∫cos(2x+1)^5dx=?

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∫cos(2x+1)^5dx=?
∫cos(2x+1)^5dx=?

∫cos(2x+1)^5dx=?
5/16 *sin(2x+1)+5/96 *sin[3(2x+1)]+1/160 *sin[5(2x+1)]

∫[cos(2x+1)]^5dx
= (1/2)∫ [cos(2x+1)]^4 d sin(2x+1)
=(1/2)∫ {1-2(sin(2x+1))^2+ [sin(2x+1)]^4} d sin(2x+1)
=(1/2)[ sin(2x+1) - 2(sin(2x+1))^3/3 + [sin(2x+1)]^5/5 ] +C

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给一个通用解法，可以对任意的∫cos^mxdx来解
首先给一个小公式：
∫cos^m(x)dx=∫cos^(m-1)(x)d(sinx)=sinxcos^(m-1)x+∫(m-1)sin^2xcos^(m-2)xdx
=sinxcos^(m-1)x+∫(m-1)(1-cos^2x)cos^(m-2)xdx
=sinxcos^(m-1)x+∫(m-1)cos^(m-2)xdx-∫(m-1)cos^m(x)dx
也就是
∫cos^m(x)dx=1/msinxcos^(m-1)x+(m-1)/m∫cos^(m-2)xdx
∫cos(2x+1)^5dx
=1/2∫cosu^5du u=2x+1
=1/10sinucos^4u+2/5∫cos^3udu （用到上面的小公式）
=1/10sinucos^4u+2/15sinucos^2u+4/15∫cosudu
=1/10sinucos^4u+2/15sinucos^2u+4/15sinu
=1/10sin(2x+1)cos^4(2x+1)+2/15sin(2x+1)cos^2(2x+1)+4/15sin(2x+1)+C

收起

∫cos(2x+1)^5dx=? ∫(2/cos^2x)dx= ∫1/(1+cos^2(x)) dx ∫(1-cos^(2)2x)dx ∫cos((x/3)-1)dx=? ∫cos(x+2)dx=多少 ∫(1+sinx) / cos^2 x dx 不定积分cos(x^1/2)dx=? 微积分1/cos^2(x)dx=?微积分1/cos^2(x)dx＝? ：求 ∫ sin^2x cos^2x dx 和 ∫ x^5根号(1-x^2) dx ∫3/(x*x*x+1)dx ∫1/(2+sin x)dx 和 ∫1/(2sin x-cos x+5)dx的解法 / (cos x + 1 ) dx cos(x^2)dx ∫cos x / ( 1 + (sinx)^2 ) dx = ∫x^3 / ( 1 + x^4 ) dx = ∫(sec x)^3 * tan x dx = ∫x^2 * e^(-2x) dx = ∫x * cos 2x dx = ∫(cos 2x)^2 dx = ∫(13x - 6) / ( x (x-2)(x+3) )dx = ∫(x^2 + 2x - 2) / ((x-2)(x+1)) dx = 请把过程写出来哈.>< 旷这个怎恶魔算∫cos x / ( 1 + (sinx)^2 ) dx = ∫x^3 / ( 1 + x^4 ) dx = ∫(sec x)^3 * tan x dx = ∫x^2 * e^(-2x) dx = ∫x * cos 2x dx = ∫(cos 2x)^2 dx = ∫(13x - 6) / ( x (x-2)(x+3) )dx = ∫(x^2 + 2x - 2) / ((x-2)(x+1)) dx = 定积分∫(x^2arctanx+cos^5x)dx ∫(1+sin^2x)/(cos^2x)dx ∫sin^3(x)cos^2(x)dx=