如今咱们对相关于∫sinx/(4-cosx)dx幕详情太令人震惊，咱们都需要剖析一下∫sinx/(4-cosx)dx，那么婷婷也在网络上收集了一些对相关于dx sinxcosx的不定积分的一些信息来分享给咱们，具体事件详情揭秘，希望能够帮到咱们哦。

∫(0,π/4)(cosx-sinx)dx =sinx+cosx|(上π/4下0) =√2-1∫(π/4,π/2)(sinx-cosx)dx =-sinx-cosx|(上π/2下π/4) =-1+.

4] / cos²x dx =∫ (1/cos²x -2 +cos²x) dx = tanx -2x +∫ cos²x dx = tanx -2x +∫ (0.5+0.5cos2x) dx = tan.

∫sinx四次方*cosx三次方*dx = ∫sinx四次方*cosx二次方*dsinx = ∫sinx四次方*(1-sinx二次方)*dsinx =∫sinx四次方*dsinx-∫sinx六次方)*.

2 ∫(cosx)^4 dx =∫(1-sinx^2)cosx^2dx =∫cosx^2dx-∫sinx^2cosx^2dx =∫(1/2)(1+cos2x)x-∫(1/4)[(1-cos4x)/2].

2]dx = (1/4)∫[1-2cos2x+(cos2x)^2]dx = (1/4)∫[3/2-2cos2x+(1/2)cos4x]dx = (1/4)∫[3x/2-sin2x+(1/8)sin4x.

∫(sinx+cosx)dx =-cosx+sinx+C

原式=∫x/(1+cosX)dx+∫sinX/(1+cosX)dx =∫xsec^2(x/2)d(x/2)-∫1/(1+cosx)d(1+cosx) =∫xd[tan(x/2)]-ln(1+cosx) =xtan(x/2)-∫tan(x/2)dx-ln(1+cosx) =xtan(x/2)+2ln∣cos(x/2)∣-ln2-ln∣cos(x/2)∣+C1 =xtan(x/2)+C

∫(sinx)^4dx= (sin4x)/32 - (sin2x)/4 + (3x/8) + C.C为常数. 解答过程如下: (sinx)^4 = (sinx^2)^2 = ((1 - cos2x)/2)^2 = (1 - 2cos2x + (cos2x)^2)/4 = 0.25 - 0.5cos2x + 0.125(1 + cos4x) = (cos4x)/8 - (cos2x)/2 + 3/8 ∫ (sinx)^4dx = ∫ ((cos4x)/8 - (cos2x)/2 + 3/8)dx = ∫ ((cos4x)/8)dx - ∫ ((cos2x)/2)dx + ∫ (3/8)dx = (1/32)∫ cos4xd4x - (1/4)∫ cos2xd2x + (3x/8) = (sin4x)/32 - (sin2x)/4 + (3x/8) + C 扩展资料: 二倍角公式 sin2α=2sinαcosα tan2α=2tanα/(1-tan^2(α)) cos2α=.

解答:解法一:万能代换! 令u=tanx/2,则sinx=2u/(1+u²),cosx=(1-u²)/(1+u²),dx=2du/(1+u²),于是得 ∫1/(sinx+cosx)=∫2/(1+2u-u²)du =√2/2∫[1/(u-(1-√2))-1/(u-(1+√2))]du =√2/2ln|(u-(1-√2))/(u-(1+√2))|+C =√2/2ln|(tanx/2-1+√2)/(tanx/2-1-√2)+C. 解法二: ∫dx/(sinx+cosx)=√2/2∫dx/(√2/2sinx+√2/2cosx)=√2/2∫dx/cos(x-π/4) =√2/2∫sec(x-π/4)d(x-π/4) =√2/2ln|sec(x-π/4)+tan(x-π/4)|+C.

y=(sinx^2+cosx^2)(sinx^2-cosx^2) =sinx^2-cosx^2 =1-2cosx^2 =cos2x 所以周期是 2π/2=π 递增区间 (π/2+kπ, π+kπ)

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