Take the 2-minute tour ×  I'm studying for my finals and I have
this integral that I'm trying to evaluate (part of a bigger
problem):  $$\int\sqrt{150^2-x^2} \cdot dx$$  I have evaluated a
few integrals of this type before so the first thought that came
to my mind was to substitute $x = \sin t$ and $dx = \cos t \cdot
dt$.  So now I have:  $$\int \sqrt{150^2-\sin^2t} \cdot \cos t
\cdot dt$$  However, here is where I'm getting stuck. Usually
instead of having $150^2$ I have $1$, and by using $1-\sin^2t =
\cos^2t \space$ I can continue, but not in this case.  How
should I go on?  share|improve this question add comment  2
Answers  up vote 5 down vote accepted  With such integration by
parts problems where $\sqrt{a^2-x^2}$ appears and a
trigonometric substitution is appropriate, the substitution to
make is $x=a\sin t$. Then we have $a^2(1-\sin^2 t)$ in the
square root, which simplifies nicely.  Likewise one would
substitute $x=a \tan t$ into the integral $\displaystyle \int
\frac{1}{a^2+x^2}\, dx$, and so on.  Fo