Question

\( \int_{-\pi}^{\pi} \frac{\sin^2 x}{1+7^x}\, dx \) is equal to

• A. \(7^\pi\)
• B. \( \pi \)
• C. \( \frac{\pi}{2} \)
• D. \( 2^\pi \)
• E. \( 7\pi \)

Hint:

Use substitution \(x \to -x\) cleverly to simplify complicated symmetric integrals.

Answer 1

The Correct Option is C

Concept: Use symmetry: \[ I = \int_{-a}^{a} f(x)\,dx \] and substitution \(x \to -x\)

Step 1: Let integral be
\[ I = \int_{-\pi}^{\pi} \frac{\sin^2 x}{1+7^x}\,dx \]

Step 2: Replace \(x\) by \(-x\)
\[ I = \int_{-\pi}^{\pi} \frac{\sin^2 x}{1+7^{-x}}\,dx \] Multiply numerator and denominator: \[ = \int_{-\pi}^{\pi} \frac{7^x \sin^2 x}{1+7^x}\,dx \]

Step 3: Add both expressions
\[ 2I = \int_{-\pi}^{\pi} \sin^2 x \left(\frac{1}{1+7^x} + \frac{7^x}{1+7^x}\right) dx \] \[ = \int_{-\pi}^{\pi} \sin^2 x \, dx \]

Step 4: Evaluate integral
\[ \int_{-\pi}^{\pi} \sin^2 x \, dx = \pi \] (using average value \(1/2\) over interval length \(2\pi\))

Step 5: Solve for \(I\)
\[ 2I = \pi \Rightarrow I = \frac{\pi}{2} \] Final Answer: \[ \boxed{\frac{\pi}{2}} \]