Q96 / 160 · TS EAMCET 2015 · Mathematics

$\int \sqrt{\frac{2+x}{2-x}} d x$ is equal to

  • A
    $2 \sin ^{-1}\left(\frac{x}{2}\right)+\sqrt{4-x^2}+C$
  • B
    $\cos ^{-1}\left(\frac{x}{2}\right)-\sqrt{4-x^2}+C$
  • C
    $\sin ^{-1}\left(\frac{x}{2}\right)-\sqrt{4-x^2}+C$
  • Correct answer
    D
    $2 \sin ^{-1}\left(\frac{x}{2}\right)-\sqrt{4-x^2}+C$

Solution

Let $$ \begin{aligned} I & =\int \sqrt{\frac{2+x}{2-x}} d x \\ & =\int \sqrt{\frac{2+x}{2-x}} \times \sqrt{\frac{2+x}{2+x}} d x=\int \frac{2+x}{\sqrt{4-x^2}} d x \\ & =2 \int \frac{1}{\sqrt{(2)^2-x^2}} d x+\int \frac{x}{\sqrt{4-x^2}} d x \\ & =2 \sin ^{-1} \frac{x}{2}-\sqrt{4-x^2}+C \end{aligned} $$

Hence, the correct answer is option: 'D'.

Relevant Questions