Closed form of a Definite Integral I attempted to integrate the following function from a practice problem in my Calculus textbook:
$$\displaystyle \int_{0}^{\frac{\pi}{2}}{\frac{1}{1+\tan^\sqrt{2}(x)}} \ {\rm d}x$$
I failed to find an indefinite integral, and I am assuming getting an indefinite integral is simply impossible. Using Wolfram|Alpha to estimate the definite integral, I got $0.785398$. I am assuming this is $\frac{\pi}{4}$, but I have no formal proof that this is the answer.
 A: I finally figured it out due to a hint by Sameer Kailasa.
$$\int_{0}^{\frac{\pi}{2}}{\frac{1}{1+\tan^\sqrt{2}(x)}} \ dx$$
$$u=\frac{\pi}{2}-x \implies du=-dx$$
$$= -\int_{\frac{\pi}{2}}^{0}{\frac{1}{1+\cot^\sqrt{2}(x)}} \ dx = \int_{0}^{\frac{\pi}{2}}{\frac{\tan^\sqrt{2}(x)}{\tan^\sqrt{2}(x)+1}} \ dx $$
$$=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\frac{1}{1+\tan^\sqrt{2}(x)}+\frac{\tan^\sqrt{2}(x)}{\tan^\sqrt{2}(x)+1}} \ dx$$
$$\frac{1}{2}\int_{0}^{\frac{\pi}{2}} \frac{1+\tan^\sqrt{2}(x)}{\tan^\sqrt{2}(x)+1} \ dx =\frac{1}{2} [{x}]_{0}^{\frac{\pi}{2}} = \frac{\pi}{4}$$
A: The $\sqrt{2}$ is a complete red herring. In fact consider 
$$I=\int_0^{\frac{\pi}{2}}\frac{1}{1+\tan^{\alpha}(x)}\,dx$$
where $\alpha$ is any nonnegative real number. Then we have
\begin{align}
I&=\int_0^{\frac{\pi}{2}}\frac{1}{1+\tan^{\alpha}(x)}\,dx
\\&=\int_0^{\frac{\pi}{2}}\frac{1}{1+\frac{\sin^{\alpha}(x)}{\cos^\alpha(x)}}\,dx
\\&=\int_0^{\frac{\pi}{2}}\frac{\cos^{\alpha}(x)}{\cos^{\alpha}(x)+\sin^{\alpha}(x)}\,dx
\\&=\int_0^{\frac{\pi}{2}}\frac{\sin^{\alpha}(y)}{\cos^{\alpha}(y)+\sin^{\alpha}(y)}\,dy
\end{align}
where in the last step we have used the substitution $y=\dfrac{\pi}{2}-x$.
Now we can combine the final two steps to get
\begin{align}
2I&=\int_0^{\frac{\pi}{2}}\frac{\cos^{\alpha}(x)}{\cos^{\alpha}(x)+\sin^{\alpha}(x)}\,dx+\int_0^{\frac{\pi}{2}}\frac{\sin^{\alpha}(x)}{\cos^{\alpha}(x)+\sin^{\alpha}(x)}\,dx
\\&=\int_0^{\frac{\pi}{2}}\frac{\cos^\alpha(x)+\sin^{\alpha}(x)}{\cos^{\alpha}(x)+\sin^{\alpha}(x)}\,dx
\\&=\int_0^{\frac{\pi}{2}} 1\,dx
\\&=\frac{\pi}{2}
\end{align}
Hence $I=\dfrac{\pi}{4}$.
