Let `I_1=int_0^(pi/2)(dt)/(1+t^6)and I_2int_0^(pi/2)(xcosxdx)/(1+(xsinx+cosx)^6),` then

Options

A. `2I_1 =I_2`
B. `I_1 =2I_2`
C. `I_1=I_2`
D. `I_1=I_2=0`

Answer

To solve the problem, we need to evaluate the integrals \( I_1 \) and \( I_2 \) and show that they are equal.

### Step 1: Define the integrals
We have:
\[ I_1 = \int_0^{\frac{\pi}{2}} \frac{dt}{1 + t^6} \]
\[ I_2 = \int_0^{\frac{\pi}{2}} \frac{x \cos x \, dx}{1 + (x \sin x + \cos x)^6} \]

### Step 2: Simplify \( I_2 \)
To simplify \( I_2 \), we will use a substitution. Let:
\[ t = x \sin x + \cos x \]
We need to find \( dt \) in terms of \( dx \).

### Step 3: Differentiate \( t \)
Differentiating \( t \) with respect to \( x \):
\[ dt = \left( \sin x + x \cos x \right) dx - \sin x \, dx = (x \cos x) \, dx \]

### Step 4: Substitute in \( I_2 \)
Now we can express \( I_2 \) in terms of \( dt \):
\[ I_2 = \int_0^{\frac{\pi}{2}} \frac{dt}{1 + t^6} \]
Here, we have replaced \( x \cos x \, dx \) with \( dt \) in the numerator and \( 1 + (x \sin x + \cos x)^6 \) with \( 1 + t^6 \) in the denominator.

### Step 5: Conclusion
Thus, we have shown that:
\[ I_2 = \int_0^{\frac{\pi}{2}} \frac{dt}{1 + t^6} = I_1 \]
This means that \( I_2 = I_1 \).

### Final Result
Therefore, we conclude that:
\[ I_1 = I_2 \]