∫tanx)^1/2dx/(1+(tanx)^1/2) limit is from 0 to pi/2

Question

Aditya Gupta · Answer

Let I be the required integral. Now replace the function f(x) inside by f(a+b-x) It becomes 1/(1+tanx^.5)dx Now adding, you get 2I= ∫1dx 2I= pi/2 Or I=π/4

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∫tanx)^1/2dx/(1+(tanx)^1/2) limit is from 0 to pi/2

∫tanx)^1/2dx/(1+(tanx)^1/2) limit is from 0 to pi/2

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∫tanx)^1/2dx/(1+(tanx)^1/2) limit is from 0 to pi/2

∫tanx)^1/2dx/(1+(tanx)^1/2) limit is from 0 to pi/2

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