How do I evaluate a definite integral involving trigonometric functions? Evaluate $$\int_0^{8\sqrt{2}} \dfrac{1}{\sqrt{256-s^2}}ds$$
I know that the antiderivative of $\dfrac{1}{\sqrt{1-x^2}}$ is $\sin^{-1}x + C$ but I am a little unsure how I would change the integrand to get hat form. Could someone give me a hint or explanation on how to evaluate this integral? 
 A: Hint to proceed with your idea: $$\frac{1}{\sqrt{256-s^{2}}}=\frac{1}{\sqrt{256\left( 1-\frac{s^{2}}{256}\right) }}=\frac{1}{16\sqrt{1-\left( \frac{s}{16}\right) ^{2}}}.$$

I haven't learned u substitution yet

Added. An antiderivative is
$$\int \frac{1}{16\sqrt{1-\left( \frac{s}{16}\right) ^{2}}}ds=\arcsin \frac{s}{16},$$
because 
$$\frac{d}{ds}\arcsin \frac{s}{16}=\frac{1}{16\sqrt{1-\left( \frac{s}{16}\right) ^{2}}}=\frac{1}{\sqrt{256-s^{2}}}.$$
A: The standard trick when faced with an integral like this is to take the terms that are being squared together with the term combining them, and to form a right-triangle out of them.  You have in the denominator of your original integral $256=16^2$ and $s^2$.  The term combining them is $\sqrt{256 - s^2}$. A right triangle with sides $16, s, \sqrt{256 - s^2}$ has legs $s$ and $\sqrt{256 - s^2},$ and hypotenuse $16$.
Let's arrange the sides so we can do a trigonometric substitution that keeps the lower limit $0$. We can do it by  taking the triangle $\triangle ABC$, where $\angle A = \theta$, $\angle B = \pi/2 - \theta$, and $\angle C = \pi/2$ and having the side lengths be $BC = s, AC = \sqrt{256 - s^2}, \text{and } AB = 16$. (I'd draw this if I could here.)
Then,
$$
\begin{aligned}
s/16 &= \sin\theta,\\
\frac{\sqrt{256 - s^2}}{16} &= \cos\theta,\\
s &= 16 \sin\theta,\\
\sqrt{256 - s^2} &= 16 \cos\theta,\\
ds &= 16 \cos\theta\,d\theta.\\
\end{aligned}$$
Also, when $s=0, \theta=0,$ and when $s=8\sqrt2, \sin\theta = 1/\sqrt2, \text{and } \theta=\pi/4.$
So,
$$
\int_0^{8\sqrt{2}} \dfrac{1}{\sqrt{256-s^2}}ds = \int_0^{\pi/4} \frac{16 \cos\theta\,d\theta}{16\cos\theta} = \frac{\pi}{4}.
$$
The typical trigonometric substitutions are $s=c\sin\theta$ when one has a constant minus $s^2$ or similar, $s = c\tan\theta$ when one has a constant plus $s^2$ or similar, and $s=c\sec\theta$ when one has $s^2$ minus a constant or similar.
A: The easiest way is to do the integral yourself. Let $s = 16 \sin(t)$. We then have $ds = 16 \cos(t) dt$. Hence, we get
$$\int_0^{8 \sqrt{2}} \dfrac{ds}{\sqrt{256-s^2}} = \int_0^{\pi/4} \dfrac{16 \cos(t) dt}{\sqrt{256-256 \sin^2(t)}} = \int_0^{\pi/4} \dfrac{16 \cos(t) dt}{16 \cos(t)}= \int_0^{\pi/4}dt = \dfrac{\pi}4$$
A: Hint: Use the substituition $s=16\cdot\sin x$.
\begin{align}
\int_0^{8\sqrt{2}} \dfrac{1}{\sqrt{256-s^2}}ds
=
&
\int_0^{8\sqrt{2}} \dfrac{1}{16\sqrt{1-\left(\frac{s}{16}\right)^2}}ds
\\
=
&
\int_{16\sin0}^{16\sin(\frac{\pi}{4})} \dfrac{1}{16\sqrt{1-\left(\frac{s}{16}\right)^2}}ds
\\
=
&
\int_{0}^{\frac{\pi}{4}} \dfrac{\cos(x)}{\sqrt{1-\left(\sin(x)\right)^2}}dx
\end{align}
