For `f(x) =x^(4) +|x|, let I_(1)= int _(0)^(pi)f(cos x) dx` and `I_(2)= int_(0)^(pi//2) f(sin x ) dx "then" (I_(1))/(I_(2))` has the value equal to

Options

A. 1
B. `1//2`
C. 2
D. 4

Answer

To solve the problem, we need to evaluate the integrals \( I_1 \) and \( I_2 \) defined as follows:

1. \( I_1 = \int_0^{\pi} f(\cos x) \, dx \)
2. \( I_2 = \int_0^{\frac{\pi}{2}} f(\sin x) \, dx \)

where \( f(x) = x^4 + |x| \).

### Step 1: Analyze the function \( f(x) \)

The function \( f(x) = x^4 + |x| \) is an even function because:
- \( x^4 \) is even (i.e., \( f(-x) = (-x)^4 + |-x| = x^4 + |x| = f(x) \)).
- \( |x| \) is also even.

### Step 2: Evaluate \( I_1 \)

Using the property of even functions, we can transform \( I_1 \):
\[ I_1 = \int_0^{\pi} f(\cos x) \, dx \]
We can use the substitution \( x = \pi - t \), which gives \( dx = -dt \). Thus, we have:
\[ I_1 = \int_{\pi}^{0} f(\cos(\pi - t)) (-dt) = \int_0^{\pi} f(-\cos t) \, dt \]
Since \( f(x) \) is even, \( f(-\cos t) = f(\cos t) \). Therefore:
\[ I_1 = \int_0^{\pi} f(\cos t) \, dt \]
This shows that \( I_1 \) can be expressed as:
\[ I_1 = 2 \int_0^{\frac{\pi}{2}} f(\cos x) \, dx \]

### Step 3: Evaluate \( I_2 \)

Now we evaluate \( I_2 \):
\[ I_2 = \int_0^{\frac{\pi}{2}} f(\sin x) \, dx \]

### Step 4: Relate \( I_1 \) and \( I_2 \)

From the symmetry and properties of the sine and cosine functions, we can relate \( I_1 \) and \( I_2 \):
\[ I_1 = 2 I_2 \]

### Step 5: Calculate \( \frac{I_1}{I_2} \)

Now we can find the ratio:
\[ \frac{I_1}{I_2} = \frac{2 I_2}{I_2} = 2 \]

### Final Answer

Thus, the value of \( \frac{I_1}{I_2} \) is:
\[ \boxed{2} \]