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# Prove the following

Question:

$\int \sqrt{\frac{a+x}{a-x}}$

Solution:

Let, $I=\int \sqrt{\frac{a+x}{a-x}} d x$

$=\int \sqrt{\frac{a+x}{a-x} \times \frac{a+x}{a+x}} d x=\int \frac{a+x}{\sqrt{(a-x)(a+x)}} d x$

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

Let's consider, $\mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}$

Now, $\mathrm{I}_{1}=\int \frac{a}{\sqrt{a^{2}-x^{2}}} d x=a \cdot \sin ^{-1} \frac{x}{a}+\mathrm{C}_{1}$

And, $\mathrm{I}_{2}=\int \frac{x}{\sqrt{a^{2}-x^{2}}} d x$

Putting, $a^{2}-x^{2}=t \Rightarrow-2 x d x=d t$

$x d x=\frac{d t}{-2}$

Hence, $I_{2}=-\frac{1}{2} \int \frac{d t}{\sqrt{t}}=-\frac{1}{2} \times 2 \sqrt{t}=-\sqrt{a^{2}-x^{2}}+C_{2}$

As, $I=I_{1}+I_{2}$

$=a \sin ^{-1} \frac{x}{a}+C_{1}-\sqrt{a^{2}-x^{2}}+C_{2}$

$\therefore \mathrm{I}=a \sin ^{-1} \frac{x}{a}-\sqrt{a^{2}-x^{2}}+\left(\mathrm{C}_{1}+\mathrm{C}_{2}\right)$

Therefore, $\mathrm{I}=a \sin ^{-1} \frac{x}{a}-\sqrt{a^{2}-x^{2}}+\mathrm{C}$    $\left[\mathrm{C}=\mathrm{C}_{1}+\mathrm{C}_{2}\right]$