If $f\left(x\right) = \frac{2- x\cos x}{2+x \cos x}$ and $ g\left(x\right) =\log_{e}x ., \left(x>0\right) $ then the value of integral $\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}} g\left(f\left(x\right)\right)dx $ is :

Question

Evaluate the integral: $ \int \sin^5 x \, dx $

Answer 1

To solve the given integral $\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}} g(f(x)) \, dx$, where $f(x) = \frac{2- x\cos x}{2+x \cos x}$ and $g(x) = \log_e x$, we need to explore the properties of the functions involved.

First, let's examine the function $f(x)$.

For $f(x)$, we have:

$$ f(x) = \frac{2-x\cos x}{2+x\cos x} $$

Notice that if we substitute $x \to -x$ in $f(x)$, we observe:

$$ f(-x) = \frac{2+x\cos x}{2-x\cos x} = \frac{1}{f(x)} $$

This relation, $f(-x) = \frac{1}{f(x)}$, shows the functional property used in definite integrals concerning limits symmetric around zero.

Next, substitute $f(x)$ into $g(x)$ to find $g(f(x))$:

$$ g(f(x)) = \log_e \left( \frac{2-x\cos x}{2+x\cos x} \right) $$

Now consider the integral:

$$ \int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}} g(f(x)) \, dx = \int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}} \log_e \left( \frac{2-x\cos x}{2+x\cos x} \right) \, dx $$

Using the property of the definite integral and symmetry of limits:

$$ \int\limits_{-a}^{a} h(x) \, dx = 0 \, \text{(if)}\, h(-x) = -h(x) $$

In our problem, set $ h(x) = g(f(x)) = \log_e \left( \frac{2-x\cos x}{2+x\cos x} \right) $.

Then, $ h(-x) = \log_e \left( \frac{2+x\cos x}{2-x\cos x} \right) = - \log_e \left( \frac{2-x\cos x}{2+x\cos x} \right) = -h(x) $.

This indicates that $ h(x) $ is an odd function. Therefore, the integral simplifies to:

$$ \int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}} h(x) \, dx = 0 $$

Thus, the value of the integral is:

$$ \log_e 1 $$

Therefore, the correct answer is: $\log_e 1$

Answer 2