Let x2≠nπ−1,n∈N.The value of ∫x2sin x2+1−sin 2x2+12sin x2+1+sin 2x2+1dx is equal to
log12secx2+1+C
log sec x2+12+C
12log sec x2+1+C
None of the above
We have, ∫x2sin x2+1−sin 2x2+12sin x2+1+sin 2x2+1dx
=∫x2sin x2+1−2sinx2+1⋅cos x2+12sin x2+1+2sinx2+1⋅cos x2+1dx=∫x1−cos x2+11+cos x2+1dx=∫xtan x2+12dx=∫tan x2+12dx2+12=log sec x2+12+C