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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Integrate the function\[\frac{\sqrt{\tan x}}{\sin x\cos x}\]

$\begin{array}{1 1}2\sqrt{\tan x}+c \\ 2\sqrt{\cot x}+c \\ 2\sqrt{\sin x \cos x}+c \\ \sqrt{2 \sin x \cos x}+c \end{array}$

Can you answer this question?
 
 

1 Answer

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Toolbox:
  • Method of substitution :
  • Given f(x)dx can be transformed into another form by changing independent variable x to t by substituting x=g(t).
  • Consider $I=\int f(x)dx.$
  • Put x=g(t) so that $\frac{dx}{dt}=g'(t).$
  • $\Rightarrow $dx=g'(t)dt.
  • Thus $I=\int f(g(t).g'(t))dt.$
Given $I=\int \frac{\sqrt {\tan x}}{\sin x\cos x}dx$.
 
Multiply and divide the denominator by cos x.
 
$I=\int\frac{\sqrt {\tan x}}{\frac{\sin x\cos x}{\cos x}}$.
 
$\;\;\;=\int\frac{\sqrt{\tan x}}{\tan x.\cos ^2x}dx.$
 
$\;\;\;=\int\frac{\sec^2x}{\sqrt {\tan x}}dx$.
 
Put $\tan x=t$.
 
$\sec^2dx=dt.$
 
Now substitute for x and dx we get,
 
$I=\int\frac{dt}{\sqrt t}.$
 
On integrating we get,
 
$I=\int t^\frac{-1}{2}dt.$
 
$\;\;\;=\frac{t^\frac{1}{2}}{\frac{1}{2}}+c.$
 
Substituting back for t we get,
 
$2\sqrt{\tan x}+c.$
 
Hence $\int \frac{\sqrt {\tan x}}{\sin x\cos x}dx=2\sqrt{\tan x}+c.$

 

answered Jan 29, 2013 by sreemathi.v
 
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