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Integrals — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIntegrals1 Mark
Question
$\int_{0}^{\frac{2}{3}} \frac{d x}{4+9 x^{2}}$ equals
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Answer

Let $I=\int_{0}^{\frac{2}{3}} \frac{d x}{4+9 x^{2}}$ 
Taking 9 common from Denominator in I
$\Rightarrow I=\frac{1}{9} \int_{0}^{\frac{2}{3}} \frac{d x}{\frac{4}{9}+x^{2}}=\frac{1}{9} \int_{0}^{\frac{2}{3}} \frac{d x}{\left(\frac{2}{3}\right)^{2}+x^{2}}$ [$\int \frac{d x}{a^{2}+x^{2}}=\frac{1}{a} \tan ^{-1} \frac{x}{a}+c$]
$\Rightarrow \mathrm{I}=\frac{1}{9} \times \frac{3}{2}\left[\tan ^{-1} \frac{\mathrm{x}}{\frac{2}{3}}\right]_{0}^{\frac{2}{3}}$ = $\frac{1}{9} \times \frac{3}{2}\left[\tan ^{-1} \frac{3 x}{2}\right]_{0}^{\frac{2}{3}}$ 
$\Rightarrow I=\frac{1}{6}\left[\tan ^{-1} \frac{3}{2} \times \frac{2}{3}-\tan ^{-1} 0\right]$ = $\frac{1}{6}\left[\tan ^{-1} 1-\tan ^{-1} 0\right]$ 
$\Rightarrow I=\frac{1}{6} \times\left(\frac{\pi}{4}-0\right)=\frac{\pi}{ 24}$
$\therefore \int_{0}^{\frac{2}{3}} \frac{d x}{4+9 x^{2}}=\frac{\pi}{24}$

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