Re: another root integral

The form $\displaystyle \int \sqrt{a^2-x^2} \,dx$ is associated with trig substitution, specifically $\displaystyle x=a \sin \theta$. Hope that helps.

Re: another root integral

Quote:

Originally Posted by

**billb91** The form $\displaystyle \int \sqrt{a^2-x^2} \,dx$ is associated with trig substitution, specifically $\displaystyle x=a \sin \theta$. Hope that helps.

Thanks. If I might ask: what integral table do you use, or how did you know that? I can't seem to find that in my integral table.

Re: another root integral

Honestly, when we went over trig substitution in my class, my teacher told us that secant, sine, and tangent were the three main ones we needed to know, and if we ever saw integrals in the forms:

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

or

$\displaystyle \int \sqrt {x^2-a^2} \,dx$

then the corresponding trig functions would be tangent (1) and secant (2), while I already showed you sine. Makes sense, as

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

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

$\displaystyle \int \frac {1}{{x^2+1}} \,dx$ = $\displaystyle \arctan x+c$

So you can see the similarities in the denominators.