Integrals — Class 12 Maths Solution

Option c is correct

Let$I = \int\limits_0^{2/3} {\cfrac{{dx}}{{4 + 9{x^2}}}} = \cfrac{1}{9}\int\limits_0^{2/3} {\cfrac{{dx}}{{{{\left( {\cfrac{2}{3}} \right)}^2} + {x^2}}}}$

$= \cfrac{1}{9} \times \cfrac{1}{{\cfrac{2}{3}}}\left[ {{{\tan }^{ - 1}}\left( {\cfrac{{3x}}{2}} \right)} \right]_0^{2/3} = \cfrac{1}{6}\left[ {{{\tan }^{ - 1}}\left( {\cfrac{{3x}}{2}} \right)} \right]_0^{2/3}$

$= \cfrac{1}{6}\left[ {{{\tan }^{ - 1}}\left( 1 \right) - {{\tan }^{ - 1}}\left( 0 \right)} \right] = \cfrac{1}{6} \times \cfrac{\pi }{4} = \cfrac{\pi }{{24}}$

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