12th NCERT INTEGRALS Exercise 7.9 Questions 22
Do or do not
There is no try

### Evaluate the definite integrals in Exercises 1 to 20

Question (1)

$I = \int_{ - 1}^1 {\left( {x + 1} \right)dx}$

Solution

$I = \int_{ - 1}^1 {x\;dx + \int_{ - 1}^1 {1\;dx} }$ $I = \left[ {\frac{{{x^2}}}{2}} \right]_{ - 1}^1 + \left[ x \right]_{ - 1}^1$ $I = \frac{1}{2}\left[ {{1^2} - {{\left( { - 1} \right)}^2}} \right] + \left[ {1 - \left( { - 1} \right)} \right]$ $I = \frac{1}{2}\left( {1 - 1} \right) + \left( {1 + 1} \right)$ = 0 + 2 =2

Question (2)

$I = \int_2^3 {\frac{1}{x}dx}$

Solution

$I = \left[ {\log x} \right]_2^3$ $I = \log 3 - \log 2$ $I = \log \left( {\frac{3}{2}} \right)$

Question (3)

$I = \int_1^2 {\left( {4{x^3} - 5{x^2} + 6x + 9} \right)dx}$

Solution

$I = 4\int_1^2 {{x^3}dx} - 5\int_1^2 {{x^2}dx + 6\int_1^2 {xdx} } + 9\int_1^2 {dx}$ $I = 4\left[ {\frac{{{x^4}}}{4}} \right]_1^2 - 5\left[ {\frac{{{x^3}}}{3}} \right]_1^2 + 6\left[ {\frac{{{x^2}}}{2}} \right] + 9\left[ x \right]_1^2$ $I = \frac{\require{cancel}\cancel{4}}{\cancel{4}}\left[ {{2^4} - {1^4}} \right] - \frac{5}{3}\left[ {{2^3} - {1^3}} \right] + \frac{\cancel{6}^3}{2}\left[ {{2^2} - {1^2}} \right] + 9\left[ {2 - 1} \right]$ $I = \left( {16 - 1} \right) - \frac{5}{3}\left( {8 - 1} \right) + 3\left( {4 - 1} \right) + 9\left( 1 \right)$ $I = 15 - \frac{{35}}{3} + 9 + 9 = 33 - \frac{{35}}{3}$ $I = \frac{{99 - 35}}{3} = \frac{{64}}{3}$

Question (4)

$I = \int_0^{\frac{\pi }{4}} {\sin 2x\;dx}$

Solution

$I = \int_0^{\frac{\pi }{4}} {\sin 2x\;dx}$ $I = \left[ { - \frac{{\cos 2x}}{2}} \right]_0^{\frac{\pi }{4}}$ $I = - \frac{1}{2}\left[ {\cos \frac{{\cancel{2}\pi }}{\cancel{4}_2} - \cos 2\left( 0 \right)} \right]$ $I = - \frac{1}{2}\left[ {\cos \frac{\pi }{2} - \cos 0} \right]$ $I = - \frac{1}{2}\left[ {0 - \left( 1 \right)} \right]$ $I = - \frac{1}{2} \times \left( { - 1} \right) = \frac{1}{2}$

Question (5)

$I = \int_0^{\frac{\pi }{2}} {\cos 2x\;dx}$

Solution

$I = \int_0^{\frac{\pi }{2}} {\cos 2x\;dx}$ $I = \left[ {\frac{{\sin 2x}}{2}} \right]_0^{\frac{\pi }{2}}$ $I = \frac{1}{2}\left[ {sin2\left( {\frac{\pi }{2}} \right) - \sin 2\left( 0 \right)} \right]$ $I = \frac{1}{2}\left[ {\sin \pi - sin0} \right]$ $I = \frac{1}{2}\left( {0 - 0} \right) = 0$

Question (6)

$I = \int_4^5 {{e^x}\;dx}$

Solution

$I = \int_4^5 {{e^x}\;dx}$ $I = \left[ {{e^x}} \right]_4^5$ $I = \left[ {{e^5} - {e^4}} \right]$ $I = {e^4}\left( {e - 1} \right)$

Question (7)

$I = \int_0^{\frac{\pi }{4}} {\tan x\;dx}$

Solution

$I = \int_0^{\frac{\pi }{4}} {\tan x\;dx}$ $I = \left[ {\log \left| {\sec x} \right|} \right]_0^{\frac{\pi }{4}}$ $I = \log \left| {\sec \frac{\pi }{4}} \right| - \log \left| {\sec 0} \right|$ $I = \log \left| {\sqrt 2 } \right| - \log 1$ Note log 1= 0 $I = \log \sqrt 2 - 0$ $I = \log {2^{\frac{1}{2}}}$ $I = \frac{1}{2}\log 2$

Question (8)

$I = \int_{\frac{\pi }{6}}^{\frac{\pi }{4}} {\cos ec\;x} \;dx$

Solution

$I = \int_{\frac{\pi }{6}}^{\frac{\pi }{4}} {\cos ec\;x} \;dx$ $I = \left[ {\log \left| {\cos ec\;x - \cot x} \right|} \right]_{\frac{\pi }{6}}^{\frac{\pi }{4}}$ $I = \log \left| {\cos ec\left( {\frac{\pi }{4}} \right) - \cot \left( {\frac{\pi }{4}} \right)} \right| - \log \left| {\cos ec\left( {\frac{\pi }{6}} \right) - \cot \left( {\frac{\pi }{6}} \right)} \right|$ $I = \log \left| {\sqrt 2 - 1} \right| - \log \left| {2 - \sqrt 3 } \right|$ $I = \log \left| {\frac{{\sqrt 2 - 1}}{{2 - \sqrt 3 }}} \right|$

Question (9)

$I = \int_0^1 {\frac{{dx}}{{\sqrt {1 - {x^2}} }}}$

Solution

$I = \int_0^1 {\frac{{dx}}{{\sqrt {1 - {x^2}} }}}$ $I = \left[ {{{\sin }^{ - 1}}x} \right]_0^1$ $I = {\sin ^{ - 1}}1 - {\sin ^{ - 1}}0$ $I = \frac{\pi }{2} - 0$ $I = \frac{\pi }{2}$

Question (10)

$I = \int_0^1 {\frac{1}{{1 + {x^2}}}dx}$

Solution

$I = \int_0^1 {\frac{1}{{1 + {x^2}}}dx}$ $I = \left[ {{{\tan }^{ - 1}}x} \right]_0^1$ $I = {\tan ^{ - 1}}1 - {\tan ^{ - 1}}0$ $I = \frac{\pi }{4} - 0$ $I = \frac{\pi }{4}$

Question (11)

$I = \int_2^3 {\frac{{dx}}{{{x^2} - 1}}}$

Solution

$I = \int_2^3 {\frac{{dx}}{{{x^2} - 1}}}$$I = \frac{1}{{2\left( 1 \right)}}\left[ {\log \left| {\frac{{x - 1}}{{x + 1}}} \right|} \right]_2^3$ $I = \frac{1}{{2\left( 1 \right)}}\left[ {\log \left| {\frac{{x - 1}}{{x + 1}}} \right|} \right]_2^3$ $I = \frac{1}{2}\left[ {\log \left| {\frac{{3 - 1}}{{3 + 1}}} \right| - \log \left| {\frac{{2 - 1}}{{2 + 1}}} \right|} \right]$ $I = \frac{1}{2}\left[ {\log \frac{\cancel {2}}{\cancel {4}_2} - \log \frac{1}{3}} \right]$ $I = \frac{1}{2}\log \frac{3}{2}$

Question (12)

$I = \int_0^{\frac{\pi }{2}} {{{\cos }^2}x\;dx}$

Solution

$I = \int_0^{\frac{\pi }{2}} {{{\cos }^2}x\;dx}$
Use formula ${\cos ^2}x = \frac{{1 + \cos 2x}}{2}$
$I = \int_0^{\frac{\pi }{2}} {\frac{{1 + \cos 2x}}{2}dx}$ $I = \frac{1}{2}\int_0^{\frac{\pi }{2}} {dx} + \frac{1}{2}\int_0^{\frac{\pi }{2}} {\cos 2x\;dx}$ $I = \frac{1}{2}\left[ x \right]_0^{\frac{\pi }{2}} + \frac{1}{2}\left[ {\frac{{\sin 2x}}{2}} \right]_0^{\frac{\pi }{2}}$ $I = \frac{1}{2}\left( {\frac{\pi }{2} - 0} \right) + \frac{1}{4}\left[ {\sin 2\left( {\frac{\pi }{2}} \right) - \sin 2\left( 0 \right)} \right]$ $I = \frac{\pi }{4} + \frac{1}{4}\left( {0 - 0} \right)$ OTHER METHOD By using property $\int_0^a {f\left( x \right)dx} = \int_0^a {f\left( {a - x} \right)dx}$ $I = \int_0^{\frac{\pi }{2}} {{{\cos }^2}x\;dx} \quad - - - (1)$ Using property $I = \int_0^{\frac{\pi }{2}} {{{\cos }^2}\left( {\frac{\pi }{2} - x} \right)dx}$ $I = \int_0^{\frac{\pi }{2}} {{{\sin }^2}x\;dx} \quad - - - \left( 2 \right)$ Add (1) and (2) we get
$2I = \int_0^{\frac{\pi }{2}} {\left( {{{\cos }^2}x + {{\sin }^2}x} \right)} dx$ $2I = \int\limits_0^{\frac{\pi }{2}} {dx}$ $2I = \left[ x \right]_0^{\frac{\pi }{2}}$ $2I = \frac{\pi }{2} - 0$ $I = \frac{\pi }{4}$

Question (13)

$I = \int_2^3 {\frac{x}{{{x^2} + 1}}dx}$

Solution

$I = \int_2^3 {\frac{x}{{{x^2} + 1}}dx}$ Let x2 + 1 = t
∴ 2x dx = dt
x dx = dt/2 When x = 2, t = 22 + 1 = 4 + 1 = 5
x = 3, t = 32 + 1 =10
$I = \int_5^{10} {\frac{{\frac{{dt}}{2}}}{t}}$ $I = \frac{1}{2}\int\limits_5^{10} {\frac{{dt}}{t}}$ $I = \frac{1}{2}\left[ {\log \;t} \right]_5^{10}$ $I = \frac{1}{2}\left[ {\log 10 - \log 5} \right]$ $I = \frac{1}{2}\log \left| {\frac{{\cancel{10}^2}}{\cancel{5}}} \right|$ $I = \frac{1}{2}\log 2$

Question (14)

$I = \int_0^1 {\frac{{2x + 3}}{{5{x^2} + 1}}dx}$

Solution

$I = \int_0^1 {\frac{{2x + 3}}{{5{x^2} + 1}}dx}$ $I = \int_0^1 {\frac{{2x}}{{5{x^2} + 1}}} dx + \int_0^1 {\frac{3}{{5{x^2} + 1}}} dx$ $I = 2\int_0^1 {\frac{x}{{5{x^2} + 1}}} dx + \frac{3}{5}\int_0^1 {\frac{1}{{{x^2} + \frac{1}{5}}}dx}$ $I = 2{I_1} + \frac{3}{5}{I_2}\; - - - 1$ ${I_1} = \int_0^1 {\frac{x}{{5{x^2} + 1}}dx}$ Let 5x2 + 1 = t
10x dx = dt
x dx = dt/10
When x =0, t = 5(0)2 + 1 = 1
x = 1 t = 5(1)2 + 1 = 5 + 1 = 6
${I_1} = \int_1^6 {\frac{{\frac{{dt}}{{10}}}}{t}}$ ${I_1} = \frac{1}{{10}}\int_1^6 {\frac{{dt}}{t}}$ ${I_1} = \frac{1}{{10}}\left[ {\log \left| t \right|} \right]_1^6$ ${I_1} = \frac{1}{{10}}\left( {\log 6 - \log 1} \right)$ ${I_1} = \frac{1}{{10}}\log 6$ ${I_2} = \int_0^1 {\frac{1}{{{x^2} + \frac{1}{5}}}} dx$ ${I_2} = \int_0^1 {\frac{1}{{{{\left( x \right)}^2} + {{\left( {\sqrt {\frac{1}{5}} } \right)}^2}}}dx}$ ${I_2} = \frac{1}{{\frac{1}{{\sqrt 5 }}}}\left[ {{{\tan }^{ - 1}}\left( {\frac{x}{{\frac{1}{{\sqrt 5 }}}}} \right)} \right]_0^1$ ${I_2} = \sqrt 5 \left[ {{{\tan }^{ - 1}}\left( {\sqrt 5 x} \right)} \right]_0^1$ ${I_2} = \sqrt 5 \left[ {{{\tan }^{ - 1}}\sqrt 5 - {{\tan }^{ - 1}}0} \right]$ ${I_2} = \sqrt 5 \left( {{{\tan }^{ - 1}}\sqrt 5 - 0} \right)$ ${I_2} = \sqrt 5 {\tan ^{ - 1}}\sqrt 5$ Substitute values of I1 and I2 in equestion 1 $I = \cancel{2}\left[ {\frac{1}{{\cancel{10}}}\log 6} \right] + \frac{3}{5}\sqrt 5 {\tan ^{ - 1}}\sqrt 5$ $I = \frac{1}{5}\log 6 + \frac{3}{{\sqrt 5 }}{\tan ^{ - 1}}\sqrt 5$

Question (15)

$I = \int_0^1 {x{e^{{x^2}}}} dx$

Solution

$I = \int_0^1 {x{e^{{x^2}}}} dx$ Let x2 = t
2x dx = dt
x dx = dt/2
Where x = 0, t = 02 = 0
x = 1, t =11 = 1 $I = \int_0^1 {\frac{{dt}}{2}{e^t}}$ $I = \frac{1}{2}\int_0^1 {{e^t}dt}$ $I = \frac{1}{2}\left[ {{e^t}} \right]_0^1$ $I = \frac{1}{2}\left[ {{e^1} - {e^0}} \right]$ $I = \frac{{e - 1}}{2}$

Question (16)

$I = \int_1^2 {\frac{{5{x^2}}}{{{x^2} + 4x + 3}}dx}$

Solution

$I = 5\int_1^2 {\frac{{{x^2}}}{{{x^2} + 4x + 3}}dx}$ $$\require{enclose} \begin{array}{r1} 1 \quad\; \\ {x^2} +4x + 3 \enclose{longdiv}{\; {x^2}+{0} +\;0}\ \\ \underline {-{x^2} \pm \ 4x\pm \ 3}\ \\ \;-4x -3\phantom{00} \\ \end{array}$$
$I = 5\int_1^2 {\left( {1 + \frac{{ - 4x - 3}}{{{x^2} + 4x + 3}}} \right)dx}$ $I = 5\int_1^2 {1\;dx - 5\int_1^2 {\frac{{4x + 3}}{{{x^2} + 4x + 3}}\;} } dx$ $I = 5\left[ x \right]_1^2 - 5{I_2}$ $I = 5\left( {2 - 1} \right) - 5{I_2}$ $I = 5 - 5{I_2}$ $I = \int_1^2 {\frac{{4x + 3}}{{{x^2} + 4x + 3}}} dx$ $Let\quad\frac{A}{{x + 3}} + \frac{B}{{x + 1}} = \frac{{4x + 3}}{{\left( {x + 3} \right)\left( {x + 1} \right)}}$ $A\left( {x + 1} \right) + B\left( {x + 3} \right) = 4x + 3$ If x = -1 ⇒ B(-1+3) = 4(-1) + 3
2B = -4 + 3 = -1
B = -1/2
If x = - 3 ⇒ A(-3+1) = 4(-3) + 3
-2A = -12 + 3
A = 9/2 ${I_2} = \int_1^2 {\left( {\frac{{\frac{9}{2}}}{{x + 3}} + \frac{{ - \frac{1}{2}}}{{x + 1}}} \right)} dx$ $I = \frac{9}{2}\int_1^2 {\frac{1}{{x + 3}}dx} - \frac{1}{2}\int_1^2 {\frac{1}{{x + 1}}} dx$ $I = \frac{9}{2}\left[ {\log \left| {x + 3} \right|} \right]_1^2 - \frac{1}{2}\left[ {\log \left| {x + 1} \right|} \right]_1^2$ $I = \frac{9}{2}\left[ {\log \left( {2 + 3} \right) - \log \left( {1 + 3} \right)} \right] - \frac{1}{2}\left[ {\log \left( {2 + 1} \right) - \log \left( {1 + 1} \right)} \right]$ $I = \frac{9}{2}\left[ {\log 5 - \log 4} \right] - \frac{1}{2}\left[ {\log 3 - \log 2} \right]$ $I = \frac{9}{2}\log \frac{5}{4} - \frac{1}{2}\log \frac{3}{2}$ I = 5 - 5I2 $I = 5 - 5\left[ {\frac{9}{2}\log \frac{5}{4} - \frac{1}{2}\log \frac{3}{2}} \right]$ $I = 5 - \frac{{45}}{2}\log \frac{5}{4} + \frac{5}{2}\log \frac{3}{2}$ $I = 5 - \frac{5}{2}\left( {9\log \frac{5}{4} - \log \frac{3}{2}} \right)$

Question (17)

$I = \int_0^{\frac{\pi }{4}} {\left( {2{{\sec }^2}x + {x^3} + 2} \right)dx}$

Solution

$I = \int_0^{\frac{\pi }{4}} {\left( {2{{\sec }^2}x + {x^3} + 2} \right)dx}$ $I = 2\int_0^{\frac{\pi }{4}} {{{\sec }^2}x\;dx + } \int_0^{\frac{\pi }{4}} {{x^3}dx + 2\int_0^{\frac{\pi }{4}} {dx} }$ $I = 2\left[ {\tan x} \right]_0^{\frac{\pi }{4}} + \left[ {\frac{{{x^4}}}{4}} \right]_0^{\frac{\pi }{4}} + 2\left[ x \right]_0^{\frac{\pi }{4}}$ $I = \left[ {\tan \frac{\pi }{4} - \tan 0} \right] + \frac{1}{4}\left[ {{{\left( {\frac{\pi }{4}} \right)}^4} - {0^4}} \right] + 2\left[ {\frac{\pi }{4}} \right]$ $I = 2\left( {1 - 0} \right) + \frac{1}{4}\left( {\frac{{{\pi ^4}}}{{256}} - 0} \right) + \frac{\pi }{2}$ $I = 2 + \frac{{{\pi ^4}}}{{1024}} + \frac{\pi }{2}$

Question (18)

$I = \int_0^\pi {\left( {{{\sin }^2}\frac{x}{2} - {{\cos }^2}\frac{x}{2}} \right)} \;dx$

Solution

$I = \int_0^\pi { - \left( {{{\cos }^2}\frac{x}{2} - {{\sin }^2}\frac{x}{2}} \right)} dx$
Use Identity cos2x - sin2x = cos 2x
$I = - \int_0^\pi {\cos \cancel{2}\left( {\frac{x}{\cancel{2}}} \right)} \;dx$ $I = - \int_0^\pi {\cos x} \;dx$ $I = - \left[ {\sin \;x} \right]_0^\pi$ $I = - \left[ {\sin \pi - sin0} \right]$ $I = - \left( {0 - 0} \right)$ $I = 0$

Question (19)

$I = \int_0^2 {\frac{{6x + 3}}{{{x^2} + 4}}} \;dx$

Solution

$I = \int_0^2 {\frac{{6x}}{{{x^2} + 4}}\;dx + \int_0^2 {\frac{3}{{{x^2} + 4}}} } \;dx$ $I = 6\int_0^2 {\frac{x}{{{x^2} + 4}}\;} dx + 3\int_0^2 {\frac{1}{{{x^2} + 4}}} \;dx$ I = 6I1 + 3I2 ${I_1} = \int_0^2 {\frac{x}{{{x^2} + 4}}\;} dx$ Let x2 + 4 = t
2x dx = dt
x dx = dt/2
If x = 0, t = 02 + 4 = 4
x = 2, t= 22 + 4 = 8 ${I_1} = \int_4^8 {\frac{{\frac{{dt}}{2}}}{t}}$ ${I_1} = \frac{1}{2}\left[ {\log t} \right]_4^8$ ${I_1} = \frac{1}{2}\left[ {\log 8 - \log 4} \right]$ ${I_1} = \frac{1}{2}\log \frac{\cancel{8}^2}{\cancel{4}}$ ${I_1} = \frac{1}{2}\log 2$ Now ${I_2} = \int_0^2 {\frac{1}{{{x^2} + 4}}} \;dx$ ${I_2} = \int_0^2 {\frac{1}{{{x^2} + {2^2}}}\;dx}$ ${I_2} = \frac{1}{2}\left[ {{{\tan }^{ - 1}}\left( {\frac{x}{2}} \right)} \right]_0^2$ ${I_2} = \frac{1}{2}\left[ {{{\tan }^{ - 1}}\left( {\frac{2}{2}} \right) - {{\tan }^{ - 1}}\left( {\frac{0}{2}} \right)} \right]$ ${I_2} = \frac{1}{2}\left[ {{{\tan }^{ - 1}}1 - {{\tan }^{ - 1}}0} \right]$ ${I_2} = \frac{1}{2}\left( {\frac{\pi }{4} - 0} \right)$ ${I_2} = \frac{\pi }{8}$ I = 6I1 + 3I2 $I = \cancel{6}^3\left[ {\frac{1}{\cancel{2}}\log 2} \right] + 3\left( {\frac{\pi }{8}} \right)$ $I = 3\log 2 + \frac{{3\pi }}{8}$

Question (20)

$I = \int_0^1 {\left( {x{e^x} + \sin \frac{{\pi x}}{4}} \right)\;dx}$

Solution

$I = \int_0^1 {x{e^x}\;} dx + \int_0^1 {\sin } \frac{{\pi x}}{4}\;dx$ I = I1 + I2 ${I_1} = \int_0^1 {x{e^x}\;} dx$ Using integration by part ${I_1} = \left[ {x\int {{e^x}dx} } \right]_0^1 - \int_0^1 {\left( {\frac{d}{{dx}}x\int {{e^x}} dx} \right)} \;dx$ $I = \left[ {x{e^x}} \right]_0^1 - \int_0^1 {1 \cdot {e^x}} dx$ $I = \left( {1{e^1} - 0{e^0}} \right) - \left[ {{e^x}} \right]_0^1$$I = e - \left( {{e^1} - {e^0}} \right)$ $I = e - \left( {{e^1} - {e^0}} \right)$ $I = e - e + 1$ $I = 1$ Now I2 ${I_2} = \int\limits_0^1 {\sin \left( {\frac{{\pi x}}{4}} \right)} \;dx$ ${I_2} = \left[ { - \frac{{\cos \left( {\frac{{\pi x}}{4}} \right)}}{{\frac{\pi }{4}}}} \right]_0^1$ $I_2 = \frac{{ - 4}}{\pi }\left[ {\cos \frac{\pi }{4}\left( 1 \right) - \cos \left( {\frac{\pi }{4}} \right)\left( 0 \right)} \right]$ $I_2 = \frac{{ - 4}}{\pi }\left[ {\cos \frac{\pi }{4} - \cos 0} \right]$ $I_2 = \frac{{ - 4}}{\pi }\left[ {\frac{1}{{\sqrt 2 }} - 1} \right]$ $I_2 = \frac{{ - 2\sqrt 2 }}{\pi } + \frac{4}{\pi }$ I = I1 + I2 $I = 1 - \frac{{2\sqrt 2 }}{\pi } + \frac{4}{\pi }$

### Choose the correct answer in Exercise 21 and 22

Question (21)

$I = \int_1^{\sqrt 3 } {\frac{{dx}}{{1 + {x^2}}}}$ $\left( A \right)\;\quad \frac{\pi }{3}$ $\left( B \right)\;\quad \frac{{2\pi }}{3}$ $\left( C \right)\;\quad \frac{\pi }{6}$ $\left( D \right)\;\quad \frac{\pi }{{12}}$

Solution

$I = \int_1^{\sqrt 3 } {\frac{{dx}}{{1 + {x^2}}}}$ $I = \left[ {{{\tan }^{ - 1}}\;x} \right]_1^{\sqrt 3 }$ $I = {\tan ^{ - 1}}\sqrt 3 - {\tan ^{ - 1}}1$ $I = \frac{\pi }{3} - \frac{\pi }{4}$ $I = \frac{\pi }{{12}}$ 'D' is the correct option

Question (21)

$I = \int_0^{\frac{2}{3}} {\frac{{dx}}{{4 + 9{x^2}}}}\qquad \text{equals}$ $\left( A \right)\;\quad \frac{\pi }{6}$ $\left( B \right)\;\quad \frac{\pi }{{12}}$ $\left( C \right)\;\quad \frac{\pi }{{24}}$ $\left( D \right)\;\quad \frac{\pi }{4}$

Solution

$I = \int_0^{\frac{2}{3}} {\frac{{dx}}{{4 + 9{x^2}}}}$ $I = \frac{1}{9}\int_0^{\frac{2}{3}} {\frac{{dx}}{{\frac{4}{9} + {x^2}}}}$ $I = \frac{1}{9}\int_0^{\frac{2}{3}} {\frac{{dx}}{{{{\left( {\frac{2}{3}} \right)}^2} + {x^2}}}}$ $I = \frac{1}{\cancel{9}_3}\left[ {\frac{1}{{\frac{2}{\cancel{3}}}}{{\tan }^{ - 1}}\left( {\frac{x}{{\frac{2}{3}}}} \right)} \right]_0^{\frac{2}{3}}$ $I = \frac{1}{6}\left[ {{{\tan }^{ - 1}}\left( {\frac{{3x}}{2}} \right)} \right]_0^{\frac{2}{3}}$ $I = \frac{1}{6}\left[ {{{\tan }^{ - 1}}\frac{3}{2} \times \frac{2}{3} - {{\tan }^{ - 1}}\frac{3}{2} \times 0} \right]$ $I = \frac{1}{6}\left[ {{{\tan }^{ - 1}}1 - {{\tan }^{ - 1}}0} \right]$ $I = \frac{1}{6}\left( {\frac{\pi }{4} - 0} \right)$ $I = \frac{\pi }{{24}}$ So 'C' is correct option