Key Concepts and Formulas

• Integration by parts: $\int u \, dv = uv - \int v \, du$
• Trigonometric identities: $\csc^2 x = 1 + \cot^2 x$, $\frac{d}{dx}(\cot x) = -\csc^2 x$, $\frac{d}{dx}(\csc x) = -\csc x \cot x$
• $\int \csc x \, dx = \ln \left| \tan \frac{x}{2} \right| + C$

Step-by-Step Solution

Step 1: Define the integral and apply integration by parts. Let $I = \int \csc^5 x \, dx$. We apply integration by parts by splitting $\csc^5 x$ into $\csc^3 x \cdot \csc^2 x$. Let $u = \csc^3 x$ and $dv = \csc^2 x \, dx$. Then $du = 3 \csc^2 x (-\csc x \cot x) \, dx = -3 \csc^3 x \cot x \, dx$ and $v = \int \csc^2 x \, dx = -\cot x$. Thus, $I = \int \csc^5 x \, dx = \int \csc^3 x \csc^2 x \, dx = -\csc^3 x \cot x - \int (-\cot x)(-3 \csc^3 x \cot x) \, dx$ $I = -\csc^3 x \cot x - 3 \int \csc^3 x \cot^2 x \, dx$

Step 2: Use the trigonometric identity $\cot^2 x = \csc^2 x - 1$ to simplify the integral. We rewrite the integral in terms of $\csc x$: $I = -\csc^3 x \cot x - 3 \int \csc^3 x (\csc^2 x - 1) \, dx$ $I = -\csc^3 x \cot x - 3 \int \csc^5 x \, dx + 3 \int \csc^3 x \, dx$ $I = -\csc^3 x \cot x - 3I + 3 \int \csc^3 x \, dx$

Step 3: Isolate the integral $I$ and define $I_1 = \int \csc^3 x \, dx$. $4I = -\csc^3 x \cot x + 3 \int \csc^3 x \, dx$ Let $I_1 = \int \csc^3 x \, dx$. Then, $4I = -\csc^3 x \cot x + 3I_1$

Step 4: Evaluate $I_1 = \int \csc^3 x \, dx$ using integration by parts. We apply integration by parts to $I_1 = \int \csc^3 x \, dx = \int \csc x \csc^2 x \, dx$. Let $u = \csc x$ and $dv = \csc^2 x \, dx$. Then $du = -\csc x \cot x \, dx$ and $v = -\cot x$. $I_1 = -\csc x \cot x - \int (-\cot x)(-\csc x \cot x) \, dx$ $I_1 = -\csc x \cot x - \int \csc x \cot^2 x \, dx$

Step 5: Use the trigonometric identity $\cot^2 x = \csc^2 x - 1$ to simplify the integral $I_1$. We rewrite the integral in terms of $\csc x$: $I_1 = -\csc x \cot x - \int \csc x (\csc^2 x - 1) \, dx$ $I_1 = -\csc x \cot x - \int \csc^3 x \, dx + \int \csc x \, dx$ $I_1 = -\csc x \cot x - I_1 + \int \csc x \, dx$ $2I_1 = -\csc x \cot x + \int \csc x \, dx$

Step 6: Evaluate $\int \csc x \, dx$ and solve for $I_1$. We know that $\int \csc x \, dx = \ln \left| \tan \frac{x}{2} \right| + C$. $2I_1 = -\csc x \cot x + \ln \left| \tan \frac{x}{2} \right|$ $I_1 = -\frac{1}{2} \csc x \cot x + \frac{1}{2} \ln \left| \tan \frac{x}{2} \right|$

Step 7: Substitute $I_1$ back into the equation for $I$. $4I = -\csc^3 x \cot x + 3I_1$ $4I = -\csc^3 x \cot x + 3 \left( -\frac{1}{2} \csc x \cot x + \frac{1}{2} \ln \left| \tan \frac{x}{2} \right| \right)$ $4I = -\csc^3 x \cot x - \frac{3}{2} \csc x \cot x + \frac{3}{2} \ln \left| \tan \frac{x}{2} \right|$

Step 8: Solve for $I$. $I = -\frac{1}{4} \csc^3 x \cot x - \frac{3}{8} \csc x \cot x + \frac{3}{8} \ln \left| \tan \frac{x}{2} \right| + C$ $I = -\frac{1}{4} \csc x \cot x \left( \csc^2 x + \frac{3}{2} \right) + \frac{3}{8} \ln \left| \tan \frac{x}{2} \right| + C$

Step 9: Identify $\alpha$ and $\beta$ and compute $8(\alpha + \beta)$. Comparing this with the given form, we have $\alpha = -\frac{1}{4}$ and $\beta = \frac{3}{8}$. Therefore, $8(\alpha + \beta) = 8 \left( -\frac{1}{4} + \frac{3}{8} \right) = 8 \left( -\frac{2}{8} + \frac{3}{8} \right) = 8 \left( \frac{1}{8} \right) = 1$.

Common Mistakes & Tips

• Remember to use the correct trigonometric identities when simplifying integrals.
• Be careful with signs when applying integration by parts.
• When dealing with powers of trigonometric functions, integration by parts and trigonometric identities are often required repeatedly.

Summary

We used integration by parts and trigonometric identities to evaluate the integral $\int \csc^5 x \, dx$. We found that $\alpha = -\frac{1}{4}$ and $\beta = \frac{3}{8}$, which gives $8(\alpha + \beta) = 1$.

Final Answer The final answer is \boxed{1}.