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Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals1 Mark
MCQ
$\int\limits^\frac{\pi}{2}_0\frac{1}{2+\cos\text{x}}\text{ dx}$ equals:
  • A
    $\frac{1}{3}\tan^{-1}\Big(\frac{1}{\sqrt{3}}\Big)$
  • ✓
    $\frac{2}{\sqrt{3}}\tan^{-1}\Big(\frac{1}{\sqrt{3}}\Big)$
  • C
    ${\sqrt{3 }}\tan^{-1}\big({\sqrt{3}}\big)$
  • D
    $2{\sqrt{3 }}\tan^{-1}{\sqrt{3}}$
✓

Answer

Correct option: B.
$\frac{2}{\sqrt{3}}\tan^{-1}\Big(\frac{1}{\sqrt{3}}\Big)$
We have,
$=\int\limits^\frac{\pi}{2}_0\frac{1}{2+\cos\text{x}}\text{ dx}$
$=\int\limits^\frac{\pi}{2}_0\frac{1}{2+\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}}\text{ dx}$
$=\int\limits^\frac{\pi}{2}_0\frac{1+\tan^2\frac{\text{x}}{2}}{2+2\tan^2\frac{\text{x}}{2}+1-\tan^2\frac{\text{x}}{2}}\text{ dx}$
$=\int\limits^\frac{\pi}{2}_0\frac{\sec^2\frac{\text{x}}{2}}{3+\tan^2\frac{\text{x}}{2}}\text{dx}$
Putting $\tan\frac{\text{x}}{2}=\text{t}$
$\Rightarrow \frac{1}{2}\sec^2\frac{\text{x}}{2}\text{dx}=\text{dt}$
$\Rightarrow \sec^2\frac{\text{x}}{2}\text{dx}=2\text{dt}$
when $\text{x}\rightarrow0;\text{ t}\rightarrow0$
and $\text{x}\rightarrow\frac{\pi}{2};\text{ t}\rightarrow1$
$\therefore\ \text{I}=\int\limits^1_0\frac{2}{3+\text{t}^2}\text{ dt}$
$=2\int\limits^1_0\frac{1}{(\sqrt{3})^2+\text{t}^2}\text{dt}$
$=\frac{2}{\sqrt{3}}\Big[\tan^{-1}\frac{\text{t}}{\sqrt{3}}\Big]^1_0$
$=\frac{2}{\sqrt{3}}\Big[\tan^{-1}\frac{1}{\sqrt{3}}-\tan^{-1}\frac{0}{\sqrt{3}}\Big]$
$=\frac{2}{\sqrt{3}}\tan^{-1}\Big(\frac{1}{\sqrt{3}}\Big)$

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