91Ó°ÊÓ

Recall that the cosecant function, \(\csc\theta\), is defined as the reciprocal of the sine function, or \(\csc\theta = \frac{1}{\sin\theta}\). Therefore, the given integral can be rewritten as: $$\int \frac{d \theta}{1-\frac{1}{\sin\theta}}$$

To simplify the function inside the integral, find a common denominator: $$\int \frac{d \theta}{\frac{\sin\theta - 1}{\sin\theta}}$$ Now, we will perform a substitution. Let \(u = \sin\theta\). Then, \(du = \cos\theta \, d\theta\). Also, notice that whenever \(\theta = 0\), \(u = 0\) and whenever \(\theta = \pi\), \(u = 0\). Thus, the integral becomes: $$\int \frac{\frac{1}{\cos\theta} \, du}{u-1}$$

To evaluate the integral, we will use a simple substitution, letting \(v = u - 1\) so \(u = v + 1\) and \(du = dv\). The integral becomes: $$\int \frac{\frac{1}{\cos\theta} \, dv}{v}$$ To find the relationship between \(v\) and \(\theta\), use the substitution we did earlier. We have: $$v = u - 1 = (\sin\theta) - 1$$ $$\Rightarrow v + 1 = \sin\theta$$ $$\Rightarrow \sqrt{1 - (\sin^2\theta)} = \sqrt{1 - (v+1)^2} = \cos\theta$$ Applying this relationship between \(v\) and \(\theta\), we get: $$\int \frac{\frac{1}{\sqrt{1 - (v+1)^2}} \, dv}{v}$$ Now, we can recognize this integral as arsinh function, $$\int \frac{1}{\sqrt{1 - (v+1)^2}} \, dv = \text{arsinh}(v) + C$$ Finally, we will substitute back the original expression of \(u\) and \(\theta\): $$y(\theta) = \text{arsinh}(u-1) + C = \text{arsinh}(\sin\theta - 1) + C$$ The final answer is: $$y(\theta) = \text{arsinh}(\sin\theta - 1) + C$$