Chapter 7: Integrals

Ans: To simplify the question let us suppose,

$  \int {\dfrac{{dx}}{{1 + {x^2}}}}  = {\tan ^{ - 1}}x $

By second fundamental theorem of calculus, we obtain

$ \int\limits_1^{\sqrt 3 } {\dfrac{{dx}}{{1 + {x^2}}}}  = F(\sqrt 3 ) - F(1) $

$  = {\tan ^{ - 1}}\sqrt 3  - {\tan ^{^{ - 1}}}1 $

$= \dfrac{\pi }{3} - \dfrac{\pi }{4} $

$   = \dfrac{\pi }{{12}} $ 

Hence, the correct Answer is $\dfrac{\pi }{{12}}$

Ans: To simplify the question let us suppose,

 3x = t

$\Rightarrow 3dx = dt$ 

$\int {\dfrac{{dx}}{{{{(2)}^2} + {{(3x)}^2}}}}  = \dfrac{1}{3}\int {\dfrac{{dt}}{{{{(2)}^2} + {{(t)}^2}}}}  $

$ = \dfrac{1}{3}\left[ {\dfrac{1}{2}{{\tan }^{ - 1}}\dfrac{t}{2}} \right] $

$ = \dfrac{1}{6}{\tan ^{ - 1}}\left( {\dfrac{{3x}}{2}} \right) $

$= F(x) $ 

By second fundamental theorem of calculus,

$\int\limits_0^{\dfrac{2}{3}} {\dfrac{{dx}}{{4 + 9{x^2}}}}  = F\left( {\dfrac{2}{3}} \right) - F(0) $

$= \dfrac{1}{6}{\tan ^{ - 1}}\left( {\dfrac{3}{2} \times \dfrac{2}{3}} \right) - \dfrac{1}{6}{\tan ^{^{ - 1}}}0 $

 $  = \dfrac{1}{6}{\tan ^{ - 1}}1 - 0 $

 $  = \dfrac{1}{6} \times \dfrac{\pi }{4} $

 $  = \dfrac{\pi }{{24}} $ 

Hence, the correct answer is C.$\dfrac{\pi }{{24}}$