# What are the mean and standard deviation of the probability density function given by p(x)=tankx^2 for  x in [pi/8,(5pi)/12], in terms of k, with k being a constant such that the cumulative density across the range of x is equal to 1?

Mean $E \left(x\right) = \mu = \frac{1}{2 k} \cdot \left[\ln \sec \left(\frac{25 {\pi}^{2} k}{144}\right) - \ln \sec \left(\frac{{\pi}^{2} k}{64}\right)\right]$

Standard deviation $\sigma \left(x\right) = \sqrt{{\int}_{\frac{\pi}{8}}^{\frac{5 \pi}{12}} {\left(x - \mu\right)}^{2} \cdot \tan k {x}^{2} \mathrm{dx}}$

#### Explanation:

Solution:

Mean $= E \left(x\right) = \mu = {\int}_{a}^{b} x \cdot p \left(x\right) \cdot \mathrm{dx}$

$\mu = {\int}_{\frac{\pi}{8}}^{\frac{5 \pi}{12}} x \cdot \tan k {x}^{2} \cdot \mathrm{dx}$

$\mu = \frac{1}{2 k} \cdot {\int}_{\frac{\pi}{8}}^{\frac{5 \pi}{12}} 2 k \cdot x \cdot \tan k {x}^{2} \cdot \mathrm{dx}$

$\mu = \frac{1}{2 k} \cdot {\int}_{\frac{\pi}{8}}^{\frac{5 \pi}{12}} \tan k {x}^{2} \cdot 2 k x \cdot \mathrm{dx}$

$\mu = \frac{1}{2 k} \cdot {\left[\ln \sec k {x}^{2}\right]}_{\frac{\pi}{8}}^{\frac{5 \pi}{12}}$

$\mu = \frac{1}{2 k} \cdot \left[\ln \sec k \left(\frac{25 {\pi}^{2}}{144}\right) - \ln \sec k \left(\frac{{\pi}^{2}}{64}\right)\right]$

Standard deviation

$\sigma \left(x\right) = \sqrt{{\int}_{a}^{b} {\left(x - \mu\right)}^{2} p \left(x\right) \mathrm{dx}}$

$\sigma \left(x\right) = \sqrt{{\int}_{\frac{\pi}{8}}^{\frac{5 \pi}{12}} {\left(x - \mu\right)}^{2} \cdot \tan k {x}^{2} \mathrm{dx}}$

God bless....I hope the explanation is useful.