Question

$\int^\frac{2}{3}_0 {\frac{dx}{4+9x^2}$
(A)\frac{\pi}{6}[/tex]
(B)\frac{\pi}{12}[/tex]
(C)\frac{\pi}{24}[/tex]
(D)\frac{\pi}{4}[/tex]

Answer 1

Answer:

option (c) is correct

Step-by-step explanation:

Concept:

1.If $\int{f(x)}\:dx=g(x)+c$, then

$\int{f(ax+b)}\;dx=\frac{1}{a}g(ax+b)+c$

2.$\int{\frac{1}{1+x^2}}\:dx=tan^{-1}x+c$

Now,

$\int\limits^{\frac{2}{3}}_0{\frac{1}{4+9x^2}}\:dx\\\\=\int\limits^{\frac{2}{3}}_0{\frac{1}{(3x)^2+2^2}}\:dx\\\\=\frac{1}{3}\frac{1}{2}[tan^{-1}(\frac{3x}{2})]^{\frac{2}{3}}_0\\\\=\frac{1}{6}[tan^{-1}(\frac{3}{2}\frac{2}{3})-tan^{-1}0]\\\\=\frac{1}{6}[tan^{-1}(1)-tan^{-1}(0)]\\\\=\frac{1}{6}[\frac{\pi}{4}-0]\\\\=\frac{\pi}{24}$