Definite Integration Question 26
Question 26
- The value of
(2001, 1M)
(a)
Show Answer
Solution:
- Let
On adding Eqs. (i) and (ii), we get
$$ \begin{aligned} 2 I & =\int_{-\pi}^{\pi} \frac{1+a^{x}}{1+a^{x}} \cos ^{2} x d x \ & =\int_{-\pi}^{\pi} \cos ^{2} x d x=2 \int_{0}^{\pi} \frac{1+\cos 2 x}{2} d x \ & =\int_{0}^{\pi}(1+\cos 2 x) d x=\int_{0}^{\pi} 1 d x+\int_{0}^{\pi} \cos 2 x d x \ & =[x]{0}^{\pi}+2 \int{0}^{\pi / 2} \cos 2 x d x=\pi+0 \ \Rightarrow \quad 2 I & =\pi \quad \Rightarrow \quad I=\pi / 2 \end{aligned} $$
