分部积分法.以上,请采纳.
∫sin^4xdx/cos^2x =∫[1-cosx]dx/cosx=∫[1-2cosx+cos^4x]dx/cosx=∫[cosx-2+1/cosx]dx=∫(1/2)(cos2x+1)-2+1/cosxdx=(1/4)sin2x-3x/2+tanx+C
-((3 x)/2) + 1/4 Sin[2 x] + Tan[x]
∫[sinxsin(3x)]dx =∫[cos(x-3x)-cos(x+3x)]dx =∫[cos(-2x)-cos(4x)]dx =∫[cos(2x)-cos(4x)]dx =∫cos(2x)dx -∫cos(4x)dx =∫cos(2x)d(2x)-∫cos(4x)d(4x) =-sin(2x)+sin(4x)+C 提示:先对
方法1:原式=∫sinx cosx =∫sinx (1 - sinx) dx =∫(sinx - sin^6x) dx = ∫sinx dx - ∫sin^6x dx 后面的看附图,自己整理吧 方法2:原式=∫sinx cosx dx=∫sinx (sinxcosx) dx=∫sinx * (sin2x / 2) dx=1/4 ∫(1 - cos2x)/2 * ( 1 - cos4x)/2
如图所示
如图 符号改了一次,自己看着办吧
∫ (sinx)^4/(cosx)^2 dx=∫ (1-(cosx)^2)^2/(cosx)^2 dx=∫ (1+(cosx)^4-2(cosx)^2)/(cosx)^2 dx=∫ 1/(cosx)^2+(cosx)^2-2 dx=∫ 1/(cosx)^2+(1+cos2x)/2-2 dx=tanx+x/2+sin2x/4-2x+C=tanx+sin(2x)/4-3x/2+C大概就是这么做,有可能算错= =
(sinx+cosx)^2=2 2cosxsinx=1 sin2x=1 x=π/4所以sin^4x-cos^4x=0分解因式sin^4x-cos^4x=(cos^2x+sin^2x)(sin^2x-cos^2x)=(sinx+cosx)(sinx-cosx)=0望采纳
不定积分sin^4xcos^4xdx=1/16*s(sin2x)^4 dx=1/64*s(1-cos4x)^2 dx=1/64*s(1-2cos4x+(cos4x)^2)dx=1/64*(sdx-s2cos4xdx+1/2*s(1+cos8x)dx)=1/64*x-1/128*sin4x+1/128*sdx+1/128*scos8xdx=x/64-1/128*sin4x+x/128+1/1024*sin8x+c=3x/128-1/128*sin4x+1/1024*sin8x+c
