Browse Questions

Integrate the function$\frac{\sqrt{\tan x}}{\sin x\cos x}$

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

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 $\int \frac{\sqrt {\tan x}}{\sin x\cos x}dx=2\sqrt{\tan x}+c.$