If `a=int_(0)^(1)(cos(sinx))/(secx)dx,` then the value of `a^(2)+cos^(2)(sin1)` is equal to

Options

A. 0
B. 1
C. `sin(1)`
D. `sin(sin1)`

Answer

To solve the problem, we need to evaluate the integral \( a = \int_{0}^{1} \frac{\cos(\sin x)}{\sec x} \, dx \) and then find the value of \( a^2 + \cos^2(\sin 1) \).

### Step 1: Simplify the integral
We start with the integral:
\[ a = \int_{0}^{1} \frac{\cos(\sin x)}{\sec x} \, dx \]
Since \( \sec x = \frac{1}{\cos x} \), we can rewrite the integral as:
\[ a = \int_{0}^{1} \cos(\sin x) \cdot \cos x \, dx \]

### Step 2: Use substitution
Let \( t = \sin x \). Then, the differential \( dx \) can be expressed as:
\[ dx = \frac{dt}{\cos x} \]
We also need to change the limits of integration. When \( x = 0 \), \( t = \sin(0) = 0 \). When \( x = 1 \), \( t = \sin(1) \).

Thus, the integral becomes:
\[ a = \int_{0}^{\sin 1} \cos(t) \cdot \frac{dt}{\cos x} \]
Since \( \cos x = \sqrt{1 - t^2} \) (from \( t = \sin x \)), we have:
\[ a = \int_{0}^{\sin 1} \cos(t) \cdot \frac{dt}{\sqrt{1 - t^2}} \]

### Step 3: Recognize the integral
The integral \( \int \cos(t) \, dt \) is straightforward:
\[ \int \cos(t) \, dt = \sin(t) \]
Thus, we compute:
\[ a = \left[ \sin(t) \right]_{0}^{\sin 1} = \sin(\sin 1) - \sin(0) = \sin(\sin 1) \]

### Step 4: Calculate \( a^2 + \cos^2(\sin 1) \)
Now we need to compute \( a^2 + \cos^2(\sin 1) \):
\[ a^2 = (\sin(\sin 1))^2 \]
Thus, we have:
\[ a^2 + \cos^2(\sin 1) = \sin^2(\sin 1) + \cos^2(\sin 1) \]
Using the Pythagorean identity:
\[ \sin^2(x) + \cos^2(x) = 1 \]
for \( x = \sin 1 \), we find:
\[ \sin^2(\sin 1) + \cos^2(\sin 1) = 1 \]

### Final Answer
Therefore, the value of \( a^2 + \cos^2(\sin 1) \) is:
\[ \boxed{1} \]