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最简单的方法:
∫tan(θ/2)*[sec(θ/2)]^2 *dθ
=∫2tan(θ/2)d[tan(θ/2)]
=[tan²(θ/2)]<π/2,π/3>
=[tan²(π/4)-tan²(π/6)]
=1-1/3=2/3
∫tan(θ/2)*[sec(θ/2)]^2 *dθ
=∫2tan(θ/2)d[tan(θ/2)]
=[tan²(θ/2)]<π/2,π/3>
=[tan²(π/4)-tan²(π/6)]
=1-1/3=2/3
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展开全部
∫tan(θ/2)*[sec(θ/2)]^2 *dθ
=2*∫tan(θ/2) * [sec(θ/2)]^2 *d(θ/2)
=2*∫tan(θ/2) * d[tan(θ/2)]
=[tan(θ/2)]^2 |θ = π/3 →π/2
=[tan(π/4)]^2 - [tan(π/6)]^2
=1 - 1/3
=2/3
=2*∫tan(θ/2) * [sec(θ/2)]^2 *d(θ/2)
=2*∫tan(θ/2) * d[tan(θ/2)]
=[tan(θ/2)]^2 |θ = π/3 →π/2
=[tan(π/4)]^2 - [tan(π/6)]^2
=1 - 1/3
=2/3
追问
请问下第二步最前面的数字2是不是从后面的dθ中提出来的?
追答
是的。可能是我变换的太快了!
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