How can one use the chain rule to integrate? I am trying to calculate the anti-derivative of $y=\sqrt{25-x^2}$, for which I believe I may need the chain rule $\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$. How I would use it, however, is a different matter entirely.  
I used this website for a tutorial, however my answer of $\left[\frac{(200-8x^2)^\frac{3}{4}}{7}\right]_0^5$ is vastly different from the actual answer $\frac{25\pi}{4}$. How exactly would someone calculate the antiderivative of a function like this?
 A: hint: $\displaystyle \int \sqrt{25-x^2}dx$. Put $x = 5\sin \theta$. The rest should not be too hard for you to continue.
A: Since $x=5\sin { t } \\ dx=5\cos { tdt } \\ $
$$\int { \sqrt { { 25-x }^{ 2 } } dx= } 25\int { \cos ^{ 2 }{ t } dt } =25\int { \frac { 1+\cos { 2t }  }{ 2 } dt=\frac { 25 }{ 2 } \left( t+\frac { \sin { 2t }  }{ 2 }  \right)  } +C=\\ =\frac { 25 }{ 2 } \left( \arcsin { \frac { x }{ 5 } +\frac { 25x\sqrt { 25-{ x }^{ 2 } }  }{ 50 }  }  \right) +C$$
A: Hint:
put $x=5cos(t)$ with $t \in [0,\pi]$.
with $dx=-5sin(t)dt$.
Remember that
$1-cos^2(t)=sin^2(t)$ ,
$\sqrt{sin^2(t)}=|sin(t)|=sin(t)$  since
$t \in [0,\pi]$,
and
$sin^2(a)=\frac{1-cos(2a)}{2}$ to finish.
A: $$\int \sqrt{25-x^2}dx$$
There is a common trick with these kind of integrals, when you do them by trigonometric substitution. The idea is that we can use the relation $\sin^2 x+ \cos^2 x = 1$ to get rid of the square root.
Start by taking the $25$ out of the integral:
$$5 \int \sqrt{1 - \left(\frac x5\right)^2}dx$$
Call $t = \frac x5 \implies dx = 5dt \implies $
$$= 25 \int \sqrt{1 - t^2} dt$$
Now call $t = \cos z$; you get $dt = -\sin z dz$ and using the relation $1 - \cos^2 z = \sin^2 z$, substituting you get 
$$= -25\int \sin^2 z \ dz = -\frac{25}2(z - \sin z \cos z)$$
Substituing back $t$ into the expression
$$= -\frac {25}2 (\arccos t - t\sqrt{1-t^2})$$
So the final answer is 
$$\int \sqrt{25-x^2}dx = \frac{25}{2}\left(\frac x5\sqrt{1 - \left(\frac x5\right)^2} - \arccos \frac x5\right) + constant$$
A: I lied in my comment - the substitution rule is what you need but it's not used in the conventional way. See, we usually sub $u = g(x)$ so we make a hard problem easier. However, we need something different here.
We start with $\int \sqrt{25 - x^2} \ dx$. Well, let's remember a trig identity: $1 - \sin^2 x = \cos^2 x.$ So what do we do? Here's a trick. Let $x = 5 \sin w$. Then $25 - x^2 = 25 - 25 \sin^2 x$. What's next? Hey, pull a 25 out and we get something nice!
Here's what we got:
\begin{align*}
\int \sqrt{25 - x^2} \ dx &= \int 5 \cos w \sqrt{25 - 25 \sin^2 w} \ dw \\
&= \int 25 \cos^2 w \ dw
\end{align*}
Now with the help of another trig identity (half angle), we'll be done! Can you take it from here, pal?
A: Hint:
By parts (integrating an implicit factor $1$),
$$I=\int\sqrt{25-x^2}dx=x\sqrt{25-x^2}+\int\frac{x^2}{\sqrt{25-x^2}}dx=x\sqrt{25-x^2}+\int\frac{25-(25-x^2)}{\sqrt{25-x^2}}dx.$$
Then
$$2I=x\sqrt{25-x^2}+\int\frac{25\,dx}{\sqrt{25-x^2}}.$$
After rescaling the variable, you will recognize the derivative of the $\arcsin$.
