Choose the correct answer in Exercise: ^2 3 _ 0 dx 4+9x^ 2 equals - Vidyadip

MCQ

Choose the correct answer in Exercise:$\int^\frac{2}{3}_{0}\frac{\text{dx}}{4+9\text{x}^{2}}\text{equals}$

A
$\frac{\pi}{6}$
B
$\frac{\pi}{12}$
✓
$\frac{\pi}{24}$
D
$\frac{\pi}{4}$

✓

Answer

Correct option: C.

$\frac{\pi}{24}$

$\frac{\pi}{24}$ $\int\frac{\text{dx}}{4+9\text{x}^{2}}=\int\frac{\text{dx}}{(2)^{2}+(3\text{x})^{2}}$$\ \text{put}\ 3\text{x}=\text{t}\Rightarrow3\text{dx}=\text{dt}$
$\therefore\int\frac{\text{dx}}{(2)^{2}+(3\text{x})^{2}}=\frac{1}{3}\int\frac{\text{dt}}{(2)^{2}+\text{t}^{2}}$
$=\frac{1}{3}\bigg[\frac{1}{2}\tan^{-1}\frac{t}{2}\bigg]$
$=\frac{1}{6}\tan^{-1}\bigg(\frac{3\text{x}}{2}\bigg)$
$=\text{F}\text{(x)}$
By second fundamental theorem of calculus, we obtain $\int\limits_{0}^{\frac{2}{3}}\frac{\text{dx}}{4+9\text{x}^{2}}=\text{F}\bigg(\frac{2}{3}\bigg)-\text{F}(0)$ $=\frac{1}{6}\tan^{-1}\bigg(\frac{3}{2}.\frac{2}{3}\bigg)-\frac{1}{6}\tan^{-1}0$ $=\frac{1}{6}\tan^{-1}1-0$ $=\frac{1}{6}\times\frac{\pi}{4}$ $=\frac{\pi}{24}$

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