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Find the following integral ∫dx/(1+(tanx)^1/2) Limit is from pi/6 to pi/3 Please reply soon.

Find the following integral
∫dx/(1+(tanx)^1/2)
Limit is from pi/6 to pi/3
Please reply soon.

Grade:12

2 Answers

Sumit Majumdar IIT Delhi
askIITians Faculty 137 Points
9 years ago
Dear student,
We have the given integral given by:
I=\int_{\frac{\pi}{6} }^{\frac{\pi}{3}}\frac{dx}{1+\sqrt{tanx}}=\int_{\frac{\pi}{6} }^{\frac{\pi}{3}}\frac{dx}{1+\sqrt{tan\left (\frac{\pi }{3}+\frac{\pi }{6}-x \right )}}=\int_{\frac{\pi}{6} }^{\frac{\pi}{3}}\frac{dx}{1+\sqrt{cotx}}Hence, if we add I twice, we get:
2I=\int_{\frac{\pi}{6} }^{\frac{\pi}{3}}\frac{dx}{1+\sqrt{tanx}}+\int_{\frac{\pi}{6} }^{\frac{\pi}{3}}\frac{dx}{1+\sqrt{cotx}}=\int_{\frac{\pi}{6} }^{\frac{\pi}{3}}\left ( \frac{dx}{1+\sqrt{tanx}}+\frac{\sqrt{tanx}dx}{1+\sqrt{tanx}} \right )=\int_{\frac{\pi}{6} }^{\frac{\pi}{3}}dx=\frac{\pi}{6}\Rightarrow I=\frac{\pi}{12}Regards
Sumit
Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 years ago
Ans:Hello student, please find answer to your question
I =\int_{\pi /6}^{\pi /3}\frac{1}{1+\sqrt{tanx}}dx
I =\int_{\pi /6}^{\pi /3}\frac{\sqrt{cosx}}{\sqrt{cosx}+\sqrt{sinx}}dx…......(1)
\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx
I =\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
I =\int_{\pi /6}^{\pi /3}\frac{\sqrt{sin(x)}}{\sqrt{sin(x)}+\sqrt{cos(x)}}dx......(2)
(1) + (2)
2I =\int_{\pi /6}^{\pi /3}\frac{\sqrt{cosx}+\sqrt{sin(x)}}{\sqrt{sin(x)}+\sqrt{cos(x)}}dx
2I =\int_{\pi /6}^{\pi /3}dx
2I = [x]_{\pi /6}^{\pi /3}
2I = \frac{\pi }{6}
I = \frac{\pi }{12}

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