Let `n ge 2` be a natural number and `0 lt theta lt (pi)/(2)`, Then, `int ((sin^(n)theta - sin theta)^(1/n) cos theta)/(sin^(n+1) theta)d theta` is equal to (where C is a constant of integration)

Options

A. `n/(n^2-1)(1-1/(sin^(n+1)theta))^((n+1)/n)+C`
B. `n/(n^2+1)(1-1/(sin^(n-1)theta))^((n+1)/n)+C`
C. `n/(n^2-1)(1-1/(sin^(n-1)theta))^((n+1)/n)+C`
D. `n/(n^2-1)(1+1/(sin^(n-1)theta))^((n+1)/n)+C`

Answer

To solve the integral

\[ I = \int \frac{(\sin^n \theta - \sin \theta)^{1/n} \cos \theta}{\sin^{n+1} \theta} \, d\theta, \]

we will use a substitution method. Let's follow the steps:

### Step 1: Substitution
Let \( t = \sin \theta \). Then, \( d\theta = \frac{dt}{\cos \theta} \) and \( \cos \theta = \sqrt{1 - t^2} \). The limits of integration will change accordingly, but since we are not given specific limits, we will keep it general.

### Step 2: Rewrite the Integral
Substituting \( t = \sin \theta \) into the integral, we have:

\[ I = \int \frac{(t^n - t)^{1/n} \sqrt{1 - t^2}}{t^{n+1}} \cdot \frac{dt}{\sqrt{1 - t^2}}. \]

The \( \sqrt{1 - t^2} \) terms cancel out:

\[ I = \int \frac{(t^n - t)^{1/n}}{t^{n+1}} \, dt. \]

### Step 3: Simplify the Integral
Now we can simplify the integrand:

\[ I = \int (t^n - t)^{1/n} t^{-(n+1)} \, dt = \int \frac{(t^n - t)^{1/n}}{t^{n+1}} \, dt. \]

### Step 4: Factor Out \( t^n \)
Factor out \( t^n \) from the term \( (t^n - t) \):

\[ I = \int \frac{t^n(1 - t^{1-n})^{1/n}}{t^{n+1}} \, dt = \int (1 - t^{1-n})^{1/n} t^{-1} \, dt. \]

### Step 5: Further Substitution
Let \( u = 1 - t^{1-n} \). Then, we differentiate to find \( dt \):

\[ du = -\frac{1-n}{n} t^{-n} dt \implies dt = -\frac{n}{1-n} t^n du. \]

### Step 6: Substitute Back into the Integral
Now substitute back into the integral:

\[ I = \int u^{1/n} \left(-\frac{n}{1-n} t^n\right) t^{-1} du = -\frac{n}{1-n} \int u^{1/n} t^{n-1} du. \]

### Step 7: Integrate
Now we can integrate \( u^{1/n} \):

\[ \int u^{1/n} du = \frac{n}{n+1} u^{(n+1)/n} + C. \]

### Step 8: Substitute Back for \( u \)
Substituting back for \( u \):

\[ I = -\frac{n}{1-n} \cdot \frac{n}{n+1} (1 - t^{1-n})^{(n+1)/n} + C. \]

### Step 9: Final Substitution
Finally, substituting back \( t = \sin \theta \):

\[ I = -\frac{n^2}{(1-n)(n+1)} \left(1 - \sin^{1-n} \theta\right)^{(n+1)/n} + C. \]

### Final Answer
Thus, the integral evaluates to:

\[ I = -\frac{n^2}{(1-n)(n+1)} \left(1 - \sin^{1-n} \theta\right)^{(n+1)/n} + C. \]