首页 >>  正文

1cost3方的积分

来源:baiyundou.net   日期:2024-09-28

谢婉矩23911.COS的X的三次方乘以SINX的平方的不定积分是多少? 2.(sint - cost)/(sint+cost)的不定积分 -
侯朗齿13473546611 ______ ∫ cos³xsin²x dx = ∫ cos²xsin²x dsinx = ∫ (1 - sin²x)sin²x dsinx = ∫ (sin²x - sin⁴x) dsinx = (1/3)sin³x - (1/5)sin⁵x + C _______________________ ∫ (sint - cost)/(sint + cost) dt = ∫ d(- cost - sint)/(sint + cost) dt = - ∫ d(sint + cost)/(sint + cost)...

谢婉矩23911/(cost的平方)*根号下(1+tant)的不定积分怎么求急,要详细解答过程! -
侯朗齿13473546611 ______[答案] ∫1/[(cos²t)√(1+tant)]dt=∫1/√(1+tant)dtant 后面知道怎么求了吧 这就简单的凑微分1/[(cos²t)dt=dtant

谢婉矩2391大一高数求积分,求(sint)的平方/(cost)的三方 •dt的积分 -
侯朗齿13473546611 ______[答案] x=sint则,dx=cotdt 原式=∫(sin²tcost)/(cos²t)²dt=∫x²/(1-x²)² dx =∫ 【1/(4 (-1 + x)^2) + 1/(4 (-1 + x)) + 1/(4 (1 + x)^2) - 1/(4 (1 + x))】dx =..

谢婉矩2391(cost+sint)^4的积分 -
侯朗齿13473546611 ______[答案] ∫[(cost+sint)^4]dt = ∫{(cost)^4+4[(cost)^3]sint+6(costsint)^2+4cost[(sint)^3]+(sint)^4}dt = …… (按三角函数积分的方法计算,留给你)

谢婉矩2391设g(x)在(0 - 3)上,g(x)=x*积分cost^3 dt,求g(x)的导数. -
侯朗齿13473546611 ______[答案] g(x)=x*积分cost^3 dt =X积分cost^2dcost =1/3*x*cost^3 g(x)'=1/3cost^3+1/3x*3cost^2*(-sint) =1/3cost^3-sint*cos^2t

谢婉矩2391大一高数求积分,谢谢! -
侯朗齿13473546611 ______ x=sint则,dx=cotdt 原式=∫(sin²tcost)/(cos²t)²dt=∫x²/(1-x²)² dx=∫ 【1/(4 (-1 + x)^2) + 1/(4 (-1 + x)) + 1/(4 (1 + x)^2) - 1/(4 (1 + x))】dx=..

谢婉矩23911 - cosx的5/2次方的积分 -
侯朗齿13473546611 ______[答案] ∫(1-cosx)^(5/2) dx =∫[2sin(x/2)^2]^(5/2) dx=2*2^(5/2)∫[sin(x/2)]^5d(x/2)---------∫(sint)^5dt =- ∫(sint)^4dcost =- ∫[1-(cost)^2]^2dcost=- ∫[1-2(cost)^2+(cost)^4]dcost=-sint+2/3(cost)^3-1/5 (cost...

谢婉矩2391(1 - cost)2转化为sint -
侯朗齿13473546611 ______[答案] 解(1-cost)2 =(1-(1-2sin^2(t/2)))2 =(2sin^2(t/2))2 =4sin^4(t/2).

谢婉矩2391sint/(1 - cost)求导 -
侯朗齿13473546611 ______[答案] f'(t)=[(sint)'(1-cost)-(sint)(1-cost)']/(1-cost)? =[cost(1-cost)-sintsint]/(1-cost)? =[cost-(cos?t+sin?t)]/(1-cost)? =(cost-1)/(1-cost)? =1/(cost-1)

谢婉矩2391密度为1的螺线,x=cost,y=sint,z=2t(0 -
侯朗齿13473546611 ______[答案] 螺线长度 的微分 dL= 根号((dx)^2 + (dy)^2 + (dz )^2) 质量的微分dm = 密度1 * dL 对于z轴的转动惯量 = 积分号(半径1的平方 * dm)

(编辑:自媒体)
关于我们 | 客户服务 | 服务条款 | 联系我们 | 免责声明 | 网站地图 @ 白云都 2024