evaluating $\int\sqrt{81-x^2} \, dx$ So I know this problem needs trig sub to solve and I put it on the side opposite $\theta$ and I got a different answer since from the key since they put it on the length and not the height of the triangle.
I got a very similar answer, the only difference is that everything except C is - and instead of arcsin I had arccos.
Integral:
$$\int\sqrt{81-x^2} \, dx$$
Answer key:
$$\frac{1}{2}x\sqrt{81-x^2}+\frac{81}{2}\arcsin\frac{x}{9}+C$$
My answer:
$$-\frac{1}{2}x\sqrt{81-x^2}-\frac{81}{2}\arccos\frac{x}{9}+C$$
To explain what I mean by opposite of $\theta$. In my triangle, $\cos\theta=\frac{x}{9}$. With the answer key's set up, $\cos\theta=\frac{\sqrt{81-x^2}}{9}$.
My question is how do you know where to put the radical on your triangle? It seems like it shouldn't matter since the triangle works.
 A: $\int \sqrt {81-x^2}\ dx$ don't forget the index of integration.
Either one of these substitutions is acceptable.
$x = 9\cos t$ or $x = 9\sin t$
Lets demonstrate both ways
$\int \sqrt {81-x^2}\ dx\\
x = 9\cos t\\
dx = -9\sin t\\
\int \sqrt {81-81\cos^2 t}(-9\sin t)\ dt\\
\int - 81 \sin^2 t dt\\
\int - \frac {81}{2} + \frac {81}2\cos 2t\ dt\\
 - \frac {81}{2}t + \frac {81}4\sin 2t\\
- \frac {81}{2}t + \frac {81}2 \sin t\cos t+C\\
t = \arccos \frac {x}{9}\\
- \frac {81}{2}\arccos \frac {x}{9} + \frac {81}2 (\sqrt {1-\frac {x^2}{81}})( \frac {x}{9})+C\\
- \frac {81}{2}\arccos \frac {x}{9} + \frac 12 x\sqrt {81-x^2}  + C 
$
$\int \sqrt {81-x^2}\ dx\\
x = 9\sin t\\
dx = 9\cos t\\
\int \sqrt {81-81\sin^2 t}(9\cos t)\ dt\\
\int 81 \cos^2 t dt\\
\int \frac {81}{2} + \frac {81}2\cos 2t\ dt\\
  \frac {81}{2}t + \frac {81}4\sin 2t\\
\frac {81}{2}t + \frac {81}2 \sin t\cos t+C\\
t = \arcsin \frac {x}{9}\\
\frac {81}{2}\arcsin \frac {x}{9} + \frac {81}2 \frac {x}{9}\sqrt {1-\frac {x^2}{81}} +C\\
\frac {81}{2}\arcsin \frac {x}{9} + \frac 12 x\sqrt {81-x^2} + C 
$
Both are fine ways to go.
$- \frac {81}{2}\arccos \frac {x}{9} + \frac 12 x\sqrt {81-x^2}  + C = \frac {81}{2}\arcsin \frac {x}{9} + \frac 12 x\sqrt {81-x^2} + D$ 
as 
$ -\frac {81}{2}\arccos \frac {x}{9}$ differs from $\frac {81}{2}\arcsin \frac {x}{9}$ by a constant.
Generally, you will see the $x = a \sin t$ substitution because it has fewer minus signs.
A: hint
Put $$x=9\cos (t) $$
with $t\in [0,\pi] $.
the integral becomes
$$-81\int \sqrt {1-\cos^2 (t)}\sin (t)dt $$
$$-81\int \frac {1-\cos (2t)}{2}dt $$
You can finish using
$$\sin (2t)=2\sin (t)\cos (t) $$
$$=2x/9.\sqrt {1-x^2/81} $$
A: ArcSin and and ArcCos add up to $ \pi/2 $ or $ 3 \pi/2$. In the presence of an arbitrary constant one integral result can be expressed/absorbed in terms of the other.
