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Question 73 of 96

Q.Prove that ∫π/6π/3dx1+cot⁡x=∫π/6π/3dx1+tan⁡x\displaystyle\int_{\pi/6}^{\pi/3} \dfrac{dx}{1+\sqrt{\cot x}} = \displaystyle\int_{\pi/6}^{\pi/3} \dfrac{dx}{1+\sqrt{\tan x}}.

Puducherry TnboardTamil Nadu HSC (DGE) Board 2019Subjective· 2mImportance★★★★★
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Applying the substitution x↦a+b−xx\mapsto a+b-x (here a+b=π/2a+b=\pi/2) to the cot-integral turns cot⁡x\cot x into tan⁡x\tan x, showing the two integrals are identical.

  1. Let I=∫π/6π/3dx1+cot⁡xI=\displaystyle\int_{\pi/6}^{\pi/3}\dfrac{dx}{1+\sqrt{\cot x}}.
  2. Use the standard property ∫abg(x) dx=∫abg(a+b−x) dx\displaystyle\int_a^b g(x)\,dx=\int_a^b g(a+b-x)\,dx, valid for any integrable gg, with a=π/6a=\pi/6, b=π/3b=\pi/3, so a+b=π/2a+b=\pi/2.
  3. Replacing xx by π/2−x\pi/2-x inside II: I=∫π/6π/3dx1+cot⁡(π/2−x)I=\displaystyle\int_{\pi/6}^{\pi/3}\dfrac{dx}{1+\sqrt{\cot(\pi/2-x)}}.
  4. Since cot⁡(π/2−x)=tan⁡x\cot(\pi/2-x)=\tan x, this becomes I=∫π/6π/3dx1+tan⁡xI=\displaystyle\int_{\pi/6}^{\pi/3}\dfrac{dx}{1+\sqrt{\tan x}}. …

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