouble with a few problems, any hints or help would be great!

1) antiderivative of 1/(81-x^2)
i ended up getting this down to (1/9)integral of (1/cos(theta)), and i think the integral of that is (1/9)(ln(sec(theta)+tan(theta))), but it was wrong. any ideas? i may have just integrated the wrong thing, im not sure

2) antiderivative of 1/((16x^2+1)^2)
i have no idea where to begin with this one, haven't seen something of this type before

3)antiderivative of 1/((2x^2+2x+7)^2)
again, unsure of where to begin. should i multiply that all out, or what?

2. 1. no need for trigonometric substituions here:

$\frac{1}
{{81 - x^2 }} = \frac{1}
{{\left( {9 - x} \right)\left( {9 + x} \right)}} = \frac{{\frac{1}
{{18}}}}
{{9 - x}} + \frac{{\frac{1}
{{18}}}}
{{9 + x}}
$

2. try the following substitution:

$
x = \frac{1}
{4}\tan u
$

3. $
\frac{1}
{{\left( {2x^2 + 2x + 7} \right)^2 }} = \frac{1}
{{4\left( {x^2 + x + \frac{7}
{2}} \right)^2 }} = \frac{1}
{{4\left[ {\left( {x + \frac{1}
{2}} \right)^2 + \frac{{13}}
{4}} \right]^2 }}
$

now try the following substitution:

$
x = \sqrt {\frac{{13}}
{4}} \tan u - \frac{1}
{2}
$