evaluate the definite integral. (int_{0}^{\frac{pi}{6}}(sec t\tan t)mathbf{i}+(\tan t)mathbf{j}+(2sin tcos…

Explanation:

Step1: Integrate each component separately

• For the $\mathbf{i}$ - component: $\int_{0}^{\frac{\pi}{6}}\sec t\tan tdt$. Recall that the derivative of $\sec t$ is $\sec t\tan t$. So, $\int_{0}^{\frac{\pi}{6}}\sec t\tan tdt=\left[\sec t\right]_{0}^{\frac{\pi}{6}}=\sec\frac{\pi}{6}-\sec0=\frac{2\sqrt{3}}{3}- 1$.
• For the $\mathbf{j}$ - component: $\int_{0}^{\frac{\pi}{6}}\tan tdt=\int_{0}^{\frac{\pi}{6}}\frac{\sin t}{\cos t}dt$. Let $u = \cos t$, then $du=-\sin tdt$. When $t = 0$, $u = 1$; when $t=\frac{\pi}{6}$, $u=\frac{\sqrt{3}}{2}$. So, $\int_{0}^{\frac{\pi}{6}}\frac{\sin t}{\cos t}dt=-\int_{1}^{\frac{\sqrt{3}}{2}}\frac{du}{u}=-\left[\ln|u|\right]_{1}^{\frac{\sqrt{3}}{2}}=-\left(\ln\frac{\sqrt{3}}{2}-\ln1\right)=\ln\frac{2}{\sqrt{3}}$.
• For the $\mathbf{k}$ - component: $\int_{0}^{\frac{\pi}{6}}2\sin t\cos tdt$. Since $2\sin t\cos t=\sin(2t)$, then $\int_{0}^{\frac{\pi}{6}}\sin(2t)dt$. Let $v = 2t$, $dv = 2dt$. When $t = 0$, $v = 0$; when $t=\frac{\pi}{6}$, $v=\frac{\pi}{3}$. So, $\int_{0}^{\frac{\pi}{6}}\sin(2t)dt=\frac{1}{2}\int_{0}^{\frac{\pi}{3}}\sin vdv=\frac{1}{2}[-\cos v]_{0}^{\frac{\pi}{3}}=\frac{1}{2}\left(-\cos\frac{\pi}{3}+\cos0\right)=\frac{1}{2}\left(-\frac{1}{2} + 1\right)=\frac{1}{4}$.

Answer:

$\left(\frac{2\sqrt{3}}{3}-1\right)\mathbf{i}+\ln\frac{2}{\sqrt{3}}\mathbf{j}+\frac{1}{4}\mathbf{k}$