Question

Evaluate the following definite integrals. $$\int_{1 / 2}^{1}\left(\frac{3}{1+2 t} \mathbf{i}-\pi \csc ^{2}\left(\frac{\pi}{2} t\right) \mathbf{k}\right) d t$$

Short answer

Question: Evaluate the following integral $$\int_{1/2}^1 \left(\frac{3}{1+2t} \mathbf{i} - \pi\csc^2\left(\frac{\pi}{2}t\right) \mathbf{k}\right)dt$$ Answer: $$\frac{3}{2}\ln\left(\frac{3}{2}\right) \mathbf{i} - \sqrt{2} \mathbf{k}$$

Step-by-step solution

Observe the given function

Observe that we are given the following function: $$\boldsymbol{F}(t)=\left(\frac{3}{1+2 t} \mathbf{i}-\pi \csc ^{2}\left(\frac{\pi}{2} t\right) \mathbf{k}\right)$$ We need to evaluate the definite integral of this function from \(t=\frac{1}{2}\) to \(t=1\).

Separating the i and k components

We can observe that the given function has two components, the i-component, and the k-component. Separate the components and write the integral: $$\int_{1/2}^1 \frac{3}{1+2t} dt\mathbf{i} - \int_{1/2}^1 \pi \csc^2\left(\frac{\pi}{2} t\right) dt \mathbf{k}$$

Evaluate the i-component integral

First, we will evaluate the i-component integral as follows: $$\int_{1/2}^1 \frac{3}{1+2t} dt$$ Using substitution, let \(u = 1 + 2t\), then \(du = 2dt\). This swaps the integral bounds for substitution, so we get: $$\frac{3}{2}\int_{2}^3 \frac{1}{u} du$$ Now, integrate the i-component integral: $$\frac{3}{2}\left[\ln(u)\right]_2^3 = \frac{3}{2}\left(\ln(3) - \ln(2)\right) = \frac{3}{2} \ln\left(\frac{3}{2}\right)$$

Evaluate the k-component integral

Next, we will evaluate the k-component integral as follows: $$\int_{1/2}^1 \pi \csc^2\left(\frac{\pi}{2} t\right) dt$$ Using substitution, let \(x = \frac{\pi}{2}t\), then \(dx = \frac{\pi}{2}dt\). As the interval changes from \(t\) to \(x\), the new integral limits will be \(\frac{\pi}{4}\) and \(\frac{\pi}{2}\). $$\int_{\pi/4}^{\pi/2} \pi\csc^2(x) \frac{2}{\pi} dx = 2\int_{\pi/4}^{\pi/2} \csc^2(x) dx$$ Now, integrate the k-component integral: $$2\left[-\cot(x)\right]_{\pi/4}^{\pi/2} = -2\big(\cot(\pi/2) - \cot(\pi/4)\big) = 2\left[-\frac{\sqrt{2}}{2}\right] = -\sqrt{2}$$

Combine the i and k components

Now, we can put the i and k components back together to get the final result of the integral: $$\left(\frac{3}{2}\ln\left(\frac{3}{2}\right) \mathbf{i}\right) - \left(\sqrt{2}\mathbf{k}\right) = \frac{3}{2}\ln\left(\frac{3}{2}\right) \mathbf{i} - \sqrt{2} \mathbf{k}$$ So, the final evaluated integral for the given exercise is: $$\int_{1/2}^1 \left(\frac{3}{1+2t} \mathbf{i} - \pi\csc^2\left(\frac{\pi}{2}t\right) \mathbf{k}\right)dt = \frac{3}{2}\ln\left(\frac{3}{2}\right) \mathbf{i} - \sqrt{2} \mathbf{k}$$