Let `f` beintegrable over `[0,a]` for any real value of `a`.
If `I_(1)=int_(0)^(pi//2)cos theta f(sin theta +cos^(2) theta) d theta` and `I_(2)=int_(0)^(pi//2) sin 2 theta f(sin theta+cos^(2) theta) d theta`, then

Options

A. `I_(1)=-2I_(2)`
B. `I_(1)=I_(2)`
C. `2I_(1)=I_(2)`
D. `I_(1)=-I_(2)`

Explanation

To solve the problem, we need to find a relationship between the integrals \( I_1 \) and \( I_2 \) defined as follows:

\[ I_1 = \int_0^{\frac{\pi}{2}} \cos \theta \, f(\sin \theta + \cos^2 \theta) \, d\theta \]

\[ I_2 = \int_0^{\frac{\pi}{2}} \sin 2\theta \, f(\sin \theta + \cos^2 \theta) \, d\theta \]

### Step 1: Rewrite \( I_1 - I_2 \)

We start by expressing \( I_1 - I_2 \):

\[ I_1 - I_2 = \int_0^{\frac{\pi}{2}} \left( \cos \theta - \sin 2\theta \right) f(\sin \theta + \cos^2 \theta) \, d\theta \]

### Step 2: Simplify \( \sin 2\theta \)

Recall that \( \sin 2\theta = 2 \sin \theta \cos \theta \). Thus, we can rewrite the expression:

\[ I_1 - I_2 = \int_0^{\frac{\pi}{2}} \left( \cos \theta - 2 \sin \theta \cos \theta \right) f(\sin \theta + \cos^2 \theta) \, d\theta \]

### Step 3: Factor out the common terms

Now, factor out \( \cos \theta \):

\[ I_1 - I_2 = \int_0^{\frac{\pi}{2}} \cos \theta \left( 1 - 2 \sin \theta \right) f(\sin \theta + \cos^2 \theta) \, d\theta \]

### Step 4: Change of Variable

Let \( t = \sin \theta + \cos^2 \theta \). We need to find \( dt \):

- The derivative \( dt = (\cos \theta - 2 \sin \theta \cos \theta) d\theta \).

### Step 5: Determine the limits of integration

When \( \theta = 0 \):
\[ t = \sin(0) + \cos^2(0) = 0 + 1 = 1 \]

When \( \theta = \frac{\pi}{2} \):
\[ t = \sin\left(\frac{\pi}{2}\right) + \cos^2\left(\frac{\pi}{2}\right) = 1 + 0 = 1 \]

### Step 6: Evaluate the integral

Thus, the limits of integration for \( t \) are from \( 1 \) to \( 1 \):

\[ I_1 - I_2 = \int_1^1 f(t) \, dt = 0 \]

### Conclusion

Since \( I_1 - I_2 = 0 \), we conclude that:

\[ I_1 = I_2 \]

### Final Answer

Thus, the relationship between \( I_1 \) and \( I_2 \) is:

\[ I_1 = I_2 \]