The value of  I=∫log⁡2log⁡3 xsin⁡x2sin⁡x2+sin⁡log⁡6−x2dx is

Put x2=t, so thatI=12∫log⁡2log⁡3 sin⁡tsin⁡t+sin⁡(log⁡6−t)dt=12∫log⁡2log⁡3 sin⁡(log⁡2+log⁡3−t)sin⁡(log⁡2+log⁡3−t)+sin⁡tdt=12∫log⁡2log⁡3 sin⁡(log⁡6−t)sin⁡t+sin⁡(log⁡6−t)dt2I=I+I=12∫log⁡2log⁡3 sin⁡t+sin⁡(log⁡6−t)sin⁡t+sin⁡(log⁡6−t)dt=12[log⁡3−log⁡2]⇒  I=14log⁡32.