xxsr.net
当前位置:首页 >> sintCost积分 >>

sintCost积分

解: ∫(sint·cost)²dt =∫(½·sin2t)²dt =1/4·∫(sin2t)²dt =1/4·∫(1-cos4t)/2 dt =1/8·∫(1-cos4t)dt =1/8·(t-1/4·sin4t)+C =t/8-1/32·sin4t+C

都是,你定积分计算有误 -0.25cos2t=-.5cos^2 t + .25 0.5(sint)^2=0.5[1-(cost)^2]=.5-.5(cost)^2 由于不定积分应该是-0.5(cost)^2 +C 所以都等价,你定积分算错了吧 因为C在定积分时最后会消去

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

百度百度

原式=∫[(sint)^3][(cost)^2]d(sint) =∫[(sint)^3][1-(sint)^2]d(sint) =∫[(sint)^3]d(sint)-∫[(sint)^5]d(sint)

=∫2sin2t√(cos²t-sin²t)(cos²t+sin²t)dt =-∫√cos2tdcos2t =-(2/3)(cos2t)^(3/2) =4/3

cost/sin2t=1/(2sint)=sint/(2sint*sint)=sint/2(1-cost*cost)=sint*(1/(1+cost)+1/(1-cost))/4. 故原不定积分=∫1/4*(1/(1+cost)+1/(1-cost))d(-cost)=1/4*[∫1/(1+cost)d(-cost)+∫1/(1-cost)d(-cost) =1/4*ln[(1-cost)/(1+cost)]

∫[(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)

网站首页 | 网站地图
All rights reserved Powered by www.xxsr.net
copyright ©right 2010-2021。
内容来自网络,如有侵犯请联系客服。zhit325@qq.com