Another Integral

We can always apply the following formula: $\int_{-a}^a f(t)\, dt = \int_0^a (f(t) + f(-t))\,dt$

Doing so with our integral gives $\begin{aligned} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\cos x}{e^{\sin^3\!x}+1}\,dx &= \int_0^{\frac{\pi}{4}} \left(\frac{\cos x}{e^{\sin^3\!x}+1} + \frac{\cos x}{e^{-\sin^3\!x}+1}\right)\,dx \\ &= \int_0^{\frac{\pi}{4}} \left(\frac{\cos x}{e^{\sin^3\!x}+1} + \frac{\cos x \, e^{\sin^3\!x}}{1+e^{\sin^3\!x}}\right)\,dx \\ &= \int_0^{\frac{\pi}{4}} \frac{\cos x\, \left(e^{\sin^3\!x}+1\right)}{e^{\sin^3\!x}+1}\,dx \\ &= \int_0^{\frac{\pi}{4}} \cos x \, dx \\ &= \sin\frac{\pi}{4} - \sin 0 \\ &= \frac{\sqrt{2}}{2} \approx \boxed{0.7071} \end{aligned}$

$\begin{aligned} I & = \int_{-\frac \pi 4}^\frac \pi 4 \frac {\cos x}{e^{\sin^3 x}+1} dx & \small \color{#3D99F6} \text{Let }u = \sin x \implies du = \cos x \ dx \\ & = \int_{-\frac 1{\sqrt 2}}^{\frac 1{\sqrt 2}} \frac 1{e^{u^3}+1} du \\ & = \int_{-\frac 1{\sqrt 2}}^{\frac 1{\sqrt 2}} \frac {e^{u^3}+1-e^{u^3}}{e^{u^3}+1} du \\ & = \int_{-\frac 1{\sqrt 2}}^{\frac 1{\sqrt 2}} du - \int_{-\frac 1{\sqrt 2}}^{\frac 1{\sqrt 2}} \frac {e^{u^3}}{e^{u^3}+1} du \\ & = \frac 2{\sqrt 2} - \int_{-\frac 1{\sqrt 2}}^{\frac 1{\sqrt 2}} \frac 1{1+e^{-u^3}} du & \small \color{#3D99F6} \text{Let }t = -u \implies dt = - du \\ & = \frac 2{\sqrt 2} - \int_{-\frac 1{\sqrt 2}}^{\frac 1{\sqrt 2}} \frac 1{1+e^{t^3}} dt \\ & = \frac 2{\sqrt 2} - I \\ & = \frac 1{\sqrt 2} \approx \boxed{0.707} \end{aligned}$