Question

求高数高手 求解答此题的不定积分

Answer 1

问：求高数高手 求解答此题的不定积分

答：您好，∫tan^4x dx=∫tan²xtan²x dx=∫(sec²x-1)tan²x dx=∫(sec²x)(tan²x) dx - ∫tan²x dx=∫tan²x d(tanx) - ∫(sec²x-1) dx=(1/3)tan³x - tanx + x + Csinxsin3x=-(cos4x-cos2x)/2∫sinxsin2xsin3xdx=(1/4)∫sin4xdx - (1/2)∫sin2xcos4xdx=-(1/16)cos4x - (1/4)∫cos4xd(cos2x)=-(1/16)cos4x - (1/4)∫(2(cos2x)^2-1)d(cos2x)=-(1/16)cos4x + (1/4)cos2x - (1/2)∫(cos2x)^2d(cos2x)=-(1/16)cos4x + (1/4)cos2x - (1/6)(cos2x)^3+c∫sinxcosx/(1+sin^4x)dx=∫sinx/(1+sin^4x)d(sinx)=1/2*∫1/(1+(sin^2x)^2)d(sin^2x)=1/2*arctan(sin^2x)+C