Indefinite Integration 1 Question 26
27. The value of $\int _{-\pi}^{\pi} \frac{\cos ^{2} x}{1+a^{x}} d x, a>0$, is
(2001, 1M)
(a) $\pi$
(b) $a \pi$
(c) $\frac{\pi}{2}$
(d) $2 \pi$
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Solution:
- Let $I=\int _{-\pi}^{\pi} \frac{\cos ^{2} x}{1+a^{x}} d x$
$$ \begin{aligned} & =\int _{\pi}^{-\pi} \frac{\cos ^{2}(-x)}{1+a^{-x}} d(-x) \\ \Rightarrow \quad I & =\int _{-\pi}^{\pi} a^{x} \frac{\cos ^{2} x}{1+a^{x}} d x \end{aligned} $$
On adding Eqs. (i) and (ii), we get
$$ \begin{aligned} 2 I & =\int _{-\pi}^{\pi} \frac{1+a^{x}}{1+a^{x}} \cos ^{2} x d x \\ & =\int _{-\pi}^{\pi} \cos ^{2} x d x=2 \int _0^{\pi} \frac{1+\cos 2 x}{2} d x \\ & =\int _0^{\pi}(1+\cos 2 x) d x=\int _0^{\pi} 1 d x+\int _0^{\pi} \cos 2 x d x \\ & =[x] _0^{\pi}+2 \int _0^{\pi / 2} \cos 2 x d x=\pi+0 \\ \Rightarrow \quad 2 I & =\pi \quad \Rightarrow \quad I=\pi / 2 \end{aligned} $$
$$ e^{\cos x} \quad \sin x, \text { for }|x| \leq 2 $$
