求定积分t/(1+cost),上限π/2,下限是2π/3
1个回答
∫(π/2→2π/3)t/(1+cost)*dt,应该是这个吧?
cos2t=2cos²t-1
1+cos2t=2cos²t
1+cost=2cos²(t/2)
∫t/(1+cost)*dt
=∫t/[2cos²(t/2)]*dt
=∫t/cos²(t/2)*d(t/2)
=∫t*d[tan(t/2)]
=t*tan(t/2)-∫tan(t/2)*dt
=t*tan(t/2)-2∫tan(t/2)*d(t/2)
=t*tan(t/2)+2ln|cos(t/2)|+C
∫(π/2→2π/3)t/(1+cost)*dt
=[t*tan(t/2)+2ln|cos(t/2)|]|(π/2→2π/3)
=[t*tan(t/2)+2ln[cos(t/2)]]|(π/2→2π/3)
=[2π/3*tan(π/3)+2ln[cos(π/3)]]-[π/2*tan(π/4)+2ln[cos(π/4)]]
=[2π/3*√3+2ln(1/2)]-[π/2*1+2ln(1/√2)]
=2π√3/3-π/2+2*(-1)ln2-2*(-1/2)ln2
=2π√3/3-π/2-ln2
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