Question

If $f(-x)=-f\left(x\right)$ then prove that ${\int }_{-a}^{a}f\left(x\right)dx=0.$ Hence evaluate ${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\log \left(\frac{2-\sin x}{2+\sin x}\right)dx.$

Answer 1

$\begin{array}{c}{\int }_{-a}^{a}f\left(x\right)dx={\int }_{-a}^{0}f\left(x\right)dx+{\int }_0^{a}f\left(x\right)dx\\ Letx=-t\\ =-{\int }_{-a}^{0}f\left(-t\right)dt+{\int }_0^{a}f\left(x\right)dx\\ ={\int }_0^{a}f\left(-t\right)dt+{\int }_0^{a}f\left(x\right)dx\\ ={\int }_0^{a}f\left(\left(-x\right)+f\left(x\right)\right)dx\left[f\left(x\right)=-f\left(x\right)\right]\\ ={\int }_0^{a}0dx=0\\ I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\log \left(\frac{2-sinx}{2+\sin x}\right)dx\\ I=={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\log \left(\frac{2-sinx}{2+\sin x}\right)dx\\ addingthem\\ 2I=={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}0\\ \Rightarrow I=0\end{array}$