Definite Integration Question 69
Question 69
- Evaluate $\int_{0}^{\pi} \frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}} d x$.
$(1999,3 \mathrm{M})$
Show Answer
Solution:
- Let $I=\int_{0}^{\pi} \frac{e^{\cos x}}{e^{\cos x}+e^{-\cos x}} d x$
$$ \begin{aligned} & =\int_{0}^{\pi} \frac{e^{\cos (\pi-x)}}{e^{\cos (\pi-x)}+e^{-\cos (\pi-x)}} d x \ & {\left[\because \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right] } \ \Rightarrow \quad I & =\int_{0}^{\pi} \frac{e^{-\cos x}}{e^{-\cos x}+e^{\cos x}} d x \end{aligned} $$
On adding Eqs. (i) and (ii), we get
$$ \begin{aligned} & =\int_{0}^{\pi} \frac{e^{\cos x}+e^{-\cos x}}{e^{\cos x}+e^{-\cos x}} d x=\int_{0}^{\pi} 1 d x=[x]_{0}^{\pi}=\pi \ \Rightarrow \quad I & =\pi / 2 \end{aligned} $$
