Let `I_1=int_0^(pi/2)(dt)/(1+t^6)and I_2=int_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 find their relationship.

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

### Step 2: Change of variable in \( I_2 \)
We will perform a substitution in \( I_2 \). Let:
\[ t = x \sin x + \cos x \]
Now, we need to differentiate \( t \) with respect to \( x \):
\[ \frac{dt}{dx} = x \cos x + \sin x - \sin x = x \cos x \]
Thus, we can express \( dx \) in terms of \( dt \):
\[ dx = \frac{dt}{x \cos x} \]

### Step 3: Substitute in \( I_2 \)
Substituting \( t \) and \( dx \) into \( I_2 \):
\[ I_2 = \int_0^{\frac{\pi}{2}} \frac{x \cos x}{1 + t^6} \cdot \frac{dt}{x \cos x} \]
The \( x \cos x \) terms cancel out:
\[ I_2 = \int_0^{\frac{\pi}{2}} \frac{dt}{1 + t^6} \]

### Step 4: Compare \( I_2 \) with \( I_1 \)
Notice that:
\[ I_2 = \int_0^{\frac{\pi}{2}} \frac{dt}{1 + t^6} = I_1 \]

### Conclusion
Thus, we find that:
\[ I_1 = I_2 \]

### Final Answer
The correct option is that \( I_1 \) is equal to \( I_2 \).

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