\\int_{\\frac{\\pi}{6}}^{\\frac{\\pi}{4}}\\tan u du

Explanation:

Step1: Recall the integral formula of $\tan u$

The integral of $\tan u$ is $-\ln|\cos u| + C$. So, $\int\tan udu=-\ln|\cos u|+C$.

Step2: Apply the fundamental theorem of calculus

By the fundamental theorem of calculus $\int_{a}^{b}f(x)dx = F(b)-F(a)$, where $F(x)$ is an antiderivative of $f(x)$. Here, $a = \frac{\pi}{6}$, $b=\frac{\pi}{4}$, and $f(u)=\tan u$, $F(u)=-\ln|\cos u|$. [ \begin{align*} \int_{\frac{\pi}{6}}^{\frac{\pi}{4}}\tan udu&=-\ln|\cos u|\big|_{\frac{\pi}{6}}^{\frac{\pi}{4}}\ &=-\ln\left(\cos\frac{\pi}{4}\right)+\ln\left(\cos\frac{\pi}{6}\right) \end{align*} ]

Step3: Simplify using the values of cosine

Since $\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}$ and $\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}$, and using the property of logarithms $\ln a-\ln b=\ln\frac{a}{b}$, we have: [ \begin{align*} -\ln\left(\frac{\sqrt{2}}{2}\right)+\ln\left(\frac{\sqrt{3}}{2}\right)&=\ln\left(\frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{2}}{2}}\right)\ &=\ln\left(\frac{\sqrt{3}}{\sqrt{2}}\right)\ &=\frac{1}{2}\ln\frac{3}{2} \end{align*} ]

Answer:

$\frac{1}{2}\ln\frac{3}{2}$