Question

Evaluate the following:

${\displaystyle {\int }_0^2\frac{\text{d}x}{{\text{e}}^x+{\text{e}}^{-x}}}$

Answer 1

Let I = ${\displaystyle {\int }_0^2\frac{\text{d}x}{{\text{e}}^x+{\text{e}}^{-x}}}$

= ${\displaystyle {\int }_0^1\frac{\text{d}x}{{\text{e}}^x+\frac{1}{{\text{e}}^x}}}$

= ${\displaystyle {\int }_0^1\frac{\text{d}x}{\frac{{\text{e}}^{2x}+1}{{\text{e}}^x}}}$

= ${\displaystyle {\int }_0^1\frac{{\text{e}}^x\text{d}x}{{\text{e}}^{2x}+1}}$

Put e^x = t

⇒ e^x dx = dt

Changing the limit, we have

When x = 0

∴ t = e⁰ = 1

When x = 1

∴ I = ${\displaystyle {\int }_1^{\text{e}}\frac{\text{dt}}{{\text{t}}^2+1}}$

= ${\displaystyle {\left[{\tan }^{-1}\text{t}\right]}_1^{\text{e}}}$

= ${\displaystyle [{\tan }^{-1}\text{e}-{\tan }^{-1}\left(1\right)]}$

= ${\displaystyle {\tan }^{-1}\text{e}-\frac{\pi }{4}}$

Hence, I = ${\displaystyle {\tan }^{-1}\text{e}-\frac{\pi }{4}}$.