12th NCERT INTEGRALS Exercise 7.7 Questions 11
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Special formulas \[\int {\sqrt {{x^2} - {a^2}} \;dx} = \frac{x}{2}\sqrt {{x^2} - {a^2}} - \frac{{{a^2}}}{2}\log \left| {x + \sqrt {{x^2} - {a^2}} } \right| + c\] \[\int {\sqrt {{x^2} + {a^2}} \;dx} = \frac{x}{2}\sqrt {{x^2} + {a^2}} + \frac{{{a^2}}}{2}\log \left| {x + \sqrt {{x^2} + {a^2}} } \right| + c\] \[\int {\sqrt {{a^2} - {x^2}} \;dx} = \frac{x}{2}\sqrt {{a^2} - {x^2}} + \frac{{{a^2}}}{2}{\sin ^{ - 1}}\left( {\frac{x}{a}} \right) + c\]

Integrate the rational functions in Exercise 1 to 9

Question (1)

\[\sqrt {4 - {x^2}} \]

Solution

\[I = \int {\sqrt {4 - {x^2}} dx} \] \[I = \int {\sqrt {{{\left( 2 \right)}^2} - {x^2}} } dx\]
Usinge formula \[\int {\sqrt {{a^2} - {x^2}} dx = \frac{x}{2}} \sqrt {{a^2} - {x^2}} + \frac{{{a^2}}}{2}{\sin ^{ - 1}}\frac{x}{a} + c\]
\[I = \frac{x}{2}\sqrt {{2^2} - {x^2}} + \frac{{{2^2}}}{2}{\sin ^{ - 1}}\frac{x}{2} + c\] \[I = \frac{x}{2}\sqrt {{2^2} - {x^2}} + 2{\sin ^{ - 1}}\frac{x}{2} + c\]

Question (2)

\[\sqrt {1 - 4{x^2}} \]

Solution

\[I = \int {\sqrt {1 - 4{x^2}} dx} \]
In the equation there is s acoefficient of x2. We have to bring it to "1". We will take 4 commom
\[I = 2\int {\sqrt {\frac{1}{4} - {x^2}} } dx\]
Usinge formula \[\int {\sqrt {{a^2} - {x^2}} dx = \frac{x}{2}} \sqrt {{a^2} - {x^2}} + \frac{{{a^2}}}{2}{\sin ^{ - 1}}\frac{x}{a} + c\]
\[I = 2\left[ {\frac{x}{2}\sqrt {\frac{1}{4} - {x^2}} + \frac{{\frac{1}{4}}}{2}{{\sin }^{ - 1}}\left( {\frac{x}{{\frac{1}{2}}}} \right)} \right] + c\] \[I = 2\left[ {\frac{x}{2}\frac{{\sqrt {1 - 4{x^2}} }}{2} + \frac{1}{8}{{\sin }^{ - 1}}\left( {2x} \right)} \right] + c\] \[I = x\frac{{\sqrt {1 - 4{x^2}} }}{2} + \frac{1}{4}{\sin ^{ - 1}}\left( {2x} \right) + c\]

Question (3)

\[\sqrt {{x^2} + 4x + 6} \]

Solution

\[I = \int {\sqrt {{x^2} + 4x + 6} } \] \[I = \int {\sqrt {{{\left( {x + 2} \right)}^2} + {{\left( {\sqrt 2 } \right)}^2}} } dx\]
Usinge formula \[\int {\sqrt {{x^2} + {a^2}} dx = \frac{x}{2}} \sqrt {{x^2} + {a^2}} + \frac{{{a^2}}}{2}\log \left| {x + \sqrt {{x^2} + {a^2}} } \right| + c\]
\[I = \frac{{x + 2}}{2}\sqrt {{{\left( {x + 2} \right)}^2} + {{\left( {\sqrt 2 } \right)}^2}} + \frac{2}{2}\log \left| {x + 2 + \sqrt {{{\left( {x + 2} \right)}^2} + {{\left( {\sqrt 2 } \right)}^2}} } \right| + c\] \[I = \frac{{x + 2}}{2}\sqrt {{x^2} + 4x + 6} + \log \left| {x + 2 + \sqrt {{x^2} + 4x + 6} } \right| + c\]

Question (4)

\[\sqrt {{x^2} + 4x + 1} \]

Solution

\[I = \int {\sqrt {{x^2} + 4x + 1} } dx\] \[I = \int {\sqrt {{x^2} + 4x + 4 - 3} } \;dx\] \[I = \int {\sqrt {{{\left( {x + 2} \right)}^2} - {{\left( {\sqrt 3 } \right)}^2}} } \;dx\]
Usinge formula \[\int {\sqrt {{x^2} - {a^2}} \;dx} = \frac{x}{2}\sqrt {{x^2} - {a^2}} - \frac{{{a^2}}}{2}\log \left| {x + \sqrt {{x^2} - {a^2}} } \right| + c\]
\[I = \frac{{x + 2}}{2}\sqrt {{{\left( {x + 2} \right)}^2} - {{\left( {\sqrt 3 } \right)}^2}} - \frac{3}{2}\log \left| {x + 2 + \sqrt {{{\left( {x + 2} \right)}^2} - {{\left( {\sqrt 3 } \right)}^2}} } \right| + c\] \[I = \frac{{x + 2}}{2}\sqrt {{x^2} + 4x + 1} - \frac{3}{2}\log \left| {x + 2 + \sqrt {{x^2} + 4x + 1} } \right| + c\]

Question (5)

\[\sqrt {1 - 4x - {x^2}} \]

Solution

\[I = \int {\sqrt {1 - 4x - {x^2}} } \;dx\] \[I = \int {\sqrt {5 - \left( {{x^2} + 4x + 4} \right)} } dx\] \[I = \int {\sqrt {{{\left( {\sqrt 5 } \right)}^2} - {{\left( {x + 2} \right)}^2}} dx} \]
Usinge formula \[\int {\sqrt {{a^2} - {x^2}} dx = \frac{x}{2}} \sqrt {{a^2} - {x^2}} + \frac{{{a^2}}}{2}{\sin ^{ - 1}}\frac{x}{a} + c\]
\[I = \frac{{x + 2}}{2}\sqrt {{{\left( {\sqrt 5 } \right)}^2} - {{\left( {x + 2} \right)}^2}} + \frac{5}{2}{\sin ^{ - 1}}\left( {\frac{{x + 2}}{{\sqrt 5 }}} \right) + c\] \[I = \frac{{x + 2}}{2}\sqrt {1 - 4x - {x^2}} + \frac{5}{2}{\sin ^{ - 1}}\left( {\frac{{x + 2}}{{\sqrt 5 }}} \right) + c\]

Question (6)

\[\sqrt {{x^2} + 4x - 5} \]

Solution

\[I = \int {\sqrt {{x^2} + 4x - 5} \;dx} \] \[I = \int {\sqrt {{x^2} + 4x + 4 - 9} \;dx} \] \[I = \int {\sqrt {{{\left( {x + 2} \right)}^2} - {{\left( 3 \right)}^2}} } \;dx\]
Usinge formula \[\int {\sqrt {{x^2} - {a^2}} \;dx} = \frac{x}{2}\sqrt {{x^2} - {a^2}} - \frac{{{a^2}}}{2}\log \left| {x + \sqrt {{x^2} - {a^2}} } \right| + c\]
\[I = \frac{{x + 2}}{2}\sqrt {{x^2} + 4x - 5} - \frac{9}{2}\log \left| {x + 2 + \sqrt {{x^2} + 4x - 5} } \right| + c\]

Question (7)

\[\sqrt {1 + 3x - {x^2}} \]

Solution

\[I = \int {\sqrt {1 + 3x - {x^2}} \;dx} \]
given equation is not in the form a2 - x2 use following formula
\[LT = \frac{{{{\left( {MT} \right)}^2}}}{{4 \times \left( {FT} \right)}}\] \[LT = \frac{{9{x^2}}}{{4{x^2}}} = \frac{9}{4}\]
\[I = \int {\sqrt {\frac{{13}}{4} - \left( {{x^2} - 3x + \frac{9}{4}} \right)} } \;dx\] \[I = \int {\sqrt {{{\left( {\frac{{\sqrt {13} }}{2}} \right)}^2} - {{\left( {x - \frac{3}{2}} \right)}^2}} } \;dx\]
Usinge formula \[\int {\sqrt {{a^2} - {x^2}} dx = \frac{x}{2}} \sqrt {{a^2} - {x^2}} + \frac{{{a^2}}}{2}{\sin ^{ - 1}}\frac{x}{a} + c\]
\[I = \frac{{2x - 3}}{4}\sqrt {1 + 3x - {x^2}} + \frac{{13}}{8}{\sin ^{ = 1}}\left( {\frac{{2x - 3}}{{\sqrt {13} }}} \right) + c\]

Question (8)

\[\sqrt {{x^2} + 3x} \]

Solution

\[I = \int {\sqrt {{x^2} + 3x} } \;dx\]
given equation is not in the form x2 - x2 use following formula
\[LT = \frac{{{{\left( {MT} \right)}^2}}}{{4 \times \left( {FT} \right)}}\] \[LT = \frac{{9{x^2}}}{{4{x^2}}} = \frac{9}{4}\]
\[I = \int {\sqrt {{x^2} + 3x + \frac{9}{4} - \frac{9}{4}} } \;dx\] \[I = \int {\sqrt {{{\left( {x + \frac{3}{2}} \right)}^2} - {{\left( {\frac{3}{2}} \right)}^2}} } \;dx\]
Usinge formula \[\int {\sqrt {{x^2} - {a^2}} \;dx} = \frac{x}{2}\sqrt {{x^2} - {a^2}} - \frac{{{a^2}}}{2}\log \left| {x + \sqrt {{x^2} - {a^2}} } \right| + c\]
\[I = \frac{{x + \frac{3}{2}}}{2}\sqrt {{{\left( {x + \frac{3}{2}} \right)}^2} - {{\left( {\frac{3}{2}} \right)}^2}} \; - \frac{9}{{4 \times 2}}\log \left| {x + \frac{3}{2} + \sqrt {{x^2} + 3x} } \right| + c\]\[I = \frac{{2x + 3}}{4}\sqrt {{x^2} + 3x} - \frac{9}{8}\log \left| {\frac{{2x + 3}}{2} + \sqrt {{x^2} + 3x} } \right| + c\]

Question (9)

\[\sqrt {1 + \frac{{{x^2}}}{9}} \]

Solution

\[I = \int {\sqrt {1 + \frac{{{x^2}}}{9}} } \;dx\] \[I = \int {\sqrt {\frac{{9 + {x^2}}}{9}} \;dx} \] \[I = \frac{1}{3}\int {\sqrt {{3^2} + {x^2}} } dx\]
Usinge formula \[\int {\sqrt {{x^2} + {a^2}} dx = \frac{x}{2}} \sqrt {{x^2} + {a^2}} + \frac{{{a^2}}}{2}\log \left| {x + \sqrt {{x^2} + {a^2}} } \right| + c\]
\[I = \frac{1}{3}\left[ {\frac{x}{2}\sqrt {9 + {x^2}} + \frac{9}{2}\log \left| {x + \sqrt {9 + {x^2}} } \right|} \right] + c\] \[I = \frac{x}{6}\sqrt {9 + {x^2}} + \frac{3}{2}\log \left| {x + \sqrt {9 + {x^2}} } \right| + c\]

Choose the correct answer in Exercise 10 to 11

Question (10)

\[\int {\sqrt {1 + {x^2}} dx} \quad \text{is equal to}\] Options
\[(A)\quad \frac{x}{2}\sqrt {1 + {x^2}} + \frac{1}{2}\log \left| {\left( {x + \sqrt {1 + x} } \right)} \right| + c\] \[(B)\quad \frac{2}{3}{\left( {1 + {x^2}} \right)^{\frac{3}{2}}} + c\] \[(C)\quad \frac{2}{3}x{\left( {1 + {x^2}} \right)^{\frac{3}{2}}} + c\] \[(D)\quad \frac{{{x^2}}}{2}\sqrt {1 + {x^2}} + \frac{1}{2}{x^2}\log \left| {x + \sqrt {1 + {x^2}} } \right| + c\]

Solution

\[I = \frac{x}{2}\sqrt {1 + {x^2}} + \frac{1}{2}\log \left| {x + \sqrt {1 + {x^2}} } \right| + c\] Option (A) correct

Question (11)

Solution

\[I = \int {\sqrt {{x^2} - 8x + 7} \;dx} \] \[I = \int {\sqrt {{x^2} - 8x + 16 - 9} \;dx} \] \[I = \int {\sqrt {{{\left( {x - 4} \right)}^2} - {3^2}} \;dx} \] \[I = \frac{{x - 4}}{2}\sqrt {{{\left( {x - 4} \right)}^2} - {3^2}} - \frac{9}{2}\log \left| {x - 4 + \sqrt {{{\left( {x - 4} \right)}^2} - {3^2}} } \right| + c\] \[I = \frac{{x - 4}}{2}\sqrt {{x^2} - 8x + 7} - \frac{9}{2}\log \left| {x - 4 + \sqrt {{x^2} - 8x + 7} } \right| + c\] Option (D) is correct
Exercise 7.6 ⇐
⇒ Exercise 7.8