Question

Evaluate the following definite integrals. $$\int_{0}^{\pi / 4}\left(\sec ^{2} t \mathbf{i}-2 \cos t \mathbf{j}-\mathbf{k}\right) d t$$

Short answer

Short Answer: The definite integral of the given vector function is given by the expression \((1\mathbf{i}-\sqrt{2}\mathbf{j}-\frac{\pi}{4}\mathbf{k})\).

Step-by-step solution

Integrate \(\sec^2 t\) with respect to \(\mathbf{i}\)

To perform this integral, recall that the indefinite integral of \(\sec^2 t\) is simply \(\tan t\). We have: $$\int_0^{\pi/4} \sec^2 t \ dt = \left[\tan t\right]_0^{\pi/4} = \tan\frac{\pi}{4} - \tan 0 = 1 - 0 = 1$$

Integrate \(-2\cos t\) with respect to \(\mathbf{j}\)

To perform this integral, recall that the indefinite integral of \(\cos t\) is \(\sin t\). We have: $$\int_0^{\pi/4} -2\cos t \ dt = -2\left[\sin t\right]_0^{\pi/4} = -2\left(\sin\frac{\pi}{4} - \sin 0\right) = -2\left(\frac{\sqrt{2}}{2} - 0\right) = -\sqrt{2}$$

Integrate \(-1\) with respect to \(\mathbf{k}\)

To perform this integral, recall that the indefinite integral of a constant is the constant multiplied by the variable. We have: $$\int_0^{\pi/4} -1 \ dt = -\left[t\right]_0^{\pi/4} = -\frac{\pi}{4} + 0 = -\frac{\pi}{4}$$

Combine the results

Now we combine the results from Steps 1-3 to get the final result: $$\int_{0}^{\pi / 4}\left(\sec ^{2} t \mathbf{i}-2 \cos t \mathbf{j}-\mathbf{k}\right) d t=(1\mathbf{i}-\sqrt{2}\mathbf{j}-\frac{\pi}{4}\mathbf{k})$$