12th NCERT INTEGRALS Exercise 7.10 Questions 22
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Evaluate the integrals in Exercise 1 to 8 using substitution

Question (1)

\[\int_0^1 {\frac{x}{{{x^2} + 1}}\;dx} \]

Solution

\[I = \int_0^1 {\frac{x}{{{x^2} + 1}}\;dx} \] Let x2 + 1 = t
2x dx = dt
x dx = dt/2
when x = 0, t= 02 + 1
x = 1, t = 12 + 1 = 1 + 1 = 2 \[I = \int_1^2 {\frac{{\frac{{dt}}{2}}}{t}} \] \[I = \frac{1}{2}\int_1^2 {\frac{{dt}}{t}} \] \[I = \frac{1}{2}\left[ {\log \left| t \right|} \right]_1^2\] \[I = \frac{1}{2}\left[ {\log 2 - \log 1} \right]\] \[I = \frac{1}{2}\left( {\log 2 - 0} \right)\] \[I = \frac{1}{2}\log 2\]

Question (2)

\[I = \int_0^{\frac{\pi }{2}} {\sqrt {\sin \phi } } {\cos ^5}\phi \;d\phi \]

Solution

\[I = \int_0^{\frac{\pi }{2}} {\sqrt {\sin \phi } } {\cos ^5}\phi \;d\phi \] Let sin Φ = t2
∴ cos Φ dΦ = 2t dt
When Φ = 0. t2 = sin 0 = 0 ⇒ 0
Φ = π/2 , t2 = sin π/2 = 1 ⇒ t = 1
\[I = \int_0^{\frac{\pi }{2}} {\sqrt {\sin \phi } } {\left( {{{\cos }^2}\phi } \right)^2}\cos \phi \;d\phi \] \[I = \int_0^{\frac{\pi }{2}} {\sqrt {\sin \phi } } {\left( {1 - {{\sin }^2}\phi } \right)^2}\cos \phi \;d\phi \] \[I = \int_0^1 {\sqrt {{t^2}} } {\left( {1 - {t^4}} \right)^2} \cdot 2t\;dt\] \[I = \int_0^1 {t\left( {1 - 2{t^4} + {t^8}} \right) \cdot 2tdt} \] \[I = 2\int_0^1 {{t^2}} \left( {1 - 2{t^4} + {t^8}} \right)\,dt\] \[I = 2\int_0^1 {\left( {{t^2} - 2{t^6} + {t^{10}}} \right)dt} \] \[I = 2\int_0^1 {{t^2}dt - 4\int_0^1 {{t^6}dt + 2\int_0^1 {{t^{10}}dt} } } \] \[I = 2\left[ {\frac{{{t^3}}}{3}} \right]_0^1 - 4\left[ {\frac{{{t^7}}}{7}} \right]_0^1 + 2\left[ {\frac{{{t^{11}}}}{{11}}} \right]_0^1\] \[I = \frac{2}{3}\left( {1 - 0} \right) - \frac{4}{7}\left( {1 - 0} \right) + \frac{2}{{11}}\left( {1 - 0} \right)\] \[I = \frac{2}{3} - \frac{4}{7} + \frac{2}{{11}}\] \[I = \frac{{154 - 132 + 42}}{{21\left( {11} \right)}}\] \[I = \frac{{64}}{{231}}\]

Question (3)

\[\int_0^1 {{{\sin }^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)\,dx} \]

Solution

\[I = \int_0^1 {{{\sin }^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)\,dx} \] Let x = tanθ
∴ dx = sec2θ dθ
When x = 0 tanθ = 0 ⇒ θ = 0
When x = 0, tanθ = 1 ⇒ θ = $\frac{\pi }{4}$
\[I = \int_0^{\frac{\pi }{4}} {{{\sin }^{ - 1}}\left( {\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}} \right)\, \cdot {{\sec }^2}\theta \,d\theta } \] \[I = \int_0^{\frac{\pi }{4}} {{{\sin }^{ - 1}}\left( {\sin 2\theta } \right)} \cdot {\sec ^2}\theta \,d\theta \] \[I = \int_0^{\frac{\pi }{4}} {{{\sin }^{ - 1}}\left( {\sin 2\theta } \right)} \cdot {\sec ^2}\theta \,d\theta \] \[I = \int_0^{\frac{\pi }{4}} {2\theta } \,{\sec ^2}\theta \;d\theta \] \[I = 2\int_0^{\frac{\pi }{4}} {\theta {{\sec }^2}\theta d\theta } \] \[I = 2\left( {\left[ {\theta \int {{{\sec }^2}\theta dx} } \right]_0^{\frac{\pi }{4}} - \int\limits_0^{\frac{\pi }{4}} {\frac{d}{{d\theta }}\theta \int {{{\sec }^2}\theta d\theta } } } \right)\] \[I = 2\left[ {\theta \tan \theta } \right]_0^{\frac{\pi }{4}} - 2\int\limits_0^{\frac{\pi }{4}} {1 \cdot \tan \theta \;d\theta } \] \[I = 2\left[ {\frac{\pi }{4}\tan \frac{\pi }{4} - 0} \right] - 2\left[ {\log \left| {\sec \theta } \right|} \right]_0^{\frac{\pi }{4}}\] \[I = 2\left( {\frac{\pi }{4}} \right) - 2\left[ {\log \left| {\sec \frac{\pi }{4}} \right| - \log \left| {\sec 0} \right|} \right]\] \[I = \frac{\pi }{2} - 2\left[ {\log \sqrt 2 - \log 1} \right]\] Note log 1 = 0 \[I = \frac{\pi }{2} - 2\log \sqrt 2 \] \[I = \frac{\pi }{2} - \require{cancel} \cancel{2} \cdot \frac{1}{\cancel{2}}\log 2\] \[I = \frac{\pi }{2} - \log 2\]

Question (4)

\[\int_0^2 {x\sqrt {x + 2} \;dx} \]

Solution

\[I = \int_0^2 {x\sqrt {x + 2} \;dx} \] Let x+2 = t2
x = t2 - 2
dx = 2tdt
When x = 0, t2= 0 + 2 = 2
t = √2
x = 2, t2 = 2 +2 =4
t = 2 \[I = \int_{\sqrt 2 }^2 {\left( {{t^2} - 2} \right) \cdot t \cdot 2tdt} \] \[I = 2\int_{\sqrt 2 }^2 {{t^2}\left( {{t^2} - 2} \right)dt} \] \[I = 2\int_{\sqrt 2 }^2 {{t^4}dt} - 4\int_{\sqrt 2 }^2 {{t^2}} dt\] \[I = \frac{2}{5}\left[ {{t^5}} \right]_{\sqrt 2 }^2 - \frac{4}{3}\left[ {{t^3}} \right]_{\sqrt 2 }^2\] \[I = \frac{2}{5}\left( {32 - 4\sqrt 2 } \right) - \frac{4}{3}\left( {8 - 2\sqrt 2 } \right)\] \[I = \frac{{64}}{5} - \frac{{8\sqrt 2 }}{5} - \frac{{32}}{3} + \frac{{8\sqrt 2 }}{3}\] \[I = 32\left( {\frac{2}{5} - \frac{1}{3}} \right) + 8\sqrt 2 \left( {\frac{1}{3} - \frac{1}{5}} \right)\] \[I = \frac{{32}}{{15}} + 8\sqrt 2 \left( {\frac{2}{{15}}} \right)\] \[I = \frac{{32}}{{15}} + \frac{{16\sqrt 2 }}{{15}}\]

Question (5)

\[\int_0^{\frac{\pi }{2}} {\frac{{\sin x}}{{1 + {{\cos }^2}x}}\;}\quad dx\]

Solution

\[I = \int_0^{\frac{\pi }{2}} {\frac{{\sin x}}{{1 + {{\cos }^2}x}}\;} dx\] Let cos x = t
∴ - sinx dx = dt
sin x dx = -dt
When x = 0, t =cos 0 =1
x = π/ 2, t = cos π/2 =0
Note \[\int\limits_a^b {f\left( x \right)dx = - \int\limits_b^a {f\left( x \right)dx} } \]
\[I = \int_1^0 {\frac{{ - dt}}{{1 + {t^2}}}} \] \[I = \left[ {{{\tan }^{ - 1}}t} \right]_0^1\] \[I = {\tan ^{ - 1}}1 - {\tan ^{ - 1}}0\] \[I = \frac{\pi }{4} - 0\] \[I = \frac{\pi }{4}\]

Question (6)

\[\int_0^2 {\frac{{dx}}{{x + 4 - {x^2}}}} \]

Solution

\[I = \int_0^2 {\frac{{dx}}{{x + 4 - {x^2}}}} \]
Denominator can be converted in a2 - x2 form by findng last term \[LT = \frac{{{{\left( {MT} \right)}^2}}}{{4 \times FT}} = \frac{{{\cancel{x^2}}}}{{4 \times {\cancel{x^2}}}} = \frac{1}{4}\]
\[I = \int_0^2 {\frac{{dx}}{{\frac{17}{4} - \left( {{x^2} - x + \frac{1}{4}} \right)}}} \] \[I = \int_0^2 {\frac{{dx}}{{{{\left( {\frac{{\sqrt {17} }}{2}} \right)}^2} - {{\left( {x - \frac{1}{2}} \right)}^2}}}} \] \[I = \frac{{ - 1}}{{2 \cdot \frac{{\sqrt {17} }}{2}}}\log \left[ {\frac{{\frac{{\sqrt {17} }}{2} - \left( {x - \frac{1}{2}} \right)}}{{\frac{{\sqrt {17} }}{2} + \left( {x - \frac{1}{2}} \right)}}} \right]_0^2\] \[ = \frac{{ - 1}}{{\sqrt {17} }}\left[ {\log \left| {\frac{{\frac{{\sqrt {17} }}{2} - \left( {2 - \frac{1}{2}} \right)}}{{\frac{{\sqrt {17} }}{2} + \left( {2 - \frac{1}{2}} \right)}}} \right| - \log \left| {\frac{{\frac{{\sqrt {17} }}{2} + \frac{1}{2}}}{{\frac{{\sqrt {17} }}{2} - \frac{1}{2}}}} \right|} \right]\] \[ = \frac{{ - 1}}{{\sqrt {17} }}\left[ {\log \left| {\frac{{\sqrt {17} - 3}}{{\sqrt {17} + 3}}} \right| - \log \left| {\frac{{\sqrt {17} + 1}}{{\sqrt {17} - 1}}} \right|} \right]\] \[ = \frac{{ - 1}}{{\sqrt {17} }}\log \left| {\frac{{\sqrt {17} - 3}}{{\sqrt {17} + 3}} \times \frac{{\sqrt {17} - 1}}{{\sqrt {17} + 1}}} \right|\] \[ = \frac{{ - 1}}{{\sqrt {17} }}\log \left| {\frac{{17 - 4\sqrt {17} + 3}}{{17 + 4\sqrt {17} + 3}}} \right|\] \[ = \frac{{ - 1}}{{\sqrt {17} }}\log \left| {\frac{{20 - 4\sqrt {17} }}{{20 + 4\sqrt {17} }}} \right|\] \[ = \frac{{ - 1}}{{\sqrt {17} }}\log \left| {\frac{{\cancel{4}\left( {5 - \sqrt {17} } \right)}}{{\cancel{4}\left( {5 + \sqrt {17} } \right)}}} \right|\] \[ = \frac{1}{{\sqrt {17} }}\log \left| {\frac{{5 + \sqrt {17} }}{{5 - \sqrt {17} }}} \right|\] \[ = \frac{1}{{\sqrt {17} }}\log \left| {\frac{{\left( {5 + \sqrt {17} } \right)\left( {5 + \sqrt {17} } \right)}}{{\left( {5 - \sqrt {17} } \right)\left( {5 + \sqrt {17} } \right)}}} \right|\] \[ = \frac{1}{{\sqrt {17} }}\log \left| {\frac{{25 + 17 + 10\sqrt {17} }}{{25 - 17}}} \right|\] \[ = \frac{1}{{\sqrt {17} }}\log \left| {\frac{{42 + 10\sqrt {17} }}{8}} \right|\] \[ = \frac{1}{{\sqrt {17} }}\log \left| {\frac{{\cancel{2}\left( {21 + 5\sqrt {17} } \right)}}{\cancel{8}_4}} \right|\] \[ = \frac{1}{{\sqrt {17} }}\log \left| {\frac{{21 + 5\sqrt {17} }}{4}} \right|\]

Question (7)

\[\int_{ - 1}^1 {\frac{1}{{{x^2} + 2x + 5}}} \;dx\]

Solution

\[I = \int_{ - 1}^1 {\frac{1}{{{x^2} + 2x + 5}}} \;dx\] \[I = \int_{ - 1}^1 {\frac{1}{{{x^2} + 2x + 1 + 4}}} \;dx\] \[I = \int_{ - 1}^1 {\frac{1}{{{{\left( {x + 1} \right)}^2} + {2^2}}}} \;dx\] \[I = \frac{1}{2}\left[ {{{\tan }^{ - 1}}\left( {\frac{{x + 1}}{2}} \right)} \right]_{ - 1}^1\] \[I = \frac{1}{2}\left[ {{{\tan }^{ - 1}}\left( {\frac{2}{2}} \right) - {{\tan }^{ - 1}}\left( {\frac{0}{2}} \right)} \right]\] \[I = \frac{1}{2}\left[ {{{\tan }^{ - 1}}\left( 1 \right) - {{\tan }^{ - 1}}\left( 0 \right)} \right]\] \[I = \frac{1}{2}\left[ {\frac{\pi }{4} - 0} \right]\] \[I = \frac{\pi }{8}\]

Question (8)

\[\int_1^2 {\left( {\frac{1}{x} - \frac{1}{{2{x^2}}}} \right){e^{2x}}\quad dx} \]

Solution

\[I = \int_1^2 {\left( {\frac{1}{x} - \frac{1}{{2{x^2}}}} \right){e^{2x}}\quad dx} \] Let 2x = t ∴ x = t/2
∴ dx = dt/2
Where x = 1, t = 2(1) = 2
x = 2, t = 2(2) = 4 \[I = \int_2^4 {\left[ {\frac{1}{{\frac{t}{2}}} - \frac{1}{{2{{\left( {\frac{t}{2}} \right)}^2}}}} \right]\quad } {e^t} \cdot \frac{{dt}}{2}\] \[I = \frac{1}{2}\int\limits_2^4 {\left[ {\frac{2}{t} - \frac{4}{{2{t^2}}}} \right]} \quad {e^t}\quad dt\] \[I = \frac{1}{2}\int\limits_2^4 {\left[ {\frac{2}{t} - \frac{2}{{{t^2}}}} \right]} \quad {e^t}\quad dt\] \[I = \frac{1}{\cancel{2}} \cdot \cancel{2}\int\limits_2^4 {\left[ {\frac{1}{t} - \frac{1}{{{t^2}}}} \right]} \quad {e^t}\quad dt\] \[let\;f\left( t \right) = \frac{1}{t}\] \[f'\left( t \right) = \frac{{ - 1}}{{{t^2}}}\] \[I = \int_2^4 {{e^t}\left[ {f\left( t \right) + f'\left( t \right)} \right]} \quad dt\] \[I = \left[ {{e^t}f\left( t \right)} \right]_2^4\] substitute a value of f(x) \[I = \left[ {{e^t}\frac{1}{t}} \right]_2^4\] \[I = \frac{{{e^4}}}{4} - \frac{{{e^2}}}{2}\] \[I = \frac{{{e^2}}}{2}\left( {\frac{{{e^2}}}{2} - 1} \right)\] \[I = \frac{{{e^2}}}{4}\left( {{e^2} - 2} \right)\] \[I = \frac{{{e^2}\left( {{e^2} - 2} \right)}}{4}\]

Choose the correct answer in Exercise 9 and 10

Question (9)

The value of the integral \[\int_{\frac{1}{3}}^1 {\frac{{{{\left( {x - {x^3}} \right)}^{\frac{1}{3}}}}}{{{x^4}}}dx} \] (A) 6    (B) 0    (C) 3    (D) 4

Solution

\[I = \int_{\frac{1}{3}}^1 {\frac{{{{\left( {x - {x^3}} \right)}^{\frac{1}{3}}}}}{{{x^4}}}dx} \] \[I = \int_{\frac{1}{3}}^1 {\frac{1}{{{x^4}}}{{\left( {\frac{1}{{{x^2}}} - 1} \right)}^{\frac{1}{3}}}x\quad dx} \] \[I = \int_{\frac{1}{3}}^1 {{{\left( {\frac{1}{{{x^2}}} - 1} \right)}^{\frac{1}{3}}}dx} \quad \frac{1}{{{x^3}}}\] \[let\quad \frac{1}{{{x^2}}} - 1 = {t^3}\] \[threfore\quad \frac{{ - 2}}{{{x^3}}}dx = dt \cdot 3{t^2}\] \[threfore\quad\frac{{dx}}{{{x^3}}} = - \frac{{dt}}{2} \cdot 3{t^2}\] \[when\quad x = \frac{1}{3}\] \[{t^3} = \frac{1}{{\frac{1}{9}}} - 1 = 8 \Rightarrow t = 2\] \[when\quad x = 1\] \[{t^3} = \frac{1}{1} - 1 = 0 \Rightarrow t = 0\] \[I = \int_2^0 {\frac{{ - 3}}{2}{t^2}dt \cdot t} \] \[I = \frac{3}{2}\int_0^2 {{t^3}dt} \] \[I = \frac{3}{2}\left[ {\frac{{{t^4}}}{4}} \right]_0^2\] \[I = \frac{3}{8}\left( {16 - 0} \right)\] \[I = \frac{3}{\cancel{8}} \times \cancel{16}\] \[I = 6\] Option (A) correct

Question (10)

\[If\quad f\left( x \right) = \int_0^x {t\sin t\;dt}, \text{then f'(x) is} \] (A) cos x + x sin x
(B) x sin x
(C) x cos x
(D) sin x + x cos x

Solution

\[f'\left( x \right) = \frac{d}{{dx}}f\left( x \right)\] \[f'\left( x \right) = \frac{d}{{dx}}\left[ {\int_0^x {tsintdt} } \right]\] \[f'\left( x \right) = \left[ {t\sin t} \right]_0^x\] \[f'\left( x \right) = x\sin x\] So (B) is correct option
Exercise 7.9 ⇐
⇒ Exercise 7.11