Problem 40 - Calculus Early Transcendentals

Question

Evaluate the following integrals. $$\int \frac{d \theta}{1-\csc \theta}$$

Step-by-Step Solution

Answer

Question: Evaluate the integral \(\int \frac{d\theta}{1-\csc{\theta}}\). Answer: \(\int \frac{d \theta}{1-\csc \theta} = \ln\left|\sin{\theta}-1 + \sqrt{-(\sin{\theta}-1)^2(\sin{\theta}+1)}\right| + C\)

Step 1: Recall the definition of cosecant function

The cosecant function, denoted by \(\csc{\theta}\), is defined as the reciprocal of the sine function, i.e. \(\csc{\theta} = \frac{1}{\sin{\theta}}\). We can use this definition to rewrite our integral as follows: $$\int \frac{d\theta}{1-\frac{1}{\sin{\theta}}}$$

Step 2: Simplify the expression inside the integral

Apply common denominator to the expression inside the integral to get the following: $$\int \frac{d\theta}{\frac{\sin{\theta} - 1}{\sin{\theta}}}$$ Now, invert and multiply the expression inside the integral: $$\int \frac{\sin{\theta}}{\sin{\theta}-1} d\theta$$

Step 3: Substitute \(u = \sin{\theta} - 1\)

Let \(u = \sin{\theta} - 1\). Then, \(\frac{du}{d\theta} = \cos{\theta}\) and \(d\theta = \frac{du}{\cos{\theta}}\). Since \(u = \sin{\theta} - 1\), we have \(\sin{\theta} = u+1\): $$\int \frac{u+1}{u} \frac{du}{\cos{\theta}}$$

Step 4: Rewrite \(\cos{\theta}\) using \(u\)

Now, \(\sin^2{\theta} + \cos^2{\theta} = 1\). Since \(\sin{\theta} = u+1\), we have \(\cos^2{\theta} = 1 - (u+1)^2 = 1 - (u^2 + 2u + 1) = -u^2 - 2u\). Hence, \(\cos{\theta} = \pm\sqrt{-u(u+2)}\). Now the integral becomes: $$\int \frac{u+1}{u} \frac{du}{\pm\sqrt{-u(u+2)}}$$

Step 5: Simplify and evaluate the integral

Simplifying the above integral: $$\int \frac{du}{\pm\sqrt{-u^2(u+2)}}$$ Join the \(\pm\) sign in the constant \(C\). $$\int \frac{du}{\sqrt{-u^2(u+2)}}$$ We can recognize it as an integral of the form \(\int \frac{du}{\sqrt{u^2-a^2}}\). The standard formula for this type of integral is given by: $$\int \frac{du}{\sqrt{u^2-a^2}} = \ln\left|u+\sqrt{u^2-a^2}\right| + C$$ Using this formula, we get: $$\int \frac{du}{\sqrt{-u^2(u+2)}} = \ln\left|u + \sqrt{-u^2(u+2)}\right| + C$$

Step 6: Substitute back \(\sin{\theta}\) and simplify

Now, substitute back \(u = \sin{\theta}-1\): $$\int \frac{d\theta}{1-\csc \theta} = \ln\left|\sin{\theta}-1 + \sqrt{-(\sin{\theta}-1)^2(\sin{\theta}+1)}\right| + C$$ Thus, we have evaluated the integral: $$\int \frac{d \theta}{1-\csc \theta} = \ln\left|\sin{\theta}-1 + \sqrt{-(\sin{\theta}-1)^2(\sin{\theta}+1)}\right| + C$$