question\nevaluate the definite integral given below.\n int_{pi/4}^{pi/3}(3\tan(x)sec(x)+2cos(x))dx \nenter…

Explanation:

Step1: Split the integral

By the property $\int_{a}^{b}(f(x)+g(x))dx=\int_{a}^{b}f(x)dx+\int_{a}^{b}g(x)dx$, we have $\int_{\pi/4}^{\pi/3}(3\tan(x)\sec(x)+2\cos(x))dx = 3\int_{\pi/4}^{\pi/3}\tan(x)\sec(x)dx+2\int_{\pi/4}^{\pi/3}\cos(x)dx$.

Step2: Integrate $\tan(x)\sec(x)$

The antiderivative of $\tan(x)\sec(x)$ is $\sec(x)$. So, $3\int_{\pi/4}^{\pi/3}\tan(x)\sec(x)dx=3[\sec(x)]_{\pi/4}^{\pi/3}=3(\sec(\frac{\pi}{3})-\sec(\frac{\pi}{4}))=3(2 - \sqrt{2})$.

Step3: Integrate $\cos(x)$

The antiderivative of $\cos(x)$ is $\sin(x)$. So, $2\int_{\pi/4}^{\pi/3}\cos(x)dx=2[\sin(x)]_{\pi/4}^{\pi/3}=2(\sin(\frac{\pi}{3})-\sin(\frac{\pi}{4}))=2(\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2})=\sqrt{3}-\sqrt{2}$.

Step4: Combine the results

$3(2 - \sqrt{2})+\sqrt{3}-\sqrt{2}=6-3\sqrt{2}+\sqrt{3}-\sqrt{2}=6+\sqrt{3}-4\sqrt{2}$.

Answer:

$6+\sqrt{3}-4\sqrt{2}$