If f(x)=2-x x2+x x and g(x)=_ex, (x0) then the value of the integral _-/4^/4g(f(x)),dx is…

Question

If $f(x)=\dfrac{2-x\cos x}{2+x\cos x}$ and $g(x)=\log_ex$, $(x>0)$ then the value of the integral $\displaystyle\int_{-\pi/4}^{\pi/4}g(f(x))\,dx$ is :

Accepted Answer

$$I=\int_{-\pi/4}^{\pi/4}\log\left(\frac{2-x\cos x}{2+x\cos x}\right)dx\ \ldots(i)$$

Using $x\to-x$ (note $x\cos x$ is odd):
$$I=\int_{-\pi/4}^{\pi/4}\log\left(\frac{2+x\cos x}{2-x\cos x}\right)dx\ \ldots(ii)$$

Adding (i) and (ii):
$$2I=\int_{-\pi/4}^{\pi/4}\log\left(\frac{2-x\cos x}{2+x\cos x}\cdot\frac{2+x\cos x}{2-x\cos x}\right)dx=\int_{-\pi/4}^{\pi/4}\log1\,dx=0$$

$$\Rightarrow I=0=\log_e1$$

Solutions · 1