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Chapter 10: Problem 52

Evaluate the definite integral. $$ \int_{0}^{\pi / 4}[(\sec t \tan t) \mathbf{i}+(\tan t) \mathbf{j}+(2 \sin t \cos t) \mathbf{k}] d t $$

Short Answer

Expert verified
The evaluated definite integral is: \((\sqrt{2} - 1)\mathbf{i} + \ln \sqrt{2}\mathbf{j} - 1/2 \mathbf{k}\)

Step by step solution

01

Split the Integral

Let's separate this integral into three parts corresponding to the terms accompanied by vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \). We can write it as: \( \int_{0}^{\pi / 4}\sec t \tan t dt \) \( \mathbf{i} \) + \( \int_{0}^{\pi / 4}\tan t dt \) \( \mathbf{j} \) + \( \int_{0}^{\pi / 4}2 \sin t \cos t dt \) \( \mathbf{k} \)
02

Evaluate the Integrals

Now evaluate these integrals separately. The integral of \( \sec t \tan t \) from 0 to \( \pi/4 \) is \( \sec t \) evaluated from 0 to \( \pi/4 \), which results in \( \sqrt{2} - 1 \). The integral of \( \tan t \) from 0 to \( \pi/4 \) is \( \ln |\sec t| \) evaluated from 0 to \( \pi/4 \), which ends up as \( \ln \sqrt{2} \). And, for the last integral, we apply the formula of \( \sin 2t = 2 \sin t \cos t \) to simplify it first, then its integration results in \( -\cos 2t/2 \) evaluated from 0 to \( \pi/4 \) yielding \( 1/2 - 1 = -1/2 \).
03

Combine the Results

Combine the results from step 2 by matching them with their corresponding vectors. So, the result will be: \((\sqrt{2} - 1)\mathbf{i} + \ln \sqrt{2}\mathbf{j} - 1/2 \mathbf{k}\)

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Most popular questions from this chapter

The graph of the vector-valued function \(\mathbf{r}(t)\) and a tangent vector to the graph at \(t=t_{0}\) are given. (a) Find a set of parametric equations for the tangent line to the graph at \(t=t_{0}\) (b) Use the equations for the tangent line to approximate \(\mathbf{r}\left(t_{0}+\mathbf{0 . 1}\right)\) $$ \mathbf{r}(t)=\left\langle t, \sqrt{25-t^{2}}, \sqrt{25-t^{2}}\right\rangle, \quad t_{0}=3 $$

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