Indefinite Integration 1 Question 7
8. If $f(x)=\frac{2-x \cos x}{2+x \cos x}$ and $g(x)=\log _e x,(x>0)$ then the value of the integral $\int _{-\pi / 4}^{\pi / 4} g(f(x)) d x$ is
(a) $\log _e 3$
(b) $\log _e e$
(c) $\log _e 2$
(d) $\log _e 1$
(2019 Main, 8 April I)
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Solution:
- The given functions are
$$ g(x)=\log _e x, x>0 \text { and } f(x)=\frac{2-x \cos x}{2+x \cos x} $$
Let
$$ I=\int _{-\pi / 4}^{\pi / 4} g(f(x)) d x $$
Then, $\quad I=\int _{-\pi / 4}^{\pi / 4} \log _e \frac{2-x \cos x}{2+x \cos x} d x$
Now, by using the property
$$ \begin{aligned} & \int _a^{b} f(x) d x=\int _a^{b} f(a+b-x) d x, \text { we get } \\ & I=\int _{-\pi / 4}^{\pi / 4} \log _e \frac{2+x \cos x}{2-x \cos x} d x \end{aligned} $$
On adding Eqs. (i) and (ii), we get
$$ \begin{aligned} 2 I & =\int _{-\pi / 4}^{\pi / 4} \log _e \frac{2-x \cos x}{2+x \cos x}+\log _e \frac{2+\cos x}{2 x \cos x} d x \\ & =\int _{-\pi / 4}^{\pi / 4} \log _e \frac{2-x \cos x}{2+x \cos x} \times \frac{2+x \cos x}{2-x \cos x} d x \\ \Rightarrow & {\left[\because \log _e A+\log _e B=\log _e A B\right] } \\ & 2 I=\int _{-\pi / 4}^{\pi / 4} \log _e(1) d x=0 \Rightarrow \quad I=0=\log _e(1) \end{aligned} $$
