integrate the given integral with respect to t on a component - by - component basis.\n int_{0}^{pi/6}(sec…

integrate the given integral with respect to t on a component - by - component basis.\n int_{0}^{pi/6}(sec t\tan t)mathbf{i}+(\tan t)mathbf{j}+(2sin tcos t)mathbf{k}dt = left(int_{0}^{pi/6}(sec t\tan t)dt\right)mathbf{i}+left(int_{0}^{pi/6}(\tan t)dt\right)mathbf{j}+left(int_{0}^{pi/6}(2sin tcos t)dt\right)mathbf{k}=leftmathbf{i}\right_{0}^{pi/6}+leftln\right_{0}^{pi/6}mathbf{j}+left\right_{0}^{pi/6}mathbf{k}

integrate the given integral with respect to t on a component - by - component basis.\n int_{0}^{pi/6}(sec t\tan t)mathbf{i}+(\tan t)mathbf{j}+(2sin tcos t)mathbf{k}dt = left(int_{0}^{pi/6}(sec t\tan t)dt\right)mathbf{i}+left(int_{0}^{pi/6}(\tan t)dt\right)mathbf{j}+left(int_{0}^{pi/6}(2sin tcos t)dt\right)mathbf{k}=leftmathbf{i}\right_{0}^{pi/6}+leftln\right_{0}^{pi/6}mathbf{j}+left\right_{0}^{pi/6}mathbf{k}

Answer

Explanation:

Step1: Integrate each component separately

We know that $\int_{0}^{\frac{\pi}{6}}\sec t\tan tdt$, $\int_{0}^{\frac{\pi}{6}}\tan tdt$ and $\int_{0}^{\frac{\pi}{6}}2\sin t\cos tdt$. For $\int_{0}^{\frac{\pi}{6}}\sec t\tan tdt$, the antiderivative of $\sec t\tan t$ is $\sec t$. 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 $\int{0}^{\frac{\pi}{6}}\tan tdt$, since $\tan t=\frac{\sin t}{\cos t}$, and the antiderivative of $\tan t$ is $-\ln|\cos t|$. So $\int_{0}^{\frac{\pi}{6}}\tan tdt=-\left[\ln|\cos t|\right]{0}^{\frac{\pi}{6}}=-\left(\ln\cos\frac{\pi}{6}-\ln\cos0\right)=-\left(\ln\frac{\sqrt{3}}{2}-\ln1\right)=\ln\frac{2}{\sqrt{3}}$. For $\int{0}^{\frac{\pi}{6}}2\sin t\cos tdt$, let $u = \sin t$, then $du=\cos tdt$. When $t = 0$, $u=0$; when $t=\frac{\pi}{6}$, $u=\frac{1}{2}$. So $\int_{0}^{\frac{\pi}{6}}2\sin t\cos tdt=\int_{0}^{\frac{1}{2}}2udu=\left[u^{2}\right]_{0}^{\frac{1}{2}}=\frac{1}{4}$.

Step2: Write the vector - valued result

The integral $\int_{0}^{\frac{\pi}{6}}[(\sec t\tan t)\mathbf{i}+(\tan t)\mathbf{j}+(2\sin t\cos t)\mathbf{k}]dt=\left(\frac{2\sqrt{3}}{3}-1\right)\mathbf{i}+\ln\frac{2}{\sqrt{3}}\mathbf{j}+\frac{1}{4}\mathbf{k}$

Answer:

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