1well, reduction formula can be used ,
\int sin^{n}x.dx=\frac{-sin^{n-1}x.cosx}{n}+\frac{n-1}{n}\int sin^{n-2}x.dx
but it'll make the prblm really really lengthy...still thinking of some ther method
1cn try it another way;
let,z be a complex number. z=cosx + isinx and 1/z = cosx - isinx
now, z-1/z = 2i sinx , (z-1/z)12 = 212 sin12x
now,by expanding (z-1/z)12,we'll get in the form of conjugates
and then putting the expansion of z, sin functions get cancelled out
and cos func. remain( De'Moivre's theorem is used for compact form).
now it cn be integrated.
106abhisek this was for others to try. dont post tuition material here please
