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Evaluate the given question of indefinite integral
limit 0 to pi by 2 dx by 1 +√tanx

Mehtab , 6 Years ago
Grade 12
anser 2 Answers
Himanshu

Last Activity: 6 Years ago

 
 
Let I be the answer
Using the idendity 
^{\int_{0}^{pi/2}}tanx=^{\int_{0}^{pi/2}}tan(pi/2-x)=^{\int_{0}^{pi/2}}cotx
so
 
^{\int_{0}^{pi/2}}2dx/(1+\sqrt{tanx})=^{\int_{0}^{pi/2}}2dx/(1+\sqrt{cotx})=^{\int_{0}^{pi/2}}2dx/(1+\sqrt{1/tanx})=^{\int_{0}^{pi/2}}2\sqrt{tanx}dx/(1+\sqrt{tanx})
I=^{\int_{0}^{pi/2}}2dx/(1+\sqrt{tanx})---eq 1
I=^{\int_{0}^{pi/2}}2\sqrt{tanx}dx/(1+\sqrt{tanx})---eq 2
Adding eq 1 and eq 2
 
2I=^{\int_{0}^{pi/2}}2dx/(1+\sqrt{tanx})+
2\sqrt{tanx}dx/(1+\sqrt{tanx})
2I=
^{\int_{0}^{pi/2}}2dx(1+\sqrt{tanx})/(1+\sqrt{tanx})
2I=^{\int_{0}^{pi/2}}2dx
I=^{\int_{0}^{pi/2}}dx
I=pi/2

Aditya Gupta

Last Activity: 6 Years ago

I= ∫dx/(1+√tanx)
Now replace f(x) by f(a+b-x)
I= ∫√tanxdx/(1+√tanx)
Adding we get
2I= ∫dx= π/2
So I= π/4.

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