Evaluation of $\int_{0}^{1} \frac{1}{x+\sqrt{1-x^2}} \space dx$ Evaluate
$$\int_{0}^{1} \frac{1}{x+\sqrt{1-x^2}} \space dx$$
My main concern is finding the indefinite integral as once i have that the rest is fairly straight forward. Please give a detailed answer with reference to why you made each substitution (what indicated that said substitution would work etc.)
My initial substitution was $x= \sin \theta$ which tidies it up a bit
$$\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\sin x + \cos x} \space dx$$
Also the answer from Wolfram Alpha is;
$$ = \frac{1}{4}\bigg( log(1-2 x^2)+2 \tanh^{-1}\bigg(\frac{x}{\sqrt{1-x^2}}\bigg)+2 \sin^{-1}x\bigg)+constant$$
I need to be able to answer questions like this in under 15 mins so any help is appreciated
 A: There's a simple solution somewhere in M.SE which goes like this...
$$I = \int_0^{\pi/2} dt \frac{\cos{t}}{\cos{t}+\sin{t}} $$
$$J = \int_0^{\pi/2} dt \frac{\sin{t}}{\cos{t}+\sin{t}} $$
$$I + J = \frac{\pi}{2}$$
$$I - J = \int_0^{\pi/2} dt \frac{\cos{t}-\sin{t}}{\cos{t}+\sin{t}} = \int_0^{\pi/2} dt \frac{(\cos{t}+\sin{t})'}{\cos{t}+\sin{t}}=[\log{(\cos{t}+\sin{t})}]_0^{\pi/2} = 0$$
Thus $I = J = \frac{\pi}{4}$.
ADDENDUM
The above technique requires the easy computation of the antiderivative, which you should see is
$$\frac12 \left [t + \log{(\cos{t}+\sin{t})} \right ] + C = \frac12 \left [\arcsin{x} + \log{(x+\sqrt{1-x^2})} \right ] + C$$ 
A: Another simple solution:
\begin{align}
&x\mapsto\sin u\\
&I=\int_0^{\pi/2} \frac{\cos u}{\sin u + \cos u}du\\
&u\mapsto \frac{\pi}2-v\\
&I=\int_0^{\pi/2} \frac{\sin v}{\sin v + \cos v}dv\\
&\therefore 2I=\int_0^{\pi/2} \frac{\sin u + \cos u}{\sin u + \cos u}du=\frac{\pi}{2}\\
&\therefore I=\frac{\pi}{4}
\end{align}
A: Since no one used it, let me consider the case of the antiderivative $$I=\int\frac{\cos x}{\sin x + \cos x} \space dx$$ Now, use the tangent half-angle substitution (Weierstrass substitution) $t=\tan(\frac x 2)$.
We so obtain $$I=\int \frac{2 \left(t^2-1\right)}{t^4-2 t^3-2 t-1}\space dt$$ But $$t^4-2 t^3-2 t-1=(t^2+1)(t-r_1)(t-r_2)\qquad r_{1,2}=1\pm \sqrt 2$$ Using partial fraction decomposition, we have $$\frac{2 \left(t^2-1\right)}{t^4-2 t^3-2 t-1}=\frac{1-t}{t^2+1}+\frac{1}{2 t-2 \sqrt{2}-2}+\frac{1}{2 t+2 \sqrt{2}-2}$$ that is to say $$\frac{2 \left(t^2-1\right)}{t^4-2 t^3-2 t-1}=\frac{1}{1+t^2}-\frac{2t}{1+t^2}+\frac{1}{2 t-2 \sqrt{2}-2}+\frac{1}{2 t+2 \sqrt{2}-2}$$ and each term is easy to integrate. After simplifications, this leads to $$I=\tan ^{-1}(t)+\frac{1}{2} \log \left(\frac{-t^2+2 t+1}{t^2+1}\right)$$ Now, if integration is from $t=0$ to $t=1$, the logarithms disappear and youare just let with $\tan ^{-1}(1)=\frac \pi 4$.
A: BIG HINT:
$$\int\frac{1}{x+\sqrt{1-x^2}}\space\text{d}x=$$

Substitute $x=\sin(u)$ and $\text{d}x=\cos(u)\space\text{d}u$.
Then $\sqrt{1-x^2}=\sqrt{1-\sin^2(u)}=\cos(u)$ and $u=\arcsin(x)$:

$$\int\frac{\cos(u)}{\sin(u)+\cos(u)}\space\text{d}u=$$
$$\int\frac{\sec^3(u)}{\sec^3(u)}\cdot\frac{\cos(u)}{\sin(u)+\cos(u)}\space\text{d}u=$$
$$\int\frac{\sec^2(u)}{\sec^2(u)+\sec^2(u)\tan(u)}\space\text{d}u=$$

Prepare to substitute $s=\tan(u)$. Rewrite $\frac{\sec^2(u)}{\sec^2(u)+\sec^2(u)\tan(u)}$ using $\sec^2(u)=1+\tan^2(u)$:

$$\int\frac{\sec^2(u)}{1+\tan(u)+\tan^2(u)+\tan^3(u)}\space\text{d}u=$$

Substitute $s=\tan(u)$ and $\text{d}s=\sec^2(u)\space\text{d}u$:

$$\int\frac{1}{s^3+s^2+s+1}\space\text{d}s=$$
$$\int\left(\frac{1-s}{2(s^2+1)}+\frac{1}{2(s+1)}\right)\space\text{d}s=$$
$$\int\frac{1-s}{2(s^2+1)}\space\text{d}s+\int\frac{1}{2(s+1)}\space\text{d}s=$$
$$\frac{1}{2}\int\frac{1-s}{s^2+1}\space\text{d}s+\frac{1}{2}\int\frac{1}{s+1}\space\text{d}s=$$
$$\frac{1}{2}\int\left(\frac{1}{s^2+1}-\frac{s}{s^2+1}\right)\space\text{d}s+\frac{1}{2}\int\frac{1}{s+1}\space\text{d}s=$$
$$-\frac{1}{2}\int\frac{s}{s^2+1}\space\text{d}s+\frac{1}{2}\int\frac{1}{s^2+1}\space\text{d}s+\frac{1}{2}\int\frac{1}{s+1}\space\text{d}s=$$

Substitute $p=s^2+1$ and $\text{d}p=2s\space\text{d}s$:

$$-\frac{1}{4}\int\frac{1}{p}\space\text{d}p+\frac{1}{2}\int\frac{1}{s^2+1}\space\text{d}s+\frac{1}{2}\int\frac{1}{s+1}\space\text{d}s=$$
$$-\frac{\ln\left|p\right|}{4}+\frac{1}{2}\int\frac{1}{s^2+1}\space\text{d}s+\frac{1}{2}\int\frac{1}{s+1}\space\text{d}s=$$
$$-\frac{\ln\left|p\right|}{4}+\frac{\arctan\left(s\right)}{2}+\frac{1}{2}\int\frac{1}{s+1}\space\text{d}s=$$

Substitute $w=s+1$ and $\text{d}w=\space\text{d}s$:

$$-\frac{\ln\left|p\right|}{4}+\frac{\arctan\left(s\right)}{2}+\frac{1}{2}\int\frac{1}{w}\space\text{d}w=$$
$$-\frac{\ln\left|p\right|}{4}+\frac{\arctan\left(s\right)}{2}+\frac{\ln\left|w\right|}{2}+\text{C}$$
A: Finding indefinite integral:
\begin{align}
&x\mapsto\sin u\\
&I=\int \frac{\cos u}{\sin u + \cos u}du\\
&=\frac12\int \frac{\cos u - \sin u + \cos u + \sin u}{\sin u + \cos u}du=\frac12 \int\frac{\cos u-\sin u}{\sin u + \cos u}du+\frac u2\\
&=\frac12 \ln (\cos u + \sin u) + \frac u2+C\\
&=\frac12 \ln (\sqrt{1-x^2} + x) + \frac {\sin^{-1}x}{2}+C\\
\end{align}
