Question

Integrate with respect to x

$\sqrt{tanx}$
​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{\displaystyle\,\int\sqrt{tan(x)}\,dx}$

$\sf{Put\,\,\, tan(x)=t^2}$

$\sf{\implies sec^2(x)\,dx=2t\,dt}$

$\sf{\implies dx=\dfrac{2t}{sec^2(x)}\,dt}$

$\sf{\implies dx=\dfrac{2t}{1+t^4}\,dt}$

So,

$\tt{\displaystyle\,\int\sqrt{t^2}\,\cdot\dfrac{2t}{1+t^4}\,dt}$

$\tt{=\displaystyle\,\int\dfrac{2t^2}{1+t^4}\,dt}$

$\tt{=\displaystyle\,\int\dfrac{t^2+t^2}{1+t^4}\,dt}$

$\tt{=\displaystyle\,\int\dfrac{t^2+1+t^2-1}{1+t^4}\,dt}$

$\tt{=\displaystyle\,\int\dfrac{t^2+1}{1+t^4}\,dt+\int\dfrac{t^2-1}{1+t^4}\,dt}$

$\tt{=\displaystyle\,\int\dfrac{1+\dfrac{1}{t^2}}{t^2+\dfrac{1}{t^2}}\,dt+\int\dfrac{1-\dfrac{1}{t^2}}{t^2+\dfrac{1}{t^2}}\,dt}$

$\tt{=\displaystyle\,\int\dfrac{1+\dfrac{1}{t^2}}{\left(t-\dfrac{1}{t}\right)^2+2}\,dt+\int\dfrac{1-\dfrac{1}{t^2}}{\left(t+\dfrac{1}{t}\right)^2-2}\,dt}$

$\sf{Put\,\,\,t-\dfrac{1}{t}=u\,\,\,\,\,\,\,and\,\,\,\,\,\,\,t+\dfrac{1}{t}=v}$

$\sf{\implies\,\left(1+\dfrac{1}{t^2}\right)\,dt=du\,\,\,\,\,\,\,and\,\,\,\,\,\,\,\left(1-\dfrac{1}{t^2}\right)\,dt=dv}$

So, the integrals become,

$\tt{=\displaystyle\,\int\dfrac{du}{u^2+2}+\int\dfrac{dv}{v^2-2}}$

$\tt{=\displaystyle\,\int\dfrac{du}{(u)^2+\left(\sqrt{2}\right)^2}+\int\dfrac{dv}{(v)^2-\left(\sqrt{2}\right)^2}}$

$\tt{=\displaystyle\,\dfrac{1}{\sqrt{2}}\,tan^{-1}\left(\dfrac{u}{\sqrt{2}}\right)+\dfrac{1}{2\sqrt{2}}ln\left|\dfrac{v-\sqrt{2}}{v+\sqrt{2}}\right|+C}$

$\tt{=\displaystyle\,\dfrac{1}{\sqrt{2}}\,tan^{-1}\left(\dfrac{t-\dfrac{1}{t}}{\sqrt{2}}\right)+\dfrac{1}{2\sqrt{2}}ln\left|\dfrac{t+\dfrac{1}{t}-\sqrt{2}}{t+\dfrac{1}{t}+\sqrt{2}}\right|+C}$

$\tt{=\displaystyle\,\dfrac{1}{\sqrt{2}}\,tan^{-1}\left(\dfrac{t^2-1}{\sqrt{2}\,t}\right)+\dfrac{1}{2\sqrt{2}}ln\left|\dfrac{t^2+1-\sqrt{2}\,t}{t^2+1+\sqrt{2}\,t}\right|+C}$

$\tt{=\displaystyle\,\dfrac{1}{\sqrt{2}}\,tan^{-1}\left(\dfrac{t^2-1}{\sqrt{2}\,t}\right)+\dfrac{1}{2\sqrt{2}}ln\left|\dfrac{t^2-\sqrt{2}\,t+1}{t^2+\sqrt{2}\,t+1}\right|+C}$

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