MCQ
The integral $\frac{24}{\pi} \int_{0}^{\sqrt{2}} \frac{\left(2-x^{2}\right) d x}{\left(2+x^{2}\right) \sqrt{4+x^{4}}}$ is equal to
  • A
    $1$
  • B
    $2$
  • C
    $4$
  • $3$

Answer

Correct option: D.
$3$
d
$\frac{24}{\pi} \int_{0}^{\sqrt{2}} \frac{\left(2-x^{2}\right)}{\left(x^{2}+2\right) \sqrt{4+x^{4}}} d x$

$\frac{24}{\pi} \int_{0}^{\sqrt{2}} \frac{x^{2}\left(\frac{2}{x^{2}}-1\right) d x}{x\left(x+\frac{2}{x}\right) \times x \sqrt{\frac{4}{x^{2}}+x^{2}}}$

$\frac{24}{\pi} \int_{0}^{\sqrt{2}} \frac{\left(\frac{2}{x^{2}}-1\right) d x}{\left(x+\frac{2}{x}\right) \sqrt{\left(x+\frac{2}{x}\right)^{2}-4}}$

$x +\frac{2}{ x }= t$

$dt =\left(1-\frac{2}{ x ^{2}}\right) dx$

$I =-\frac{24}{\pi} \int \frac{ dt }{ t \sqrt{ t ^{2}-4}}$

$=-\frac{24}{\pi} \times \frac{1}{2} \sec ^{-1}\left[\frac{ x +\frac{2}{ x }}{2}\right)^{\sqrt{2}}$

$=-\frac{12}{\pi}\left[\sec ^{-1}\left(\frac{2 \sqrt{2}}{2}\right)-\sec ^{-1}(\infty)\right]$

$=-\frac{12}{\pi}\left[\frac{\pi}{4}-\frac{2 \pi}{2 \times 2}\right]=-\frac{12}{\pi}\left[-\frac{\pi}{4}\right]$

$=3$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free