# Thread: Integration with square roots in the denominator

1. ## Integration with square roots in the denominator

$\int \frac {dx}{\sqrt {x^2+a^2}}$

In my solution book it says

Let $t=x+\sqrt {x^2+a^2}$

How do you figure what to use as substitution

I'm too lazy to type this out, so here's a picture.

In the circled part, how do they get from that trig expression to that?

2. $\left( \frac{1+sinx}{cosx}\right)^2=\left(\frac{1}{cosx} +\frac{sinx}{cosx}\right)$

for your integration use the sub

let $atan(u)=x$ ..... $asec^2(u) du=dx$

$tanu=\frac{x}{a}$ as you can see in the attachment

$sec(u)=\frac{\sqrt{a^2+x^2}}{a}$ the rest for you

3. Go this website
Type in the exact expression: integrate dx/sqrt[x^2 + a^2]
Click the equal sign at the end. What do you get?
Ckick the show steps button.

4. I think you misunderstood my question.

I asked how $\frac {1}{\cos x} + \tan x$ = $\sqrt {x^2+1} +x$

5. Originally Posted by chengbin
I think you misunderstood my question.

I asked how $\frac {1}{\cos x} + \tan x$ = $\sqrt {x^2+1} +x$
At the start of the question it states let $x = \tan{\theta}$.

Can you work out the $\frac{1}{\cos}$ part?

6. Thanks!

7. OK, how about the first question?

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