Question

using properties of definite integral evaluate definite integral of cosx dx/(1+e^x) from -pi/2 to pi/2

Answer 1

Dear Student,

We know ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(a+b-x\right)dx$
I = ${\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\frac{\cos x}{1+e^x}dx$ .................... (1)
I = ${\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\frac{\cos \left({\displaystyle \frac{\mathrm{\pi }}{2}}-{\displaystyle \frac{\mathrm{\pi }}{2}}-x\right)}{1+e^{\frac{\mathrm{\pi }}{2}-\frac{\mathrm{\pi }}{2}-x}}dx$
I = ${\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\frac{\cos \left(-x\right)}{1+e^{-x}}dx$
I = ${\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\frac{\cos \left(x\right)}{1+{\displaystyle \frac{1}{e^x}}}dx$
I = ${\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\frac{e^x\left(\cos x\right)}{1+e^x}dx$ .................... (2)
Adding (1) and (2) $\Rightarrow$
2I = ${\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\frac{\cos x}{1+e^x}dx$ + ${\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\frac{e^x\left(\cos x\right)}{1+e^x}dx$
2I = ${\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\frac{\left(1+e^x\right)\cos x}{\left(1+e^x\right)}dx$
2I = ${\int }_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}\cos xdx$
2I = ${\left[\sin x\right]}_{-\frac{\mathrm{\pi }}{2}}^{\frac{\mathrm{\pi }}{2}}$
2I = 1 - (-1)
2I = 2
I = 1

Regards