Question

integrate x * sqrt((1 - x ^ 2)/(1 + x ^ 2)) dx from 0 to 1 = (pi - 2)/4

Answer 1

Answer:

Step-by-step explanation:

We have,

$\displaystyle\int^{1}_{0}x\cdot\sqrt{\dfrac{1-{x}^{2}}{1+{x}^{2}}}\,dx$

$\bf{Put\,\,1+{x}^{2}={t}^{2}}$

$\bf{\implies\,2{x}\,dx=2{t}\,dt}$

$\bf{\implies\,{x}\,dx={t}\,dt}$

Limits stay the same, so,

$\displaystyle\int^{1}_{0}\sqrt{\dfrac{1-\left({t}^{2}-1\right)}{{t}^{2}}}\cdot\,t\,dt$

$\displaystyle=\int^{1}_{0}\dfrac{\sqrt{1-{t}^{2}+1}}{t}\cdot\,t\,dt$

$\displaystyle=\int^{1}_{0}\sqrt{2-{t}^{2}}\,dt$

$\displaystyle=\int^{1}_{0}\sqrt{\left(\sqrt{2}\right)^{2}-\left(t\right)^{2}}\,dt$

We know,

$\boxed{\displaystyle\bf{\int\sqrt{{a}^{2}-{x}^{2}}\,dx=\dfrac{x}{2}\cdot\sqrt{{a}^{2}-{x}^{2}}+\dfrac{{a}^{2}}{2}\cdot\,sin^{-1}\left(\dfrac{x}{a}\right)+c}}$

So,

$\displaystyle=\left[\dfrac{t}{2}\cdot\sqrt{\left(\sqrt{2}\right)^{2}-\left(t\right)^{2}}+\dfrac{\left(\sqrt{2}\right)^{2}}{2}\cdot\sin^{-1}\left(\dfrac{t}{\sqrt{2}}\right)\right]^{1}_{0}$

$\displaystyle=\dfrac{1}{2}\cdot\sqrt{\left(\sqrt{2}\right)^{2}-\left(1\right)^{2}}+\dfrac{2}{2}\cdot\sin^{-1}\left(\dfrac{1}{\sqrt{2}}\right)-\dfrac{0}{2}\cdot\sqrt{\left(\sqrt{2}\right)^{2}-\left(0\right)^{2}}-\dfrac{2}{2}\cdot\sin^{-1}\left(\dfrac{0}{\sqrt{2}}\right)$

$\displaystyle=\dfrac{1}{2}\cdot\sqrt{2-1}+1\cdot\dfrac{\pi}{4}-0$

$\displaystyle=\dfrac{1}{2}\cdot1+\dfrac{\pi}{4}$

$\displaystyle=\dfrac{1}{2}+\dfrac{\pi}{4}$