# Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \sqrt{\frac{1-x}{1+x}} d x$

Solution:

Given $I=\int \frac{\sqrt{1-x}}{\sqrt{1+x}} d x$

Rationalizing the denominator,

$\Rightarrow \int \sqrt{\frac{1-x}{1+x}} d x=\int \sqrt{\frac{1-x}{1+x} \times \frac{1-x}{1-x}} d x$

$=\int \frac{1-x}{\sqrt{1-x^{2}}} d x$

Integral is of form $\int \frac{p x+q}{\sqrt{a x^{2}+b x+c}} d x$

Writing numerator as $\mathrm{px}+\mathrm{q}=\lambda\left\{\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}\right)\right\}+\mu$

$\Rightarrow p x+q=\lambda(2 a x+b)+\mu$

$\Rightarrow-x+1=\lambda(-2 x)+\mu$

$\therefore \lambda=1 / 2$ and $\mu=1$

$\Rightarrow \int \frac{1-\mathrm{x}}{\sqrt{1-\mathrm{x}^{2}}} \mathrm{dx}=\int\left(\frac{-2 \mathrm{x}}{2 \sqrt{1-\mathrm{x}^{2}}}+\frac{1}{\sqrt{1-\mathrm{x}^{2}}}\right) \mathrm{dx}$

$=-\int \frac{\mathrm{x}}{\sqrt{1-\mathrm{x}^{2}}} \mathrm{dx}+\int \frac{1}{\sqrt{1-\mathrm{x}^{2}}} \mathrm{dx}$

Let $-x+1=1 / 2(-2 x)+1$ and split,

$\Rightarrow \int \frac{1-x}{\sqrt{1-x^{2}}} d x=\int\left(\frac{-2 x}{2 \sqrt{1-x^{2}}}+\frac{1}{\sqrt{1-x^{2}}}\right) d x$

$=-\int \frac{x}{\sqrt{1-x^{2}}} d x+\int \frac{1}{\sqrt{1-x^{2}}} d x$

Consider $\int \frac{x}{\sqrt{1-x^{2}}} d x$

Let $u=1-x^{2} \rightarrow d x=\frac{-1}{2 x} d u$

$\Rightarrow \int \frac{\mathrm{x}}{\sqrt{1-\mathrm{x}^{2}}} \mathrm{dx}=\int \frac{-1}{2 \sqrt{\mathrm{u}}} \mathrm{du}$

$=\frac{-1}{2} \int \frac{1}{\sqrt{\mathrm{u}}} \mathrm{du}$

We know that $\int x^{n} d x=\frac{x^{n+1}}{n+1}+c$

$\Rightarrow \frac{1}{2} \int \frac{1}{\sqrt{u}} d u=\frac{1}{2}(2 \sqrt{u})$

$=\sqrt{u}=-\sqrt{1-x^{2}}$

Consider $\int \frac{1}{\sqrt{1-x^{2}}} d x$

We know that $\int \frac{1}{\sqrt{1-x^{2}}} d x+c=\sin ^{-1} x+c$

$\Rightarrow \int \frac{1}{\sqrt{1-x^{2}}} d x=\sin ^{-1}(x)$

Then,

$\Rightarrow \int \frac{1-x}{\sqrt{1-x^{2}}} d x=-\int \frac{x}{\sqrt{1-x^{2}}} d x+\int \frac{1}{\sqrt{1-x^{2}}} d x$

$=\sqrt{1-x^{2}}+\sin ^{-1}(x)+c$

$\therefore \mathrm{I}=\int \sqrt{\frac{1-\mathrm{x}}{1+\mathrm{x}}} \mathrm{dx}=\sqrt{1-\mathrm{x}^{2}}+\sin ^{-1}(\mathrm{x})+\mathrm{c}$
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