# \nint_{\frac{pi}{6}}^{\frac{pi}{4}}\tan u du
# \nint_{\frac{pi}{6}}^{\frac{pi}{4}}\tan u du
Answer
Explanation:
Step1: Recall the integral formula for (\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 logarithmic and trigonometric properties
We know that (\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}), (\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}), and (\ln a-\ln b=\ln\frac{a}{b}). [ \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})