The correct option is B π24 ∫_0^2/3dx4 + 9x^2 = ∫_0^2/3dx(2)^2 + #160;(3x)^2 = 1/2 . 1/3 ( tan^ 13x2 )_0^2/3 = 16 [ π4 ] = π24 ...

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Property 2

∫02/3dx/4 + 9...

Question

$\int_{0}^{\frac{2}{3}} \frac{d x}{4 + 9 x^{2}} =$

\frac{π}{12}

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\frac{π}{24}

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\frac{π}{4}

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Solution

The correct option is B $\frac{π}{24}$
$\int_{0}^{2 / 3} \frac{d x}{4 + 9 x^{2}} = \int_{0}^{2 / 3} \frac{d x}{(2)^{2} + (3 x)^{2}}$
$= \frac{1}{2} . \frac{1}{3} {(t a n^{- 1} \frac{3 x}{2})}_{0}^{- 1} = \frac{1}{6} [\frac{π}{4}] = \frac{π}{24}$

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