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∫Cos^4Dx

=∫(cos²x)²dx =∫[(1+cos2x)/2]²dx =∫(cos²2x+2cos2x+1)/4 dx =1/4*∫cos²2xdx+1/2*∫cos2xdx+∫dx/4 =1/4*∫(1+cos4x)/2 dx+1/4*sin2x+x/4 =x/8+1/32*sin4x+1/4*sin2x+x/4+C =1/32*sin4x+1/4*sin2x+3x/8+C

一楼的答案好像错了呀 ∫cos^4xdx =∫((1+cos2x)/2)²dx =1/4∫(1+2cos2x+(1+cos4x)/2)dx =∫(3/8+cos2x/2+cos4x/8)dx =3x/8+sin2x/4+sin4x/32+C

(cosx)^2=1/2 cos2x+ 1/2 所以 (cosx)^4=(1/2 cos2x+ 1/2)^2 =1/4 *(cos2x)^2 +1/2 *cos2x +1/4 =1/8 *cos4x + 1/2 *cos2x +3/8 因此得到 ∫ (cosx)^4 dx = ∫ 1/8 *cos4x + 1/2 *cos2x +3/8 dx = 1/32 *sin4x + 1/4 *sin2x +3x/8 +C,C为常数

欢迎采纳,不要点错答案哦╮(╯◇╰)╭

先降次 原式=1/8∫(sin2x)^2·(1+cos2x)dx =1/8∫(sin2x)^2dx+1/8∫(sin2x)^2·cos2xdx =1/16∫(1-cos4x)dx+1/16∫(sin2x)^2d(sin2x) =x/16-1/64·sin4x+1/48·(sin2x)^3+C

(sinx)^2(cosx)^4=[1-(cosx)^2](cosx)^4=(cosx)^4-(cosx)^6 =[(1+cos2x)/2]^2-[(1+cos2x)/2]^3 =[(cos2x)^2+2cosx+1]/4-[(cos2x)^3+3(cos2x)^2+3cos2x+1]/8 ={[(1+cos4x)/2]+2cosx+1]/4-{(cos6x+3cos2x)/4+3[(1+cos4x)/2]+3cos2x+1}/8 =(cos4x+4...

(sinx)^6(cosx)^4 =sin²x(sin²xcos²x)² =¼sin²xsin²(2x) =(¼)(¼)[cos(2x-x)-cos(2x+x)]² =(1/16)[cosx-cos(3x)]² =(1/16)[cos²x-2cosxcos(3x)+cos²(3x)] =(1/16)cos²x-...

J = 4∫_(-π/2)^(π/2) cos⁴x dx =4 ∫_(-π/2)^(π/2) [(1 + cos2x)/2]² dx = ∫_(-π/2)^(π/2) (1 + 2cos2x + cos²2x) dx = ∫_(-π/2)^(π/2) (1 + 2cos2x) dx +(1/2)∫_(-π/2)^(π/2) (1 + cos4x) dx = (x + sin2x)_(-π/2)^(π/2) + (1/2...

利用三角恒等式和分部积分 ∫x(sinx)^4dx = (3/8)∫xdx - (1/2)∫x*cos(2x)dx + (1/8)∫x*cos(4x)dx = (3/16)x^2 - (1/2)*(1/2)[x*sin(2x)-∫sins(2x)dx] + (1/8)*(1/4)[x*sin(4x)-∫sin(4x)dx] = (3/16)*x^2 - (1/4)x*sin(2x)-(1/8)*cos(2x) + (1/32)...

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