Integral Calculus> Find the following integral ∫dx/(1+(tanx)...

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

Lovey , 11 Years ago

Grade 12

2 Answers

Sumit Majumdar

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

Last Activity: 11 Years ago

Jitender Singh

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)
$2I =\int_{\pi /6}^{\pi /3}\frac{\sqrt{cosx}+\sqrt{sin(x)}}{\sqrt{sin(x)}+\sqrt{cos(x)}}dx$
$2I =\int_{\pi /6}^{\pi /3}dx$
$2I = [x]_{\pi /6}^{\pi /3}$
$2I = \frac{\pi }{6}$
$I = \frac{\pi }{12}$