If `I_(1)=int_(0)^(pi//4)sin^(2)xdx and I_(2)=int_(0)^(pi//4)cos^(2)xdx`, then

Options

A. `I_(1)=I_(2)`
B. `I_(1) lt I_(2)`
C. `I_(1) gt I_(2)`
D. `I_(2)=I_(1)+pi//4`

Answer

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

\[ I_1 = \int_{0}^{\frac{\pi}{4}} \sin^2 x \, dx \]
\[ I_2 = \int_{0}^{\frac{\pi}{4}} \cos^2 x \, dx \]

### Step 1: Evaluate \( I_1 \)

Using the identity \( \sin^2 x = \frac{1 - \cos(2x)}{2} \), we can rewrite \( I_1 \):

\[ I_1 = \int_{0}^{\frac{\pi}{4}} \sin^2 x \, dx = \int_{0}^{\frac{\pi}{4}} \frac{1 - \cos(2x)}{2} \, dx \]

### Step 2: Simplify the Integral

Now, we can separate the integral:

\[ I_1 = \frac{1}{2} \int_{0}^{\frac{\pi}{4}} (1 - \cos(2x)) \, dx = \frac{1}{2} \left( \int_{0}^{\frac{\pi}{4}} 1 \, dx - \int_{0}^{\frac{\pi}{4}} \cos(2x) \, dx \right) \]

### Step 3: Evaluate Each Part

1. **Evaluate \( \int_{0}^{\frac{\pi}{4}} 1 \, dx \)**:
\[ \int_{0}^{\frac{\pi}{4}} 1 \, dx = \left[ x \right]_{0}^{\frac{\pi}{4}} = \frac{\pi}{4} \]

2. **Evaluate \( \int_{0}^{\frac{\pi}{4}} \cos(2x) \, dx \)**:
\[ \int \cos(2x) \, dx = \frac{1}{2} \sin(2x) \]
Thus,
\[ \int_{0}^{\frac{\pi}{4}} \cos(2x) \, dx = \left[ \frac{1}{2} \sin(2x) \right]_{0}^{\frac{\pi}{4}} = \frac{1}{2} \left( \sin\left(\frac{\pi}{2}\right) - \sin(0) \right) = \frac{1}{2}(1 - 0) = \frac{1}{2} \]

### Step 4: Combine Results for \( I_1 \)

Substituting back into the equation for \( I_1 \):

\[ I_1 = \frac{1}{2} \left( \frac{\pi}{4} - \frac{1}{2} \right) = \frac{\pi}{8} - \frac{1}{4} \]

### Step 5: Evaluate \( I_2 \)

Using the identity \( \cos^2 x = \frac{1 + \cos(2x)}{2} \), we can rewrite \( I_2 \):

\[ I_2 = \int_{0}^{\frac{\pi}{4}} \cos^2 x \, dx = \int_{0}^{\frac{\pi}{4}} \frac{1 + \cos(2x)}{2} \, dx \]

### Step 6: Simplify the Integral for \( I_2 \)

Similar to \( I_1 \):

\[ I_2 = \frac{1}{2} \int_{0}^{\frac{\pi}{4}} (1 + \cos(2x)) \, dx = \frac{1}{2} \left( \int_{0}^{\frac{\pi}{4}} 1 \, dx + \int_{0}^{\frac{\pi}{4}} \cos(2x) \, dx \right) \]

### Step 7: Evaluate Each Part for \( I_2 \)

Using the results from \( I_1 \):

1. **Evaluate \( \int_{0}^{\frac{\pi}{4}} 1 \, dx \)**:
\[ \int_{0}^{\frac{\pi}{4}} 1 \, dx = \frac{\pi}{4} \]

2. **Evaluate \( \int_{0}^{\frac{\pi}{4}} \cos(2x) \, dx \)**:
\[ \int_{0}^{\frac{\pi}{4}} \cos(2x) \, dx = \frac{1}{2} \]

### Step 8: Combine Results for \( I_2 \)

Substituting back into the equation for \( I_2 \):

\[ I_2 = \frac{1}{2} \left( \frac{\pi}{4} + \frac{1}{2} \right) = \frac{\pi}{8} + \frac{1}{4} \]

### Step 9: Compare \( I_1 \) and \( I_2 \)

Now we have:

\[ I_1 = \frac{\pi}{8} - \frac{1}{4} \]
\[ I_2 = \frac{\pi}{8} + \frac{1}{4} \]

### Conclusion

Since \( I_2 \) is greater than \( I_1 \):

\[ I_2 > I_1 \]