Definite Integral $\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\frac{(\cos(x))^{\arcsin(x)+1}}{(\cos(x))^{\arctan(x)}+(\cos(x))^{\arcsin(x)}}dx$ How can I prove that 

$${\large\int_{-\pi/2}^{\pi/2}}\frac{(\cos(x))^{\arcsin(x)+1}}{(\cos(x))^{\arctan(x)}+(\cos(x))^{\arcsin(x)}}dx=1$$

 A: Same trick as in the other one you posted. Let your integral be $I$. Letting $x\mapsto -x:$
$$I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos^{\arctan x+1} x}{\cos^{\arctan x}x+\cos^{\arcsin x}x}\,dx$$
Add both,
$$I=\frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos x\,dx=1$$
A: HINT:
Let $\displaystyle \arcsin x=\theta$
$\displaystyle\implies(i) x=\sin\theta$
and $\displaystyle(ii) -\frac\pi2\le\theta\le \frac\pi2$ based the definition of the principal value of inverse sine function 
$\displaystyle\implies -x=-\sin\theta=\sin(-\theta)$ and  $\displaystyle -\frac\pi2\le-\theta\le \frac\pi2$
$\displaystyle\implies -\theta=\arcsin(-x)\implies -\arcsin x=\arcsin(-x)$
Similarly, $\displaystyle -\arctan x=\arctan(-x)$
$$\text{Now,}\int_a^bf(x)dx=\int_a^bf(a+b-x)dx\implies \int_{-b}^bf(x)dx=\int_{-b}^bf(-x)dx$$
$$\implies2I=\int_{-b}^bf(x)dx+\int_{-b}^bf(-x)dx=\int_{-b}^b\left(f(x)+f(-x)\right)dx$$
A: $
I=\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\frac{(\cos(x))^{\arcsin(x)+1}}{(\cos(x))^{\arctan(x)}+(\cos(x))^{\arcsin(x)}}dx=$
$ \int_{\frac{-\pi}{2}}^{0}\frac{(\cos(x))^{\arcsin(x)+1}}{(\cos(x))^{\arctan(x)}+(\cos(x))^{\arcsin(x)}}dx +\int_{0}^{\frac{\pi}{2}}\frac{(\cos(x))^{\arcsin(x)+1}}{(\cos(x))^{\arctan(x)}+(\cos(x))^{\arcsin(x)}}dx$
In the former integral, by change of variable:
$x=-t$,
$\int_{\frac{-\pi}{2}}^{0}\frac{(\cos(x))^{\arcsin(x)+1}}{(\cos(x))^{\arctan(x)}+(\cos(x))^{\arcsin(x)}}dx= 
\int_{0}^{\frac{\pi}{2}}\frac{(\cos(x))^{\arctan(x)+1}}{(\cos(x))^{\arctan(x)}+(\cos(x))^{\arcsin(x)}}dx$
Adding new one with the latter we have :$I=
\int_{0}^{\frac{\pi}{2}}cosxdx=1$
