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

Question: Evaluate the definite integral $$\int_{0}^{\pi / 4}\left(\sec ^{2} t \mathbf{i}-2 \cos t \mathbf{j}-\mathbf{k}\right) d t.$$ Answer: The definite integral is equal to $$1\mathbf{i} - \sqrt{2}\mathbf{j} - \frac{\pi}{4}\mathbf{k}$$

Step-by-step solution

Identify the integrals for each component

We need to calculate the definite integral for i, j, and k components separately. So, we have three different integrals to evaluate: 1. $$\int_{0}^{\pi / 4} (\sec^2 t) dt$$ for the i-component. 2. $$\int_{0}^{\pi / 4} (-2\cos t) dt$$ for the j-component. 3. $$\int_{0}^{\pi / 4} (-1) dt$$ for the k-component.

Evaluate the integral for the i-component

To find the integral for the i-component, we will evaluate the following integral: $$\int_{0}^{\pi / 4} (\sec^2 t) dt$$ The antiderivative of sec^2(t) is tan(t). So, we get: $$\tan(t) \Big |_{0}^{\pi / 4} = \tan (\pi / 4) - \tan(0) = 1$$ So, our i-component is: $$1\mathbf{i}$$

Evaluate the integral for the j-component

To find the integral for the j-component, we will evaluate the following integral: $$\int_{0}^{\pi / 4} (-2\cos t) dt$$ The antiderivative of -2cos(t) is -2sin(t). So, we get: $$-2\sin(t) \Big |_{0}^{\pi / 4} = -2\sin (\pi / 4) + 2\sin(0) = -\sqrt{2}$$ So, our j-component is: $$-\sqrt{2}\mathbf{j}$$

Evaluate the integral for the k-component

To find the integral for the k-component, we will evaluate the following integral: $$\int_{0}^{\pi / 4} (-1) dt$$ The antiderivative of -1 is -t. So, we get: $$-t \Big |_{0}^{\pi / 4} = -(\pi / 4)$$ So, our k-component is: $$-\frac{\pi}{4}\mathbf{k}$$

Assemble the final result

Now that we have evaluated the integrals for each component, we can put our result together into one vector. The final result for the given definite integral is: $$\int_{0}^{\pi / 4}\left(\sec ^{2} t \mathbf{i}-2 \cos t \mathbf{j}-\mathbf{k}\right) d t = 1\mathbf{i} - \sqrt{2}\mathbf{j} - \frac{\pi}{4}\mathbf{k}$$