Integrate 1/(1+(tanx)^(1/3)) from π/6 to π/3................

Integrate 1/(1+(tanx)^(1/3)) from π/6 to π/3................


2 Answers

25758 Points
3 years ago
Dear Mihir
I have atta hed the image containing solution. If you find any difficulty, please feel free to ask  again.
I = \frac{\pi }{12}2I = \frac{\pi }{6}2I = [x]_{\pi /6}^{\pi /3}2I =\int_{\pi /6}^{\pi /3}dx2I =\int_{\pi /6}^{\pi /3}\frac{\sqrt{cosx}+\sqrt{sin(x)}}{\sqrt{sin(x)}+\sqrt{cos(x)}}dx......(2)(1) + (2)I =\int_{\pi /6}^{\pi /3}\frac{\sqrt{sin(x)}}{\sqrt{sin(x)}+\sqrt{cos(x)}}dxI =\int_{\pi /6}^{\pi /3}\frac{\sqrt{cos(\frac{\pi }{2}-x)}}{\sqrt{cos(\frac{\pi }{2}-x)}+\sqrt{sin(\frac{\pi }{2}-x)}}dx\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx…......(1)I =\int_{\pi /6}^{\pi /3}\frac{\sqrt{cosx}}{\sqrt{cosx}+\sqrt{sinx}}dxI =\int_{\pi /6}^{\pi /3}\frac{1}{1+\sqrt{tanx}}dx 
Aditya Gupta
2081 Points
3 years ago
actually aruns ans is wrong coz the power in ques is 1/3
let J= ∫1/(1+(tanx)^(1/3))dx from  π/6 to π/3.......(1)
now we know the property that f(x) can be replaced by f(a+b – x). here a+b=  π/6 + π/3= pi/2.
also, tan(pi/2 – x)= cotx
so J= 1/(1+(cotx)^(1/3))dx from  π/6 to π/3
or J= (tanx)^(1/3)/(1+(tanx)^(1/3))dx from  π/6 to π/3........(2)
adding 1 and 2
2J= (1+(tanx)^(1/3))/(1+(tanx)^(1/3))dx from  π/6 to π/3.
2J= ∫dx from  π/6 to π/3.
2J= pi/6
or J= π/12
kindly approve :)

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