Needed a detailed solution with explanation. The answer is $2$.

  • 1
    $\begingroup$ If this is homework, please say so. It would help you if you show your working so far too :) $\endgroup$ – Shaun Nov 15 '13 at 19:02
  • $\begingroup$ @Shaun- I had come till putting values tanx=u^2. $\endgroup$ – Swetank Nov 15 '13 at 19:07
  • $\begingroup$ @shaun- i am currently not familiar with Mathjax input. So I will be giving proper solution after completing it. $\endgroup$ – Swetank Nov 15 '13 at 19:19


$$\frac{\sqrt{\tan x}}{\sin x\cos x}=\frac{\sqrt{\tan x}}{\frac{\sin x\cos x}{\cos^2x}}\frac1{\cos^2x}=\frac{\sqrt{\tan x}}{\tan x}\sec^2x$$

Put $\sqrt{\tan x}=u\implies \tan x=u^2 $

  • $\begingroup$ getting 2du after that? $\endgroup$ – Swetank Nov 15 '13 at 18:49
  • $\begingroup$ @Swetank, getting $$\int_0^1\frac{u}{u^2}2u du$$ $\endgroup$ – lab bhattacharjee Nov 15 '13 at 18:51
  • $\begingroup$ yes but, upper limit is pi/4... $\endgroup$ – Swetank Nov 15 '13 at 18:53
  • $\begingroup$ @Swetank, that upper limit is for $x\implies u=+\sqrt{\tan x}=??$ $\endgroup$ – lab bhattacharjee Nov 15 '13 at 18:58
  • $\begingroup$ Ohh yeah! that upper limit is for x. therefore sqrt tanx will be 1. $\endgroup$ – Swetank Nov 15 '13 at 19:03

I'd use the result that has been pointed out by Bhattacharjee. $$ \begin{align} \int_0^{\Large\frac{\pi}{4}}\frac{\sqrt{\tan x}}{\sin x\cos x}dx&=\int_0^{\Large\frac{\pi}{4}}\frac{\sqrt{\tan x}}{\tan x}\sec^2x\;dx\\ &=\int_0^{\Large\frac{\pi}{4}}\tan^{-\frac{1}{2}} x\;d(\tan x)\\ &=2\left.\sqrt{\tan x}\right|_0^{\Large\frac{\pi}{4}}\\ &=2. \end{align} $$

$$\Large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$

  • 1
    $\begingroup$ Changing $dx$ to $d(\tan x)$ is, in fact the act of substituting. $\endgroup$ – user49685 Mar 21 '14 at 10:45

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