Evaluation of $\int_{0}^{a} \frac{\sqrt{a+x}}{\sqrt{a-x}} dx$ Evaluate : 
$\int_{0}^{a} \frac{\sqrt{a+x}}{\sqrt{a-x}} dx$ 
My approach : I multiplied both sides by $\sqrt{a+x}$ and after simplification it comes down to : 
$\int_{0}^{a} \frac{a}{\sqrt{a^{2}-x^{2}}} dx + \int_{0}^{a} \frac{x}{\sqrt{a^{2}-x^{2}}} dx$, let's denote them by $I_1$ and $I_2$ respectively.
$I_1$ can be easily solved to $a \sin ^{-1} \frac{x}{a}$. 
If I write $I_2$ as $ \frac{-1}{2}\int_{0}^{a} \frac{-2x}{\sqrt{a^{2}-x^{2}}} dx$, I get the solution as $\frac{a}{2}(\pi+1)$ but the answer given is $\frac{a}{2}(\pi+2)$. Where did I go wrong?
 A: $$I_2=\frac{-1}{2} \int_0^a \frac{d(a^2-x^2)}{\sqrt{a^2-x^2}}$$
let $a^2-x^2=t$. If $x=0,$ then $t=a^2$ and if $x=a,$ then $t=0$.
Hence $$I_2=\frac{-1}{2} \int_{a^2}^0 \frac{dt}{\sqrt{t}}$$
$$\Rightarrow I_2= \int_{0}^{a^2} \frac{dt}{2\sqrt{t}}$$
$$\Rightarrow I_2=a-0=a$$
A: Hint: Put $x=a \cos 2 \theta$.   
A: Let $a^2-x^2=t$, then $-2xdx=dt$. Thus
$$
\int_0^a \frac{x}{\sqrt{a^2-x^2}}dx=-\frac{1}{2}\int_{a^2}^0 \frac{1}{\sqrt{t}}dt =\left[\sqrt{t}\right]_0^{a^2}=a.
$$
A: So for your solution: 
$$I_2 = \int_{0}^{a} \frac{x}{\sqrt{a^2 - x^2}} dx$$
Let $u = a^2 - x^2, du = -2x dx, dx = -\frac{1}{2x} du$ then it is clear that 
$$ I_2 = - \frac{1}{2} \int \frac{1}{\sqrt{u}} du = -  \sqrt{u} $$
Now evaluating back at $x$, 
$$ = - \sqrt{a^2 - x^2} | _0^a = - \sqrt{0} + \sqrt{a^2} = a$$
I would recommend against, the expression
$$ - \frac{1}{2} \int_{0}^{a} \frac{-2x}{\sqrt{a^2 - x^2 }} dx $$ 
Since it doesn't really make it any easier to integrate.But if we insist, again making the substitution $u = a^2 - x^2, dx = -\frac{1}{2x} du$ we get:
$$ - \frac{1}{2} \int \frac{1}{\sqrt{u}} dx $$ 
Which is evaluated the same way.
A: Assume $a\in\mathbb{R}^+$
$$\int_{0}^{a}\frac{\sqrt{a+x}}{\sqrt{a-x}}\space\text{d}x=$$

Substitute $u=\sqrt{a+x}$ and $\text{d}u=\frac{1}{2\sqrt{a+x}}\space\text{d}x$.
THis gives a new lower bound $u=\sqrt{a+0}=\sqrt{a}$ and upper bound $u=\sqrt{a+a}=\sqrt{2a}$:

$$\int_{\sqrt{a}}^{\sqrt{2a}}\frac{2u^2}{\sqrt{2a-u^2}}\space\text{d}u=2\int_{\sqrt{a}}^{\sqrt{2a}}\frac{u^2}{\sqrt{2a-u^2}}\space\text{d}u=$$

Substitute $u=\sqrt{2a}\sin(s)$ and $\text{d}u=\sqrt{2a}\cos(s)\space\text{d}s$.
Then $\sqrt{2a-u^2}=\sqrt{2a-2a\sin^2(s)}=\sqrt{2a}\cos(s)$ and $s=\arcsin\left(\frac{u}{\sqrt{2a}}\right)$.
THis gives a new lower bound $s=\arcsin\left(\frac{\sqrt{a}}{\sqrt{2a}}\right)=\frac{\pi}{4}$ and upper bound $s=\arcsin\left(\frac{\sqrt{2a}}{\sqrt{2a}}\right)=\frac{\pi}{2}$:

$$4a\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\sin^2(s)\space\text{d}s=4a\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\left[\frac{1}{2}-\frac{\cos(2s)}{2}\right]\space\text{d}s=$$
$$4a\left[\frac{1}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}1\space\text{d}s-\frac{1}{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\cos(2s)\space\text{d}s\right]=\frac{a(2+\pi)}{2}$$
