If `f(x)= int_0^(sinx) cos^(-1)t dt +int_(0)^(cosx) sin^(-1)t dt, 0 lt x lt (pi)/(2)` then ` f(pi//4)` is equal to

Options

A. `(pi)/(sqrt(2))`
B. `1+(pi)/(2sqrt(2))`
C. 1
D. none of these

Explanation

To solve the problem, we need to evaluate the function \( f(x) \) defined as:

\[ f(x) = \int_0^{\sin x} \cos^{-1} t \, dt + \int_0^{\cos x} \sin^{-1} t \, dt \]

We are tasked with finding \( f\left(\frac{\pi}{4}\right) \).

### Step 1: Substitute \( x = \frac{\pi}{4} \)

First, we substitute \( x = \frac{\pi}{4} \) into the function:

\[ f\left(\frac{\pi}{4}\right) = \int_0^{\sin\left(\frac{\pi}{4}\right)} \cos^{-1} t \, dt + \int_0^{\cos\left(\frac{\pi}{4}\right)} \sin^{-1} t \, dt \]

### Step 2: Calculate \( \sin\left(\frac{\pi}{4}\right) \) and \( \cos\left(\frac{\pi}{4}\right) \)

We know that:

\[ \sin\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]

Thus, we can rewrite the function as:

\[ f\left(\frac{\pi}{4}\right) = \int_0^{\frac{1}{\sqrt{2}}} \cos^{-1} t \, dt + \int_0^{\frac{1}{\sqrt{2}}} \sin^{-1} t \, dt \]

### Step 3: Combine the Integrals

Since both integrals have the same limits, we can combine them:

\[ f\left(\frac{\pi}{4}\right) = \int_0^{\frac{1}{\sqrt{2}}} \left(\cos^{-1} t + \sin^{-1} t\right) dt \]

### Step 4: Use the Identity

We use the identity:

\[ \sin^{-1} t + \cos^{-1} t = \frac{\pi}{2} \]

Thus, we can simplify the integral:

\[ f\left(\frac{\pi}{4}\right) = \int_0^{\frac{1}{\sqrt{2}}} \frac{\pi}{2} \, dt \]

### Step 5: Evaluate the Integral

Now we can evaluate the integral:

\[ f\left(\frac{\pi}{4}\right) = \frac{\pi}{2} \int_0^{\frac{1}{\sqrt{2}}} dt \]

The integral \( \int_0^{\frac{1}{\sqrt{2}}} dt \) is simply:

\[ \int_0^{\frac{1}{\sqrt{2}}} dt = \left[ t \right]_0^{\frac{1}{\sqrt{2}}} = \frac{1}{\sqrt{2}} - 0 = \frac{1}{\sqrt{2}} \]

### Step 6: Final Calculation

Substituting back, we get:

\[ f\left(\frac{\pi}{4}\right) = \frac{\pi}{2} \cdot \frac{1}{\sqrt{2}} = \frac{\pi}{2\sqrt{2}} \]

Thus, the final answer is:

\[ f\left(\frac{\pi}{4}\right) = \frac{\pi}{2\sqrt{2}} \]