Instead of solving the specific integral, it will be simpler to derive a recursion formula for the general integral
#color(red)(I_(2n) equiv int sin^(2n) x dx)#
we will also suppress the constant of integration till the end.
Try integration by parts with #sin^(2n-1) x# as the first function, and #sin x# as the second.
#I_{2n} equiv int sin^(2n) x dx = sin^(2n-1)x int sin x dx#
#qquad -int [(d/dx sin^(2n-1) x)times int sin x dx]dx#
#qquad = -sin^(2n-1) x cos x#
#qquad -int [((2n-1)sin^(2n-2)xcos x)times (-cos x)] dx#
#qquad = -sin^(2n-1) x cos x+int (2n-1)sin^(2n-2)xcos^2 x dx#
#qquad = -sin^(2n-1) x cos x#
#qquad +int (2n-1)sin^(2n-2)x(1-sin^2 x) dx#
#qquad = -sin^(2n-1) x cos x+ (2n-1)(I_(2n-2)-I_(2n))#
This can be rearranged to write
#2n I_(2n) = -sin^(2n-1) x cos x+ (2n-1)I_(2n-2)#
so that
#I_(2n) = -1/(2n) sin^(2n-1) x cos x+ (2n-1)/(2n) I_(2n-2)#
This gives a recursion relation that can be used to find #I_(2n)#. Note that #I_0 = int sin^0 x dx = x#
In our case, we are looking for #I_6#
#I_6 = -1/6sin^5xcosx+5/6I_4#
#qquad =-1/6sin^5xcosx+5/6(-1/4sin^3xcos x+3/4I_2)#
#qquad =-1/6sin^5xcosx-5/24sin^3xcos x #
#qquad quad +5/8(-1/2sin x cos x+1/2x)#
#qquad = -1/6sin^5xcosx-5/24sin^3xcos x #
#qquad qquad -5/16 sin x cos x+5/16 x#