Now we know that,∫baf(x).dx=∫baf(a+b−x).dx
⇒I=∫π2−π2(sin4(−x)).(1+log(2+sin(−x)2−sin(−x))).dx
=∫π2−π2(sin4x).(1+log(2−sinx2+sinx)).dx
=∫π2−π2sin4x.(1−log(2+sinx2−sinx)).dx ------ii
Adding equation i and ii we get,
2I=2∫π2−π2sin4x.dx
2I=4∫π20sin4x.dx
I=2∫π20sin4x.dx
Use the reduction formula
∫sinmx.dx=−cosx.sinm−1(x)m+m−1m∫sin−2+mx.dx
Here,2∫π20sin4x.dx=−2cosx.sin3x4∣∣π20+3.24∫π20sin2x.dx (As −cosx.sin3x∣∣π20=0)
⇒64∫π201−cos2x2.dx
=68∫π201.dx−68∫π20cos2x.dx
After putting the limits we get,
=3π8 As(∫π20cos2x.dx)=0
⇒3π8