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 the 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: Change of Variables for \( I_1 \)

Let's perform a substitution in \( I_1 \). Let:

\[ t = \sin \theta + \cos^2 \theta \]

To find \( dt \), we differentiate \( t \):

\[ dt = \cos \theta \, d\theta - 2\sin \theta \cos \theta \, d\theta = \cos \theta (1 - 2\sin \theta) \, d\theta \]

### Step 2: Determine the Limits of Integration

Now, we need to find the limits of integration when \( \theta = 0 \) and \( \theta = \frac{\pi}{2} \):

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

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

Thus, both limits are 1, which means the integral evaluates to 0.

### Step 3: Evaluate \( I_1 \)

Now substituting back into the integral, we have:

\[ I_1 = \int_0^{\frac{\pi}{2}} \cos \theta \, f(t) \, d\theta = \int_1^1 f(t) \, dt = 0 \]

### Step 4: Change of Variables for \( I_2 \)

Now let's analyze \( I_2 \):

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

Using the same substitution \( t = \sin \theta + \cos^2 \theta \), we have:

\[ \sin 2\theta = 2 \sin \theta \cos \theta \]

Thus, the differential \( d\theta \) can be expressed as:

\[ d\theta = \frac{dt}{\cos \theta (1 - 2\sin \theta)} \]

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

Substituting into \( I_2 \):

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

Using the limits we found earlier, we also see that:

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

### Conclusion

Since both integrals evaluate to zero, we have:

\[ I_1 - I_2 = 0 \implies I_1 = I_2 \]

Thus, the final result is:

\[ I_1 = I_2 \]