Question

The value of \(\displaystyle\int_{\frac{\pi}{8}}^{\frac{3\pi}{8}}\frac{sin^4x}{sin^4x+cos^4x}dx\) is equal to

• A. \(\frac{\pi}{4}\)
• B. \(\frac{\pi}{8}\)
• C. \(\frac{\pi}{16}\)
• D. \(\frac{\pi}{2}\)
• E. 1

Answer 1

The Correct Option is B

We are asked to find the value of the integral \( \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} \frac{\sin^4 x}{\sin^4 x + \cos^4 x} \, dx \).

We observe that the integrand has symmetry, so we can use a substitution technique to simplify the calculation. Let:

\( I = \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} \frac{\sin^4 x}{\sin^4 x + \cos^4 x} \, dx \). 

By the symmetry of sine and cosine, we can make a substitution: \( x' = \frac{\pi}{2} - x \). This transforms the integral as follows:

\( I = \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} \frac{\cos^4 x}{\sin^4 x + \cos^4 x} \, dx \).

Now, adding the two integrals together, we get:

\( 2I = \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} \frac{\sin^4 x + \cos^4 x}{\sin^4 x + \cos^4 x} \, dx = \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} 1 \, dx \).

The integral of 1 with respect to \( x \) is simply \( x \), so we have:

\( 2I = \left[ x \right]_{\frac{\pi}{8}}^{\frac{3\pi}{8}} = \frac{3\pi}{8} - \frac{\pi}{8} = \frac{\pi}{4} \).

Thus, \( I = \frac{\pi}{8} \).

The correct answer is \( \frac{\pi}{8} \).