Question

Evaluate the integrals in Exercises 40–44.

Short answer

The value of the given vector function is.

Step-by-step solution

Step 1. Given Information

Given that,

Let us assume

Step 2. Evaluating the given integral

Evaluation,

Step 3. Find the value of the vector∫tantdt

Evaluate$\int \tan tdt$separately.

$\int \tan tdt=\int \frac{\sin t}{\cos t}dt$

Let us take$u=\cos t$.

$du=-\sin tdt$.

Then,

$$\begin{gathered} \int \frac{\sin t}{\cos t}dt=\int \frac{-du}{u} \\ =-\ln \left|u\right|+C \end{gathered}$$

Undo the substitution

$$\begin{gathered} \int \tan tdt=-\ln \left|\cos t\right|+C \\ =\ln \left|{\left(\cos t\right)}^{-1}\right|+C \\ =\ln \left|sect\right|+C \end{gathered}$$

Step 4. Substitution method

Substitute the values.

So, the value of the given vector function is