Question

Integrate the function ​

Answer 1

We're given to solve,

$\displaystyle\longrightarrow\sf{\int\sqrt{\dfrac{1+x}{1-x}}\ dx=\,?}$

Let,

$\displaystyle\longrightarrow\sf{x=\cos(2\theta)\quad\implies\quad2\theta=\cos^{-1}(x)}$

$\displaystyle\longrightarrow\sf{dx=-\sin(2\theta)\times2\,d\theta}$

$\displaystyle\longrightarrow\sf{dx=-2\sin(2\theta)\,d\theta}$

Hence,

$\displaystyle\longrightarrow\sf{\int\sqrt{\dfrac{1+x}{1-x}}\ dx=-2\int\sqrt{\dfrac{1+\cos(2\theta)}{1-\cos(2\theta)}}\cdot\sin(2\theta)\,d\theta}$

But,

• $\sf{1+\cos(2\theta)=2\cos^2\theta}$

• $\sf{1-\cos(2\theta)=2\sin^2\theta}$

Then,

$\displaystyle\longrightarrow\sf{\int\sqrt{\dfrac{1+x}{1-x}}\ dx=-2\int\sqrt{\dfrac{2\cos^2\theta}{2\sin^2\theta}}\cdot\sin(2\theta)\,d\theta}$

$\displaystyle\longrightarrow\sf{\int\sqrt{\dfrac{1+x}{1-x}}\ dx=-2\int\sqrt{\dfrac{\cos^2\theta}{\sin^2\theta}}\cdot\sin(2\theta)\,d\theta}$

$\displaystyle\longrightarrow\sf{\int\sqrt{\dfrac{1+x}{1-x}}\ dx=-2\int\dfrac{\cos\theta}{\sin\theta}\cdot\sin(2\theta)\,d\theta}$

Since $\sf{\sin(2\theta)=2\sin\theta\cos\theta,}$

$\displaystyle\longrightarrow\sf{\int\sqrt{\dfrac{1+x}{1-x}}\ dx=-4\int\dfrac{\cos\theta}{\sin\theta}\cdot\sin\theta\cos\theta\,d\theta}$

$\displaystyle\longrightarrow\sf{\int\sqrt{\dfrac{1+x}{1-x}}\ dx=-4\int\cos^2\theta\,d\theta}$

Taking $\sf{\cos^2\theta=\dfrac{1+\cos(2\theta)}{2},}$

$\displaystyle\longrightarrow\sf{\int\sqrt{\dfrac{1+x}{1-x}}\ dx=-2\int\left(1+\cos(2\theta)\right)\,d\theta}$

$\displaystyle\longrightarrow\sf{\int\sqrt{\dfrac{1+x}{1-x}}\ dx=-2\int d\theta-2\int\cos(2\theta)\,d\theta}$

But,

$\displaystyle\longrightarrow\sf{\dfrac{d(2\theta)}{d\theta}=2}$

$\displaystyle\longrightarrow\sf{d\theta=\dfrac{d(2\theta)}{2}}$

Then,

$\displaystyle\longrightarrow\sf{\int\sqrt{\dfrac{1+x}{1-x}}\ dx=-2\int d\theta-\int\cos(2\theta)\,d(2\theta)}$

$\displaystyle\longrightarrow\sf{\int\sqrt{\dfrac{1+x}{1-x}}\ dx=-2\theta-\sin(2\theta)+c}$

Undoing $\sf{2\theta=\cos^{-1}(x),}$

$\displaystyle\longrightarrow\sf{\int\sqrt{\dfrac{1+x}{1-x}}\ dx=-\cos^{-1}(x)-\sin(2\theta)+c}$

But,

$\displaystyle\longrightarrow\sf{\sin(2\theta)=\sqrt{1-\cos^2(2\theta)}}$

$\displaystyle\longrightarrow\sf{\sin(2\theta)=\sqrt{1-x^2}}$

Then,

$\displaystyle\longrightarrow\underline{\underline{\sf{\int\sqrt{\dfrac{1+x}{1-x}}\ dx=-\cos^{-1}(x)-\sqrt{1-x^2}+c}}}$