<s>[INST] Correctness of the definite integral Consider the integral
\begin{eqnarray*}
I & = & \int_{-1}^{1}\frac{dx}{\sqrt{1-x^{2}}(1+\sqrt{1-x^{2}})}\\
 & = & \int_{-1}^{0}\frac{dx}{\sqrt{1-x^{2}}(1+\sqrt{1-x^{2}})}+\int_{0}^{1}\frac{dx}{\sqrt{1-x^{2}}(1+\sqrt{1-x^{2}})}
\end{eqnarray*}
Use the substitutions $u=-\sqrt{1-x^{2}}$ and $v=\sqrt{1-x^{2}}$
into first and second integrals respectively.
\begin{eqnarray*}
= & -{\displaystyle \int_{-1}^{0}}\frac{du}{(1+u)^{\frac{1}{2}}(1-u)^{\frac{3}{2}}} & +{\displaystyle \int}_{0}^{1}\frac{dv}{(1+v)^{\frac{3}{2}}(1-v)^{\frac{1}{2}}}
\end{eqnarray*}
Noticing that
\begin{eqnarray*}
-\int_{-1}^{0}\frac{du}{(1+u)^{\frac{1}{2}}(1-u)^{\frac{3}{2}}}=-\int_{-1}^{1}\frac{du}{(1+u)^{\frac{1}{2}}(1-u)^{\frac{3}{2}}}+\int_{0}^{1}\frac{du}{(1+u)^{\frac{1}{2}}(1-u)^{\frac{3}{2}}}
\end{eqnarray*}
and
\begin{eqnarray*}
\int_{0}^{1}\frac{dv}{(1+v)^{\frac{3}{2}}(1-v)^{\frac{v}{2}}}= & {\displaystyle \int_{-1}^{1}}\frac{dv}{(1+v)^{\frac{3}{2}}(1-v)^{\frac{1}{2}}} & -\int_{0}^{1}\frac{dv}{(1+v)^{\frac{1}{2}}(1-v)^{\frac{3}{2}}}
\end{eqnarray*}
Finally leading to  the result
\begin{eqnarray}
\int_{-1}^{1}\frac{dx}{\sqrt{1-x^{2}}(1+\sqrt{1-x^{2}})} & ={\displaystyle \int}_{-1}^{1}\frac{dx}{(1+x)^{\frac{3}{2}}(1-x)^{\frac{1}{2}}} & -\int_{-1}^{1}\frac{dx}{(1+x)^{\frac{1}{2}}(1-x)^{\frac{3}{2}}}
\end{eqnarray}
The value of the integral is 2. But the R.H.S. seem to be difference of two divergent integrals. Can someone suggest whether the derivation is still consistent and valid?
 [/INST] Let $$\displaystyle I = \int_{-1}^{1}\frac{dx}{\sqrt{1-x^{2}}(1+\sqrt{1-x^{2}})}dx= \int_{0}^{1}\frac{dx}{\sqrt{1-x^{2}}(1+\sqrt{1-x^{2}})}dx$$
Above we Used $\displaystyle \bullet \int_{-a}^{a}f(x)dx = 2\int_{0}^{a}f(x)dx\;,$ If $f(-x) = f(x)$
Now Put $x=\sin \phi\;,$ Then $dx = \cos\phi d\phi$ and Changing limit, We get
So $$\displaystyle I = 2\int_{0}^{\frac{\pi}{2}}\frac{\cos \phi}{\cos \phi\cdot (1+\cos \phi)}d\phi = \int_{0}^{\frac{\pi}{2}}\frac{1-\cos \phi}{\sin^2 \phi}d\phi$$
So $$\displaystyle I = 2\int_{0}^{\frac{\pi}{2}}\csc^2 \phi-\int_{0}^{\frac{\pi}{2}}\csc\phi\cdot \cot \phi d\phi = -2\left[\cot \phi\right]_{0}^{\frac{\pi}{2}}+2\left[\csc \phi \right]_{0}^{\frac{\pi}{2}}$$
So $$\displaystyle I = 2\cdot \lim_{a\rightarrow 0}\cot (a)+2-2\cdot \lim_{b\rightarrow 0}\csc (b)$$
Yes the given Integral is Divergent.
By applying L'Hospital's rule , we evaluate the limit $$\lim_{a \rightarrow 0 } \left( \cot a - \csc a \right) = 0 $$, 
So we get
$$ I = 2 $$
</s>