$\int_{0}^{\frac{2}{3}} \frac{d x}{4+9 x^{2}}$ बराबर है:
Exercise-7.9-22
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$\int_{0}^{2 / 3} \frac{1}{4+9 x^{2}} d x$ $=\frac{1}{9} \int_{0}^{2 / 3} \frac{1}{\frac{4}{9}+x^{2}} d x$ $=\frac{1}{9} \int_{0}^{2 / 3} \frac{1}{\left(\frac{2}{3}\right)^{2}+x^{2}} d x$
$=\frac{1}{9} \cdot \frac{1}{\frac2 3}\left[\tan ^{-1}\left(\frac{x}{\frac2 3}\right)\right]_{0}^{\frac2 3}$ $\left(\because \int \frac{d x}{a^{2}+x^{2}}=\frac{1}{a} \tan ^{-1} \frac{x}{a}\right)$
$=\frac{1}{6}\left[\tan ^{-1}\left(\frac{3 x}{2}\right)\right]_{0}^{\frac2 3}$ $=\frac{1}{6}\left[\tan ^{-1}\left(\frac{3}{2} \cdot \frac{2}{3}\right)-\tan ^{-1} 0\right]$ $=\frac{1}{6}\left(\tan ^{-1} 1-0\right)=\frac{1}{6} \cdot \frac{\pi}{4}=\frac{\pi}{24}$
$=\frac{1}{9} \cdot \frac{1}{\frac2 3}\left[\tan ^{-1}\left(\frac{x}{\frac2 3}\right)\right]_{0}^{\frac2 3}$ $\left(\because \int \frac{d x}{a^{2}+x^{2}}=\frac{1}{a} \tan ^{-1} \frac{x}{a}\right)$
$=\frac{1}{6}\left[\tan ^{-1}\left(\frac{3 x}{2}\right)\right]_{0}^{\frac2 3}$ $=\frac{1}{6}\left[\tan ^{-1}\left(\frac{3}{2} \cdot \frac{2}{3}\right)-\tan ^{-1} 0\right]$ $=\frac{1}{6}\left(\tan ^{-1} 1-0\right)=\frac{1}{6} \cdot \frac{\pi}{4}=\frac{\pi}{24}$
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