Question

integrate (sqrt(cot x) - sqrt(tan x)) dx

Answer 1

Answer:

Step-by-step explanation:

We have,

$\displaystyle\rm{\int\left(\sqrt{cot(x)}-\sqrt{tan(x)}\right)dx}$

$\displaystyle\rm{=\int\left(\sqrt{\dfrac{cos(x)}{sin(x)}}-\sqrt{\dfrac{sin(x)}{cos(x)}}\right)dx}$

$\displaystyle\rm{=\int\dfrac{cos(x)-sin(x)}{\sqrt{sin(x)\,cos(x)}}\,dx}$

$\displaystyle\rm{=\sqrt{2}\int\dfrac{cos(x)-sin(x)}{\sqrt{2\,sin(x)\,cos(x)}}\,dx}$

$\displaystyle\rm{=\sqrt{2}\int\dfrac{cos(x)-sin(x)}{\sqrt{-1+1+2\,sin(x)\,cos(x)}}\,dx}$

$\displaystyle\rm{=\sqrt{2}\int\dfrac{cos(x)-sin(x)}{\sqrt{sin^{2}(x)+cos^{2}(x)+2\,sin(x)\,cos(x)-1}}\,dx}$

$\displaystyle\rm{=\sqrt{2}\int\dfrac{cos(x)-sin(x)}{\sqrt{\left(sin(x)+cos(x)\right)^{2}-1}}\,dx}$

$\bf{Put\,\,\,sin(x)+cos(x)=t}$

$\bf{\implies\,\left(cos(x)-sin(x)\right)dx=dt}$

So,

$\displaystyle\rm{=\sqrt{2}\int\dfrac{dt}{\sqrt{{t}^{2}-1}}}$

$\displaystyle\rm{=\sqrt{2}\cdot\dfrac{1}{2\cdot(1)}\ln\left|\dfrac{t-1}{t+1}\right|+C}$

$\displaystyle\rm{=\dfrac{1}{\sqrt{2}}\ln\left|\dfrac{sin(x)+cos(x)-1}{sin(x)+cos(x)+1}\right|+C}$

$\displaystyle\rm{=\dfrac{1}{\sqrt{2}}\ln\left|\dfrac{2\,sin\left(\dfrac{x}{2}\right)\,cos\left(\dfrac{x}{2}\right)-2\,sin^{2}\left(\dfrac{x}{2}\right)}{2\,sin\left(\dfrac{x}{2}\right)\,cos\left(\dfrac{x}{2}\right)+2\,cos^{2}\left(\dfrac{x}{2}\right)}\right|+C}$

$\displaystyle\rm{=\dfrac{1}{\sqrt{2}}\ln\left|\dfrac{2\,sin\left(\dfrac{x}{2}\right)\left\{cos\left(\dfrac{x}{2}\right)-sin\left(\dfrac{x}{2}\right)\right\}}{2\,cos\left(\dfrac{x}{2}\right)\left\{sin\left(\dfrac{x}{2}\right)+cos\left(\dfrac{x}{2}\right)\right\}}\right|+C}$

$\displaystyle\rm{=\dfrac{1}{\sqrt{2}}\ln\left|\dfrac{cos\left(\dfrac{x}{2}\right)-sin\left(\dfrac{x}{2}\right)}{sin\left(\dfrac{x}{2}\right)+cos\left(\dfrac{x}{2}\right)}\right|+C}$

$\displaystyle\rm{=\dfrac{1}{\sqrt{2}}\ln\left|\dfrac{1-tan\left(\dfrac{x}{2}\right)}{tan\left(\dfrac{x}{2}\right)+1}\right|+C}$

$\displaystyle\rm{=\dfrac{1}{\sqrt{2}}\ln\left|\dfrac{1-tan\left(\dfrac{x}{2}\right)}{1+tan\left(\dfrac{x}{2}\right)}\right|+C}$

$\displaystyle\rm{=\dfrac{1}{\sqrt{2}}\ln\left|tan\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)\right|+C}$