$\displaystyle\int_0^{2/3} \dfrac{dx}{4+9x^2}$ equals
A$\dfrac{\pi}{6}$
B$\dfrac{\pi}{12}$
C$\dfrac{\pi}{24}$
D$\dfrac{\pi}{4}$
Answer & Solution
Correct answer: C. $\dfrac{\pi}{24}$
1. Factor: $4+9x^2=9\left(x^2+\tfrac{4}{9}\right)$, with $a=\tfrac{2}{3}$.
2. $\int \dfrac{dx}{9(x^2+a^2)}=\dfrac{1}{9}\cdot\dfrac{1}{a}\tan^{-1}\dfrac{x}{a}=\dfrac{1}{6}\tan^{-1}\dfrac{3x}{2}$.
3. Evaluate $0$ to $\tfrac{2}{3}$: $\dfrac{1}{6}\left(\tan^{-1}1-0\right)=\dfrac{1}{6}\cdot\dfrac{\pi}{4}$.
4. $=\dfrac{\pi}{24}$.
_Source: NCERT Class 12 Mathematics Ch 7 "Integrals", p.338_
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