Mathematics

Evaluate $$\int \sqrt{\dfrac{1-\sqrt{x}}{1+\sqrt{x}}}dx$$.


SOLUTION
$$\displaystyle I= \int \sqrt{\dfrac{1-\sqrt{x}}{1+\sqrt{x}}}dx $$
Let $$\displaystyle \sqrt{x} = cost \Rightarrow  x = cos^{2} t \Rightarrow  dx = -2 cos\, t\ sint\, dt $$
$$\displaystyle I = \int  \sqrt{\frac{1-cos t}{1+cost}} \left ( -2 sint \,cos\, t  \right )dt $$
$$\displaystyle = \int \sqrt{\frac{2 sin^{2}t/2}{2 cos^{2}t/2}} (-2 sint\, cos t)dt $$
$$\displaystyle = \int \dfrac{sin t/2}{cost/2} (-2 \times 2 sin t/2 cos 1/2) cos\, t\, dt $$
$$\displaystyle = -4 \int sin^{2}t /2\, cost\, t dt $$
$$\displaystyle = -4\int \frac{(1-cos\,t)}{2}cos\,t\,dt $$ 
$$\displaystyle = -4 \int \dfrac{cost}{2}dt+4 \int \dfrac{cos^{2}t}{2}dt$$
$$\displaystyle = -2 sin t + 2\int  \frac{(1-cos t)}{2} dt + c$$
$$\displaystyle = -2 sin t + t + \dfrac{sint}{2} + c$$
$$\displaystyle = -2\sqrt{1+-cos^{2}t}+t + sin\, t\, cos t  +c$$
$$\displaystyle = -2 \sqrt{1-cos^{2}t} + t + \sqrt{1-cos^{2} t} cos\, t + c $$
Replacing cos t = $$ \sqrt{x}$$
$$\displaystyle \boxed{ I = -2\sqrt{1-x}+cos^{-1}\sqrt{x} + \sqrt{x} \sqrt{1-x}+c}$$
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Subjective Medium Published on 17th 09, 2020
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