Not sure how to get this one going exactly, we just started getting into Integrals that really require manipulation before solving them and I believe this is definitely one of those! Thanks in advance.
$\displaystyle \int{9dx/\sqrt{81-x^4}}$
Not sure how to get this one going exactly, we just started getting into Integrals that really require manipulation before solving them and I believe this is definitely one of those! Thanks in advance.
$\displaystyle \int{9dx/\sqrt{81-x^4}}$
Right I'll post working as I go along... Haven't checked this yet so just making it up as I go...
Let $\displaystyle x = 3\sin(u)$.
then $\displaystyle dx = 3\cos(u)du$.
So out integral becomes...
$\displaystyle \int \frac{9 \cdot 3\cos(u)}{\sqrt{81 - 81\sin^4(u)}}du$
$\displaystyle = \int \frac{27\cos(u)}{\sqrt{81(1 - \sin^4(u))}}du$
$\displaystyle = \int \frac{27\cos(u)}{9\sqrt{cos^4(u)}}du$
$\displaystyle = \int \frac{27\cos(u)}{9\cos^2(u)}du$
$\displaystyle = \int \frac{3}{\cos(u)}du$
$\displaystyle = 3\int \frac{1}{\cos(u)}du$
$\displaystyle = 3\int \sec(u) du$.
$\displaystyle = 3\ln|\sec(u) + \tan(u)|$. (proof http://math2.org/math/integrals/more/sec.htm)