如图,这个微积分怎么算? 20
展开全部
换元t=x-π/2,
设定积分=I,则有
I=∫(-π/2到π/2)(t+π/2)²/(1+cos²t)dt
=∫(π/2到-π/2)(-u+π/2)²/(1+cos²u)d(-u)
I+I=∫(-π/2到π/2)2(t²+π²/4)/(1+cos²t)dt
I=2∫(0到π/2)(t²+π²/4)/(1+cos²t)dt
=2∫t²/(tan²t+2)dtant+π²/2∫1/(tan²t+2)dtant
=√2∫t²darctan(tant/√2)+π²/2*arctan(tant/√2)
设定积分=I,则有
I=∫(-π/2到π/2)(t+π/2)²/(1+cos²t)dt
=∫(π/2到-π/2)(-u+π/2)²/(1+cos²u)d(-u)
I+I=∫(-π/2到π/2)2(t²+π²/4)/(1+cos²t)dt
I=2∫(0到π/2)(t²+π²/4)/(1+cos²t)dt
=2∫t²/(tan²t+2)dtant+π²/2∫1/(tan²t+2)dtant
=√2∫t²darctan(tant/√2)+π²/2*arctan(tant/√2)
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询