Answer
The solution is $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos x}{1+\sin^2x}=\frac{\pi}{2}.$$
Work Step by Step
To solve the integral $$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos x}{1+\sin^2x}$$ we will use substitution $t=sinx$ which gives us $cosxdx=dt$ and for bounds of integration we get $t=−1$ and $t=1$.
Now we have:
$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos x}{1+\sin^2x}=\int_{-1}^{1}\frac{dt}{1+t^2}=\left.\arctan t \right|_{-1}^{1}=\arctan1-\arctan(-1)=\frac{\pi}{4}-\left(-\frac{\pi}{4}\right)=\frac{\pi}{2}.$$
