sint+3dt的不定积分

暴于支2614如图,求不定积分 -
盖新桑15567455596 ______ 公式很难打,告诉你方法吧:首先,令x=sint,则 dx=costdt 则原积分式化为有理式的不定积分:∫(x³-8x²-1)/(x+3)(x²-4x+5) dx 而x³-8x²-1=x³-4x²+5x - 4x²-5x-1 =x(x²-4x+5)-(4x²-5x-1) 所以原积分式进一步化为 ∫ x/(x+3)dx - ∫(4x²-5x-1)/(x+3)(x²-4x+5)dx

暴于支2614用万能公式解cost/(sint+cost)的不定积分 -
盖新桑15567455596 ______[答案] 干嘛非用万能公式不可 cost=1/2[(sint+cost)+(cost-sint)] ∫cost/(sint+cost)dt =1/2∫[1+(cost-sint)/(sint+cost)]dt =1/2[t+ln|sint+cost|]+C

暴于支2614如何求这两个不定积分 -
盖新桑15567455596 ______ 积分:arcsinx/x^2dx 令arcsinx=t 则:dx=costdt 原式=积分:tcost/sin^2tdt=积分:t/sin^2td(sint)=-积分:td(1/sint)=-t/sint+积分:cscdt=-t/sint-ln|csct+cott|+C=-arcsinx/x-ln|(1+根号(1-x^2))/x|+C(C是常数) 积分:dx/[x^3*根号(x^2+1)] 令x=1/t dx=-1/t^...

暴于支2614∫x∧3arcsinx/√(1 - x∧2) -
盖新桑15567455596 ______[答案] 设x=sint√(1-x^2)=cost dx=costdt则原式化为∫(sint)^3arcsin(sint) *cost dt/cost=∫t(sint)^3dt=-∫t(sint)^2dcost=-∫t(1-(cost)^2 dcost=-∫tdcost+∫t(cost)^2dcost=-tcost+∫costdt+1/3∫td(cost)^3=-tcost+s...

暴于支2614分子是x的3次方乘以arccosx,分母是根号下1减x的平方求不定积分 -
盖新桑15567455596 ______ 原式=-∫{x^3arccosx/[-√(1-x^2)]}dx =-∫x^3arccosxd(arccosx) =-(1/2)∫x^3d[(arccosx)^2] =-(1/2)x^3(arccosx)^2+(1/2)∫(arccosx)^2d(x^3) =-(x^3/2)(arccosx)^2+(3/2)∫x^2(arccosx)^2dx.令arccosx=t,则:x=cost,dx=-sintdt.∴原式=-(x^3/2)(...

暴于支2614用第一换元积分公式求不定积分∫sin3xdx -
盖新桑15567455596 ______ 解 ∫sin3xdx =1/3*∫sin3xd(3x) =-1/3*cos3x+c 第一部就是凑 把d后面凑成3x 但是(3x)'=3 还原回去会发现现在的式子比原式多了3倍 所以前面乘1/3 用来配平系数

暴于支2614求不定积分1.∫xe^arctanx/(1+x^2)^3/2dx
盖新桑15567455596 ______ 1、 令arctanx=t,则x=tant,dx=(sect)^2dt, ∫xe^arctanx/(1+x^2)^3/2 dx=∫[tant*e^t/(sect)^3*(sect)^2]dt=∫e^t*sintdt=1/2*e^t(sint-cost)+C=1/2*e^arctanx*(x-1)/√(1+x^2)+C 2、 令...

暴于支2614(sint)^3+(cost)^3的不定积分怎么算 -
盖新桑15567455596 ______[答案] 原式=∫[(sint)^3][(cost)^2]d(sint) =∫[(sint)^3][1-(sint)^2]d(sint) =∫[(sint)^3]d(sint)-∫[(sint)^5]d(sint)

暴于支2614求不定积分,上限为π,下限为0,根号下(sint - sint的立方)dt, -
盖新桑15567455596 ______[答案] 是求定积分! I=∫√[sint-(sint)^3]dt=∫√{sint[1-(sint)2]}dt =∫|cost|√sintdt =∫cost√sintdt+∫(-cost)√sintdt =∫√sintdsint-∫√sintdsint =[(2/3)(sint)^(3/2)]-[(2/3)(sint)^(3/2)] =2/3-(-2/3)=4/3.

暴于支2614不定积分∫1/[1+(sint)^2]dt -
盖新桑15567455596 ______[答案] 原式=∫sec方t/(sec方t+tan方t)dt =∫1/[2(tant)^2+1] dtant =1/√2 ∫1/[2(tant)^2+1] d√2tant =√2/2 *arctan(√2tant)+c