4
$\begingroup$

$\displaystyle\int_0^{\pi/4}\frac{\sqrt{\tan(x)}}{\sin(x)\cos(x)}\,dx$

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

$\endgroup$
  • 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
7
$\begingroup$

HINT:

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

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

$\endgroup$
  • $\begingroup$ 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
5
$\begingroup$

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


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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.