Question

积分[cosx/(sin^2x-6sinx+12)]dx

Answer 1

令a=sinx

则原式=∫dsinx/(sin²a-6sina+12)

=∫da/[(a-3)²+3]

=1/3*∫da/[(a-3)²/3+1]

=1/3*√3*∫d(a/√3-√3)/[(a/√3-√3)²+1]

=√3/3*arctan(a/√3-√3)+C

=√3/3*arctan(sinx/√3-√3)+C