write sin2x as
(sinx+cosx)2=1+sin2x
sin2x=(sinx+cosx)2−1
∫√8−[(sinx+cosx)2−1)]cosx−sinx
∫√9−(sinx+cosx)2cosx−sinx
put sinx + cosx = t
cosx - sinx dx = dt
∫√9−(t)2dt→sin−1(3t)
∫√8−sin2xcosx−sinxdx
ans-----> sin−1(3sinx+cosx)+c
write sin2x as
(sinx+cosx)2=1+sin2x
sin2x=(sinx+cosx)2−1
∫√8−[(sinx+cosx)2−1)]cosx−sinx
∫√9−(sinx+cosx)2cosx−sinx
put sinx + cosx = t
cosx - sinx dx = dt
∫√9−(t)2dt→sin−1(3t)