Question

∫（tanx）^a dx =？

Answer 1

解：设定积分值为w
w=[0,π/2]∫2/(1+(tanx)^a)dx /**/方括号表示积分限
= [0,π/2]∫[2/(tanx)^a]/[1/(tanx)^a+1]dx
= [0,π/2]∫2*(cotx)^a/[(cotx)^a+1]dx

作变量代换 u=π/2 -t ==> t= π/2 -u, 积分式化为：
w= [π/2,0]∫2*[cot(π/2-u]^a/[(cot(π/2-u)^a+1](-du) /**/注意积分上下限的变化
= [0, π/2]∫2*(tanu)^a/[(tanu)^a+1]du
= [0, π/2]∫2*[1- 1/((tanu)^a+1)]du
= π- ∫2/[(tanu)^a+1]du = π-w
从而 w= π/2
即： [0,π/2]∫2/(1+(tanx)^a)dx = π/2
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