Question

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

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

Step-by-step solution

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: \( \int_{0}^{\pi / 4}\sec t \tan t dt \) \( \mathbf{i} \) + \( \int_{0}^{\pi / 4}\tan t dt \) \( \mathbf{j} \) + \( \int_{0}^{\pi / 4}2 \sin t \cos t dt \) \( \mathbf{k} \)

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 \( 1/2 - 1 = -1/2 \).

Combine the Results

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