Question

$\int_{0}^{1}\frac{\sin x}{\sin x+\sin(1-x)}\,dx$ is equal to:

• A. \(6\)
• B. \(4\)
• C. \(2\)
• D. \(\frac{1}{2}\)
• E. \(\frac{1}{4}\)

Hint:

For definite integrals involving \(f(x)\) and \(f(a-x)\), always try substitution \(x \to a-x\) and add both expressions.

Answer 1

The Correct Option is D

Step 1: Let the integral be \(I\).
\[ I=\int_{0}^{1}\frac{\sin x}{\sin x+\sin(1-x)}\,dx \]

Step 2: Use the property of definite integrals.
Replace \(x\) by \(1-x\), then: \[ I=\int_{0}^{1}\frac{\sin(1-x)}{\sin(1-x)+\sin x}\,dx \]

Step 3: Add the two expressions of \(I\).
\[ 2I=\int_{0}^{1}\left[\frac{\sin x}{\sin x+\sin(1-x)}+\frac{\sin(1-x)}{\sin x+\sin(1-x)}\right]dx \]

Step 4: Simplify the integrand.
\[ \frac{\sin x+\sin(1-x)}{\sin x+\sin(1-x)}=1 \] Thus, \[ 2I=\int_{0}^{1}1\,dx \]

Step 5: Evaluate the integral.
\[ 2I=1-0=1 \]

Step 6: Solve for \(I\).
\[ I=\frac{1}{2} \]

Step 7: Final answer.
\[ \boxed{\frac{1}{2}} \] which matches option \((4)\).