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

As a short answer: The definite integral of the given vector function from 1/2 to 1 is given by \(\frac{3}{2} \ln\left(\frac{3}{2}\right) \mathbf{i} - \mathbf{k}\).

Step-by-step solution

Break down the integral into separate scalar integrals

The given integral can be broken down into two integrals, each for one component of the vector function: $$\int_{1 / 2}^{1}\left(\frac{3}{1+2 t} \mathbf{i}-\pi \csc^{2}\left(\frac{\pi}{2} t\right) \mathbf{k}\right) d t = \int_{\frac{1}{2}}^{1} \frac{3}{1+2t} dt \mathbf{i} - \int_{\frac{1}{2}}^{1} \pi \csc^2\left(\frac{\pi}{2}t\right) dt \mathbf{k}$$ Now we can solve these two scalar integrals separately.

Evaluate the i-component scalar integral

To evaluate the i-component integral, we first solve the scalar part: $$\int_{\frac{1}{2}}^{1} \frac{3}{1+2t} dt$$ Note that the denominator can be written as a linear function of \(t\). So, we can use the substitution method to solve this integral: Let \(u=1+2t\), then \(\frac{du}{dt}=2\) and \(dt = \frac{1}{2}du\). When \(t = \frac{1}{2}\), \(u = 2\); and when \(t = 1\), \(u = 3\). So, the integral becomes: $$\int_{2}^3 \frac{3}{u} \cdot \frac{1}{2} du = \frac{3}{2} \int_{2}^3 \frac{du}{u}$$ Now we can evaluate the integral: $$\frac{3}{2} \left[\ln |u|\right]_{2}^3 = \frac{3}{2} (\ln 3 - \ln 2) = \frac{3}{2} \ln\left(\frac{3}{2}\right)$$ Hence, the i-component of the vector is: $$\frac{3}{2} \ln\left(\frac{3}{2}\right) \mathbf{i}$$

Evaluate the k-component scalar integral

Now we need to evaluate the k-component integral: $$\int_{\frac{1}{2}}^{1} \pi \csc^2\left(\frac{\pi}{2}t\right) dt$$ To evaluate this integral, notice that the derivative of the function \(\cot(\frac{\pi}{2}t)\) is exactly \(-\pi \csc^2(\frac{\pi}{2}t)\). Using this fact, we can evaluate the integral: $$\int_{\frac{1}{2}}^{1} \pi \csc^2\left(\frac{\pi}{2}t\right) dt = \left[-\cot\left(\frac{\pi}{2}t\right)\right]_{\frac{1}{2}}^1 = \cot\left(\frac{\pi}{4}\right) - \cot(0) = 1 - 0 = 1$$ So, the k-component of the vector is \(\mathbf{k}\).

Combine the components to find the resultant vector

Now that we've found the i-component and the k-component of the integral, we can combine them to form the final vector: $$\int_{1/2}^1 \left(\frac{3}{1+2t} \mathbf{i} - \pi \csc^2\left(\frac{\pi}{2}t\right) \mathbf{k}\right) dt = \frac{3}{2} \ln\left(\frac{3}{2}\right) \mathbf{i} - \mathbf{k}$$ So the definite integral of the given vector function is given by: $$\frac{3}{2} \ln\left(\frac{3}{2}\right) \mathbf{i} - \mathbf{k}$$