∫cos+t+2+dt

昌章金1807∫sin √x dx -
方芳瞿19234293893 ______ 令√x=t, 则x=t^2 ∫sin √x dx =2∫tsin t dt =-2∫tdcos t =-2tcost+2∫cos t dt =-2tcost+2sint+c ==-2√xcos√x+2sin√x+c

昌章金1807∫(π/2,0)cos^2tdt= -
方芳瞿19234293893 ______[答案] ∫(cost)^2dt=∫(cos2t+1)/2dt=[∫(cos2t)dt+∫dt]/2=(sin2t)/4+t/2+C 所以∫(π/2,0)cos^2tdt=π/4

昌章金1807∫[1/(3+cosx)]dx= -
方芳瞿19234293893 ______ 令t=tanx/2 x=2arctant dx=2/(1+t^2)dt cosx=(1-t^2)/(1+t^2) 代入得:∫1/(3+cosx)dx=∫1/(3+(1-t^2)/(1+t^2))*2/(1+t^2)dt=∫1/(2+t^2)dt=(1/√2)arctan(t/√2)+C=(1/√2)arctan(tan(x/2)/√2)+C 扩展资料:求函数f(x)的不定积分,就是要求出f(x)的所有的原函...

昌章金1807∫dx/(1+√(1 - x^2)) -
方芳瞿19234293893 ______ 答: 设x=sint 原式 =∫ [1/(1+cost)]d(sint) =∫ [cost/(1+cost)]dt =∫ dt -∫ 1/(1+cost) dt =t -∫ 1/[cos(t/2)]^2 d(t/2) =t - tan(t/2)+C =t- 2sin(t/2)cos(t/2)/2[cos(t/2)]^2+C =t-sint/(1+cost)+C =arcsinx-x/[1+√(1-x^2)]+C

昌章金1807cos^2 2x的微分,求详细步骤 -
方芳瞿19234293893 ______ 设x=sint,则: ∫√(1-x^2)dx =∫√(1-sin^2t)dsint =∫cost*costdt =∫(1+cos2t)/2dt =(1/2)∫dt+(1/4)∫cos2td2t =t/2+(1/4)sin2t+c. =(1/2)arcsinx+(1/2)x√(1-x^2)+c.

昌章金1807∫dt/(1+cost) -
方芳瞿19234293893 ______ ∫ 1/(1 + cost) dt,cos2t = 2cos²t - 1 ==> cost = 2cos²(t/2) - 1= ∫ 1/[2cos²(t/2)] dt= ∫ sec²(t/2) d(t/2)= tan(t/2) + C= sint/(1 + cost) + C

昌章金1807求不定积分:∫dx/(2x^2+1)(x^2+1)^(1/2) -
方芳瞿19234293893 ______ 解: x=tant,dx=sec²tdt ∫dx/[(2x^2+1)(x^2+1)^(1/2) ] =∫sec²tdt/[(2tan²t+1)sect] =∫dt/[cost((2sin²t/cos²t)+1)] =∫costdt/[((2sin²t+cost²)] =∫[1/(1+sin²t)]d(sint) =arctan(sint)+C 三角替换有sint=x/√(1+x²) 所以原不定积分 ∫dx/(2x^2+1)(x^2+1)^(1/2) =arctan[x/√(1+x²)]+C

昌章金1807∫dx/(1+(1 - x^2)^(1/2)) -
方芳瞿19234293893 ______ ∫ 1/[1+√(1-x²)] dx 令x=sinu,√(1-x²)=cosu,dx=cosudu=∫ cosu/(1+cosu) du=∫ (cosu+1-1)/(1+cosu) du=∫ 1 du - ∫ 1/(1+cosu) du=u - ∫ 1/[2cos²(u/2)] du=u - ∫ sec²(u/2)] d(u/2)=u - tan(u/2) + C=arcsinx - tan[(1/2)arcsinx] + C 【数学之美】团队为您解答,若有不懂请追问,如果解决问题请点下面的“选为满意答案”.

昌章金1807证明曲线y=sinx x在区间【0,2π】的弧长等于椭圆x^2+2y^2=2的周长 -
方芳瞿19234293893 ______ 弧长积分公式:一般方程:s=∫√(1+y'^2)dx 参数方程x=x(t),y=y(t):s=∫(√[x'(t)^2+y'(t)^2])dt1)对曲线y=sinx,x∈[0,2π],y'=cosx 由于正弦曲线的对称性,可先求其弧长的1/4,即s/4,x∈[0,π/2] ∴s/4=∫(0,π/2)√(1+(cosx)^2)dx2)对椭圆x^2+2y^2=2,即x^2/...

昌章金1807求tant^3+1/sect^2的不定积分 -
方芳瞿19234293893 ______[答案] ∫(tan³t+1)/(sec²t) dt = ∫sin³t / cost dt +∫cos²tdt=∫(1 - cos²t) sint / cost dt+1/2 ∫2cos²t-1+1dt= - ∫(1 - cos²t) / cost dcost +sin2t/4+t/2 = - ∫1/cost dcost + ∫cost dcost +sin2t/4+t/2= - ln|cost| + cos²t/2 +sin2t/4+t/2+C