Evaluate `int((cosx)^(n-1))/((sinx)^(n+1))dx=`
(A) `-cot^n x/n+c`
(B) `-cot^n x/(n+1)+c`
(C) `cot^n x/n+c`
(D) `cot^n x/(n+1)+c`

Options

A. `(cot^(n)x)/(n)`
B. `(-cot^(n-1)x)/(n-1)`
C. `(-cot^(n)x)/(n)`
D. `(cot^(n-1)x)/(n-1)`

Explanation

To evaluate the integral

\[ I = \int \frac{(\cos x)^{n-1}}{(\sin x)^{n+1}} \, dx, \]

we can follow these steps:

### Step 1: Rewrite the Integral
We can rewrite the integral in a more manageable form. We know that

\[ \frac{1}{(\sin x)^{n+1}} = \frac{1}{\sin^2 x} \cdot \frac{1}{(\sin x)^{n-1}}. \]

Thus, we can express the integral as:

\[ I = \int \frac{(\cos x)^{n-1}}{\sin^2 x} \cdot \frac{1}{(\sin x)^{n-1}} \, dx. \]

### Step 2: Use Trigonometric Identities
We can use the identity \(\frac{\cos^2 x}{\sin^2 x} = \cot^2 x\) to rewrite the integral:

\[ I = \int (\cos x)^{n-1} \cdot \frac{1}{\sin^2 x} \cdot \frac{1}{(\sin x)^{n-1}} \, dx = \int (\cot x)^{n-1} \cdot \frac{1}{\sin^2 x} \, dx. \]

### Step 3: Substitution
Let \(t = \cot x\). Then, we have:

\[ dt = -\csc^2 x \, dx \quad \Rightarrow \quad dx = -\frac{dt}{\csc^2 x} = -\frac{dt}{1 + t^2}. \]

### Step 4: Change of Variables
Substituting \(t\) into the integral, we get:

\[ I = -\int t^{n-1} \, dt. \]

### Step 5: Integrate
Now we can integrate:

\[ I = -\frac{t^n}{n} + C. \]

### Step 6: Substitute Back
Substituting back \(t = \cot x\):

\[ I = -\frac{\cot^n x}{n} + C. \]

### Final Result
Thus, the evaluated integral is:

\[ I = -\frac{\cot^n x}{n} + C. \]

### Conclusion
The correct answer is option (A):

\[ -\frac{\cot^n x}{n} + C. \]