What is `intsin^(-1)(cosx)` dx equal to ?

Options

A. `(xpi)/(2)-(x^(2))/(2)+k`
B. `(pi)/(2)+(x^(2))/(2)+k`
C. `-(xpi)/(2)-(x^(2))/(2)+k`
D. `(pi)/(2)-(pi)^(2)/(2)+k`

Answer

To solve the integral \( \int \sin^{-1}(\cos x) \, dx \), we can follow these steps:

### Step 1: Rewrite the integral using a trigonometric identity
We know that:
\[ \sin\left(\frac{\pi}{2} - \theta\right) = \cos(\theta) \]
Thus, we can express \( \cos x \) in terms of sine:
\[ \sin^{-1}(\cos x) = \frac{\pi}{2} - x \]
This allows us to rewrite the integral:
\[ \int \sin^{-1}(\cos x) \, dx = \int \left(\frac{\pi}{2} - x\right) \, dx \]

### Step 2: Separate the integral
Now we can separate the integral into two parts:
\[ \int \left(\frac{\pi}{2} - x\right) \, dx = \int \frac{\pi}{2} \, dx - \int x \, dx \]

### Step 3: Integrate each term
Now we can integrate each term separately:
1. The integral of a constant:
\[ \int \frac{\pi}{2} \, dx = \frac{\pi}{2} x \]
2. The integral of \( x \):
\[ \int x \, dx = \frac{x^2}{2} \]

### Step 4: Combine the results
Combining the results from the two integrals gives us:
\[ \int \sin^{-1}(\cos x) \, dx = \frac{\pi}{2} x - \frac{x^2}{2} + C \]
where \( C \) is the constant of integration.

### Final Answer
Thus, the final answer is:
\[ \int \sin^{-1}(\cos x) \, dx = \frac{\pi}{2} x - \frac{x^2}{2} + C \]