# Thread: Integration with square roots in the denominator

1. ## Integration with square roots in the denominator

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

In my solution book it says

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

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

for your integration use the sub

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

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

$\displaystyle sec(u)=\frac{\sqrt{a^2+x^2}}{a}$ 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 $\displaystyle \frac {1}{\cos x} + \tan x$ = $\displaystyle \sqrt {x^2+1} +x$

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

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

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

6. Thanks!

7. OK, how about the first question?

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# integration formula with square root in denominator

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