Key Concepts and Formulas

• Indefinite Integration: The process of finding a function whose derivative is a given function.
• Substitution Method: A technique to simplify integrals by substituting a part of the integrand with a new variable.
• Power Rule for Integration: $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ where $n \neq -1$.

Step-by-Step Solution

Step 1: Simplify the integrand. We begin by factoring out $\sin \theta$ from the term inside the parenthesis in the numerator: $\int {{{{{\left( {{{\sin }^n}\theta - \sin \theta } \right)}^{1/n}}\cos \theta } \over {{{\sin }^{n + 1}}\theta }}} \,d\theta = \int {{{{{\left( {\sin \theta \left( {{{\sin }^{n - 1}}\theta - 1} \right)} \right)}^{1/n}}\cos \theta } \over {{{\sin }^{n + 1}}\theta }}} \,d\theta$ = \int {{{{\left( {\sin \theta } \right)}^{1/n}}{{\left( {{{\sin }^{n - 1}}\theta - 1} \right)}^{1/n}}\cos \theta } \over {{{\sin }^{n + 1}}\theta }}} \,d\theta $= \int {{{{{\left( {{{\sin }^{n - 1}}\theta - 1} \right)}^{1/n}}\cos \theta } \over {{{\sin }^{n + 1 - {1 \over n}}}\theta }}} \,d\theta$

Step 2: Further simplification of the integrand. We can rewrite the integrand by multiplying both numerator and denominator by $\sin \theta$: $\int {{{{{\left( {{{\sin }^n}\theta - \sin \theta } \right)}^{1/n}}\cos \theta } \over {{{\sin }^{n + 1}}\theta }}} \,d\theta = \int {{{{{\left( {{{\sin }^n}\theta - \sin \theta } \right)}^{1/n}}\cos \theta } \over {{{\sin }^{n + 1}}\theta }}} \,d\theta = \int {{{{{\left( {\sin \theta \left( {{{\sin }^{n - 1}}\theta - 1} \right)} \right)}^{1/n}}\cos \theta } \over {{{\sin }^{n + 1}}\theta }}} \,d\theta$ = \int {{{\sin \theta {{\left( {{{\sin }^{n - 1}}\theta - 1} \right)}^{1/n}}\cos \theta } \over {{{\left( {{{\sin }^n}\theta } \right)}^{1 + {1 \over n}}}}} \,d\theta $= \int {{{{{\left( {{{\sin }^{n - 1}}\theta - 1} \right)}^{1/n}}\cos \theta } \over {{{\sin }^{n + 1 - {1 \over n}}}\theta }}} \,d\theta = \int {{{{{\left( {{{\sin }^{n - 1}}\theta - 1} \right)}^{1/n}}\cos \theta } \over {{{\sin }^{n + 1}}\theta }} \sin^{1/n}\theta} \,d\theta$ $= \int {{\frac{\sin \theta (\sin^{n-1}\theta - 1)^{1/n} \cos \theta}{\sin^{n+1}\theta}} d\theta} = \int {{\frac{(\sin^{n-1}\theta - 1)^{1/n} \cos \theta}{\sin^{n}\theta}} d\theta}$ $= \int {{{\left( { - \left( {1 - {{\sin }^{n - 1}}\theta } \right)} \right)}^{1/n}}{{\cos \theta } \over {{{\sin }^n}\theta }}d\theta } = \int {{{{{\left( { - 1} \right)}^{1/n}}{{\left( {1 - {{\sin }^{n - 1}}\theta } \right)}^{1/n}}\cos \theta } \over {{{\sin }^n}\theta }}d\theta }$

Step 3: Perform the substitution. Let $t = 1 - \frac{1}{{{\sin }^{n - 1}}\theta }$ Then, differentiating with respect to $\theta$, we get $\frac{dt}{d\theta} = - (n-1) (\sin \theta)^{-n} (-\cos \theta) = \frac{(n-1)\cos \theta}{\sin^n \theta}$ So, $dt = \frac{(n-1)\cos \theta}{\sin^n \theta} d\theta$ $\frac{dt}{n-1} = \frac{\cos \theta}{\sin^n \theta} d\theta$

Step 4: Rewrite the integral in terms of t. Substituting $t$ and $dt$ into the integral, we have: $\int {{{{{\left( {{{\sin }^n}\theta - \sin \theta } \right)}^{1/n}}\cos \theta } \over {{{\sin }^{n + 1}}\theta }}} \,d\theta = \int {{{\left( {1 - {1 \over {{{\sin }^{n - 1}}\theta }}} \right)}^{1/n}}{{\cos \theta } \over {{{\sin }^n}\theta }}d\theta } = \int {t^{1/n} \frac{dt}{n-1}} = \frac{1}{n-1} \int t^{1/n} dt$

Step 5: Evaluate the integral. Using the power rule for integration, we have: $\frac{1}{n-1} \int t^{1/n} dt = \frac{1}{n-1} \cdot \frac{t^{\frac{1}{n} + 1}}{\frac{1}{n} + 1} + C = \frac{1}{n-1} \cdot \frac{t^{\frac{n+1}{n}}}{\frac{n+1}{n}} + C = \frac{n}{(n-1)(n+1)} t^{\frac{n+1}{n}} + C = \frac{n}{n^2 - 1} t^{\frac{n+1}{n}} + C$

Step 6: Substitute back for t. Substituting $t = 1 - \frac{1}{{{\sin }^{n - 1}}\theta }$ back into the expression, we get: $\frac{n}{n^2 - 1} \left(1 - \frac{1}{{{\sin }^{n - 1}}\theta }\right)^{\frac{n+1}{n}} + C$ However, notice that the correct answer provided has $1 + \frac{1}{\sin^{n-1}\theta}$. Let's re-examine the original integral. $\int {{{{{\left( {{{\sin }^n}\theta - \sin \theta } \right)}^{1/n}}\cos \theta } \over {{{\sin }^{n + 1}}\theta }}} \,d\theta = \int {{{{{\left( {{{\sin }^n}\theta - \sin \theta } \right)}^{1/n}}\cos \theta } \over {{{\sin }^{n + 1}}\theta }}} \,d\theta$ $= \int {{{{{\left( {{{\sin }^n}\theta \left( {1 - {1 \over {{{\sin }^{n - 1}}\theta }}} \right)} \right)}^{1/n}}\cos \theta } \over {{{\sin }^{n + 1}}\theta }}} \,d\theta = \int {{{\sin \theta {{\left( {1 - {1 \over {{{\sin }^{n - 1}}\theta }}} \right)}^{1/n}}\cos \theta } \over {{{\sin }^{n + 1}}\theta }}} \,d\theta$ $= \int {{{{{\left( {1 - {1 \over {{{\sin }^{n - 1}}\theta }}} \right)}^{1/n}}\cos \theta } \over {{{\sin }^n}\theta }}} \,d\theta$ There must be an error in the sign. Let's try substituting $t = 1 + \frac{1}{{{\sin }^{n - 1}}\theta }$. Then $dt = -\frac{(n-1) \cos \theta}{\sin^n \theta}d\theta$ So, $-\frac{dt}{n-1} = \frac{\cos \theta}{\sin^n \theta}d\theta$ Substituting this into the integral, we get $\int {\left(1-\frac{1}{\sin^{n-1}\theta}\right)^{1/n} \frac{\cos\theta}{\sin^n \theta}d\theta}$ This substitution won't work either. Let's re-examine the question and the given answer. There must be an error in the original answer key. Let's try another approach. Let $u = \frac{1}{\sin^{n-1}\theta}$. Then $du = \frac{-(n-1)\cos\theta}{\sin^n \theta}d\theta$. So, $\frac{-du}{n-1} = \frac{\cos\theta}{\sin^n \theta}d\theta$. Then the integral becomes $\int (1-\frac{1}{\sin^{n-1}\theta})^{1/n} \frac{\cos\theta}{\sin^n\theta}d\theta = \int (1-u)^{1/n} \frac{-du}{n-1} = \frac{-1}{n-1} \int (1-u)^{1/n} du$ $= \frac{-1}{n-1} \frac{(1-u)^{\frac{1}{n}+1}}{\frac{1}{n}+1} (-1) + C = \frac{1}{n-1} \frac{(1-u)^{\frac{n+1}{n}}}{\frac{n+1}{n}} + C = \frac{n}{(n-1)(n+1)} (1-u)^{\frac{n+1}{n}} + C$ $= \frac{n}{n^2-1} (1-\frac{1}{\sin^{n-1}\theta})^{\frac{n+1}{n}} + C$. The answer must be C, but the provided correct answer is A.

Common Mistakes & Tips

• Be careful with algebraic manipulations, especially when dealing with fractional exponents.
• Double-check the signs when performing substitutions.
• Always remember the constant of integration, C.

Summary

We started by simplifying the integrand through factoring. Then, we used the substitution method to evaluate the integral. After performing the integration and substituting back, we obtained the final result. Upon review, the correct answer is option C and not A.

Final Answer

The final answer is $\boxed{\frac{n}{n^2 - 1} \left(1 - \frac{1}{{{\sin }^{n - 1}}\theta }\right)^{\frac{n+1}{n}} + C}$, which corresponds to option (C).