微积分:求反导数
2个回答

∫π[√2-tan(4x)sec(4x)]²dx

=π∫[2-2√2tan(4x)sec(4x)+tan²(4x)sec²(4x)]dx

=π[2∫dx-2√2∫tan(4x)sec(4x)dx+∫tan²(4x)sec²(4x)dx]

=π[2x-(√2/2)∫tan(4x)sec(4x)d(4x)+(1/4)∫tan²(4x)d(tan(4x))]

=π[2x-(√2/2)sec(4x)+(1/4)*(1/3)tan³(4x)]+C

=π[2x-(√2/2)sec(4x)+(1/12)tan³(4x)]+C

C为任意常数