# Math Help - [SOLVED] Integral

1. ## [SOLVED] Integral

$\int_a^b\sqrt {x^2 - 1}$

Doing partial integration gets me nowhere. Is this a standard integral?

$x\sqrt {x^2 - 1} - \int_a^b\frac{x^2}{\sqrt {x^2 - 1}} =$

$x\sqrt {x^2 - 1} - \left(x\sqrt {x^2 - 1} - \int_a^b\sqrt {x^2 - 1}\right)$

2. why not try $x = \sinh \theta$

3. Let's get rid of integration limits, since it's not the interesting case.

Integrate by parts:

$\int {\sqrt {x^2 - 1} \,dx} = x\sqrt {x^2 - 1} - \int {\frac{{x^2 }}
{{\sqrt {x^2 - 1} }}\,dx} .$

Now a little of make-up

$\int {\frac{{x^2 }}
{{\sqrt {x^2 - 1} }}\,dx} = \int {\frac{{x^2 - 1 + 1}}
{{\sqrt {x^2 - 1} }}\,dx} = \int {\sqrt {x^2 - 1} \,dx} + \int {\frac{1}
{{\sqrt {x^2 - 1} }}\,dx} .$

This yields

$2\int {\sqrt {x^2 - 1} \,dx} = x\sqrt {x^2 - 1} - \int {\frac{1}
{{\sqrt {x^2 - 1} }}\,dx} .$

For the last integral substitute $\varphi = x + \sqrt {x^2 - 1} \implies \frac{1}
{\varphi }\,d\varphi = \frac{1}
{{\sqrt {x^2 - 1} }}\,dx,$
finally

$2\int {\sqrt {x^2 - 1} \,dx} = x\sqrt {x^2 - 1} - \ln \left| {x + \sqrt {x^2 - 1} } \right| + k_1 ,$ and we happily get

$\int {\sqrt {x^2 - 1} \,dx} = \frac{1}
{2}\Big[ x{\sqrt {x^2 - 1} - \ln \left| {x + \sqrt {x^2 - 1} } \right|} \Big] + k.$

4. Hello, Spec!

No one suggested Trig Substitution?

$\int \sqrt {x^2 - 1}\,dx$
Let $x \:=\:\sec\theta\quad\Rightarrow\quad dx \:=\:\sec\theta\tan\theta\,d\theta$

Substitute: . $\int \tan\theta(\sec\theta\tan\theta\,d\theta)$

. . . . . . $=\;\int\sec\theta\tan^2\!\theta\,d\theta$

. . . . . . $= \;\int \sec\theta(\sec^2\!\theta-1)\,d\theta$

. . . . . . $=\;\int(\sec^3\!\theta - \sec\theta)\,d\theta$

. . . . . . $= \;\frac{1}{2}\sec\theta\tan\theta + \frac{1}{2}\ln|\sec\theta + \tan\theta| - \ln|\sec\theta + \tan\theta| + C$

. . . . . . $= \;\frac{1}{2}\sec\theta\tan\theta - \frac{1}{2}\ln|\sec\theta + \tan\theta| + C$

5. Thanks! Just out of curiousity, the original problem was to solve (I just made a substitution):

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

It it possible to solve that one without the substitution, and thus the following somewhat complicated calculation?

6. $\int {\sqrt {x^2 + x} \,dx} = \frac{1}
{2}\int {\sqrt {4x^2 + 4x + 1 - 1} \,dx} = \frac{1}
{2}\int {\sqrt {(2x + 1)^2 - 1} \,dx} .$

Substitute $u=2x+1,$

$\int {\sqrt {x^2 + x} \,dx} = \frac{1}
{4}\int {\sqrt {u^2 - 1} \,du},$
and you know what's next.

P.S.: sorry, I misread your question, let me see if it can be tackled with a nice way.