The value of $\displaystyle\int_{0}^{2/3}\dfrac{dx}{\sqrt{4-9x^2}}$ is : (a) $\pi$ (b) $\dfrac{\pi}{6}$ (c) $\dfrac{\pi}{2}$ (d) $\dfrac{\pi}{4}$

> [!TLDR] > Rewriting the integrand in the standard $\int dx/\sqrt{a^2-x^2}$ form and evaluating the limits gives $\dfrac{\pi}{6}$. > > 1. We need $\displaystyle\int_0^{2/3}\dfrac{dx}{\sqrt{4-9x^2}}$. > 2. Factor out $9$ from inside the square root: $4-9x^2=9\left(\dfrac49-x^2\right)$, so $\sqrt{4-9x^2}=3\sqrt{\left(\dfrac23\right)^2-x^2}$. > 3. So the integral becomes $\displaystyle\int_0^{2/3}\dfrac{dx}{3\sqrt{(2/3)^2-x^2}}=\dfrac13\int_0^{2/3}\dfrac{dx}{\sqrt{(2/3)^2-x^2}}$. > 4. Use the standard result $\displaystyle\int\dfrac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\!\left(\dfrac{x}{a}\right)+C$ with $a=\dfrac23$: $\dfrac13\left[\sin^{-1}\!\left(\dfrac{x}{2/3}\right)\right]_0^{2/3}=\dfrac13\left[\sin^{-1}\!\left(\dfrac{3x}{2}\right)\right]_0^{2/3}$. …

The value of ∫₀2/3dfracdx√(4-9x²) is : (a) π (b) (π)/(6) (c) (π)/(2) (d) (π)/(4)

Question 76 of 96

\displaystyle\int_{0}^{2/3}\dfrac{dx}{\sqrt{4-9x^2}}

(a)

\pi

(b)

\dfrac{\pi}{6}

(c)

\dfrac{\pi}{2}

(d)

\dfrac{\pi}{4}

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Rewriting the integrand in the standard $\int dx/\sqrt{a^2-x^2}$ form and evaluating the limits gives $\dfrac{\pi}{6}$ .

We need $\displaystyle\int_0^{2/3}\dfrac{dx}{\sqrt{4-9x^2}}$ .
Factor out $9$ from inside the square root: $4-9x^2=9\left(\dfrac49-x^2\right)$ , so $\sqrt{4-9x^2}=3\sqrt{\left(\dfrac23\right)^2-x^2}$ .
So the integral becomes $\displaystyle\int_0^{2/3}\dfrac{dx}{3\sqrt{(2/3)^2-x^2}}=\dfrac13\int_0^{2/3}\dfrac{dx}{\sqrt{(2/3)^2-x^2}}$ .
Use the standard result $\displaystyle\int\dfrac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\!\left(\dfrac{x}{a}\right)+C$ with $a=\dfrac23$ : $\dfrac13\left[\sin^{-1}\!\left(\dfrac{x}{2/3}\right)\right]_0^{2/3}=\dfrac13\left[\sin^{-1}\!\left(\dfrac{3x}{2}\right)\right]_0^{2/3}$ . …

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Q.The value of $\displaystyle\int_{0}^{2/3}\dfrac{dx}{\sqrt{4-9x^2}}$ is :

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Q.The value of $\displaystyle\int_{0}^{2/3}\dfrac{dx}{\sqrt{4-9x^2}}$ is :

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