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Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals1 Mark
MCQ
Choose the correct answer in Exercise:
$\int\frac{\text{dx}}{\sqrt{9\text{x}-4\text{x}^2}}\text{equals}$
  • A
    $\frac{1}{9}\sin^{-1}\bigg(\frac{9\text{x}-8}{8}\bigg)+\text{C}$
  • ✓
    $\frac{1}{2}\sin^{-1}\bigg(\frac{8\text{x}-9}{9}\bigg)+\text{C}$
  • C
    $\frac{1}{3}\sin^{-1}\bigg(\frac{9\text{x}-8}{8}\bigg)+\text{C}$
  • D
    $\frac{1}{2}\sin^{-1}\bigg(\frac{9\text{x}-8}{9}\bigg)+\text{C}$
✓

Answer

Correct option: B.
$\frac{1}{2}\sin^{-1}\bigg(\frac{8\text{x}-9}{9}\bigg)+\text{C}$
$\text{Let I}=\int\frac{\text{dx}}{\sqrt{9\text{x}-4\text{x}^2}}$
$=\int\frac{1}{\sqrt{-4\text{x}^2+9\text{x}}}\text{ dx}$
$=\int\frac{1}{-4\bigg(\text{x}^2-\frac{9}{4}\text{x}\bigg)}\text{ dx}$
$=\int\frac{1}{-4\Bigg[\text{x}^2-\frac{9}{4}\text{x}+\bigg(\frac{9}{8}\bigg)^2-\bigg(\frac{9}{8}\bigg)^2\Bigg]}\text{ dx}$
$=\int\frac{1}{-4\Bigg[\bigg(\text{x}-\frac{9}{8}\bigg)^2+\bigg(\frac{9}{8}\bigg)^2\Bigg]}\text{ dx}$
$=\int\frac{1}{4\Bigg[\bigg(\frac{9}{8}\bigg)^2-\bigg(\text{x}-\frac{9}{8}\bigg)^2\Bigg]}\text{ dx}$
$=\frac{1}{2}\int\frac{1}{\Bigg[\bigg(\frac{9}{8}\bigg)^2-\bigg(\text{x}-\frac{9}{8}\bigg)^2\Bigg]}\text{ dx}$
$=\frac{1}{2}\sin^{-1}\frac{\text{x}-\frac{9}{8}}{\frac{9}{8}}+\text{C}$
$=\frac{1}{2}\sin^{-1}\bigg(\frac{8\text{x}-9}{9}\bigg)+\text{C}$
Therefore, option (B) is correct.

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