Single Variable Calculus, Chapter 5, 5.4, Section 5.4, Problem 68

Find the integral $\displaystyle \int^{10}_{-10} \frac{2e^x}{\sin h x + \cos hx} dx$


$
\begin{equation}
\begin{aligned}

& \int^{10}_{-10} \frac{2e^x}{\sin h x + \cos hx} dx = \int^{10}_{-10} \frac{2e^x}{e^x} dx
&& \text{Apply sum of $\cos h$ and $\sin h (\cos hx + \sin hx = e^x)$}
\\
\\
& \int^{10}_{-10} \frac{2e^x}{\sin h x + \cos hx} dx = \int^{10}_{-10} 2 dx
&&
\\
\\
& \int^{10}_{-10} \frac{2e^x}{\sin h x + \cos hx} dx = \left. 2x \right|^{10}_{-10}
&&
\\
\\
& \int^{10}_{-10} \frac{2e^x}{\sin h x + \cos hx} dx = 2(10) - 2 (-10)
&&
\\
\\
& \int^{10}_{-10} \frac{2e^x}{\sin h x + \cos hx} dx = 20 + 20
&&
\\
\\
& \int^{10}_{-10} \frac{2e^x}{\sin h x + \cos hx} dx = 40
&&

\end{aligned}
\end{equation}
$