`int (d ( cos theta))/(sqrt(1-cos^(2) theta))`

Options

A. `cos^(-1) theta + c`
B. ` theta + c`
C. `sin^(-1) theta + c`
D. None of these

Answer

To solve the integral \( I = \int \frac{d(\cos \theta)}{\sqrt{1 - \cos^2 \theta}} \), we can follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{d(\cos \theta)}{\sqrt{1 - \cos^2 \theta}} \]
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we can rewrite \( \sqrt{1 - \cos^2 \theta} \) as \( \sin \theta \):
\[ I = \int \frac{d(\cos \theta)}{\sin \theta} \]

### Step 2: Change of Variable
Let \( x = \cos \theta \). Then, the differential \( d(\cos \theta) \) becomes \( dx \):
\[ I = \int \frac{dx}{\sqrt{1 - x^2}} \]

### Step 3: Recognize the Integral
The integral \( \int \frac{dx}{\sqrt{1 - x^2}} \) is a standard integral that evaluates to:
\[ \sin^{-1}(x) + C \]

### Step 4: Substitute Back
Now we substitute back \( x = \cos \theta \):
\[ I = \sin^{-1}(\cos \theta) + C \]

### Step 5: Simplify the Result
Using the identity \( \sin^{-1}(\cos \theta) = \frac{\pi}{2} - \theta \) (since \( \cos \theta = \sin\left(\frac{\pi}{2} - \theta\right) \)):
\[ I = \frac{\pi}{2} - \theta + C \]

### Final Result
Thus, the final result of the integral is:
\[ I = \frac{\pi}{2} - \theta + C \]

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