# NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.7

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## Chapter 7 Ex.7.7 Question 1

Integrate $$\sqrt {4 - {x^2}}$$

### Solution

Let $$I = \int {\sqrt {4 - {x^2}} dx = \int {\sqrt {{{\left( 2 \right)}^2} - {{\left( x \right)}^2}} } dx}$$

Since, $$\sqrt {{a^2} - {x^2}} dx = \frac{x}{2}\sqrt {{a^2} - {x^2}} + \frac{{{a^2}}}{2}{\sin ^{ - 1}}\frac{x}{a} + C$$

\begin{align}\therefore I &= \frac{x}{2}\sqrt {4 - {x^2}} + \frac{4}{2}{\sin ^{ - 1}}\frac{x}{2} + C\\&= \frac{x}{2}\sqrt {4 - {x^2}} + 2{\sin ^{ - 1}}\frac{x}{2} + C\end{align}

## Chapter 7 Ex.7.7 Question 2

Integrate $$\sqrt {1 - 4{x^2}}$$

### Solution

\begin{align}{\text{Let}},\;I &= \int {\sqrt {1 - 4{x^2}} dx} = \int {\sqrt {{{\left( 1 \right)}^2} - {{\left( {2x} \right)}^2}} } dx\\{\text{Put, }}2x &= t \Rightarrow 2dx = dt\\\therefore I &= \frac{1}{2}\int {\sqrt {{{\left( 1 \right)}^2} - {{\left( t \right)}^2}} }\end{align}

Since, $$\sqrt {{a^2} - {x^2}} dx = \frac{x}{2}\sqrt {{a^2} - {x^2}} + \frac{{{a^2}}}{2}{\sin ^{ - 1}}\frac{x}{a} + C$$

\begin{align}&\Rightarrow \; I = \frac{1}{2}\left[ {\frac{t}{2}\sqrt {1 - {t^2}} + \frac{1}{2}{{\sin }^{ - 1}}t} \right] + C\\&= \frac{t}{4}\sqrt {1 - {t^2}} + \frac{1}{4}{\sin ^{ - 1}}t + C\\&= \frac{{2x}}{4}\sqrt {1 - 4{x^2}} + \frac{1}{4}{\sin ^{ - 1}}2x + C\\&= \frac{x}{2}\sqrt {1 - 4{x^2}} + \frac{1}{4}{\sin ^{ - 1}}2x + C\end{align}

## Chapter 7 Ex.7.7 Question 3

Integrate $$\sqrt {{x^2} + 4x + 6}$$

### Solution

\begin{align}{\text{Let}}\;I & = \int {\sqrt {{x^2} + 4x + 6} } dx\\&= \int {\sqrt {{x^2} + 4x + 4 + 2} } dx\\&= \int {\sqrt {\left( {{x^2} + 4x + 4} \right) + 2} } dx\\&= \int {\sqrt {{{\left( {x + 2} \right)}^2} + {{\left( {\sqrt 2 } \right)}^2}} } dx\\{\text{Since, }}\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{{\left( {x + 2} \right)}}{2}\sqrt {{x^2} + 4x + 6} + \frac{2}{2}\log \left| {\left( {x + 2} \right) + \sqrt {{x^2} + 4x + 6} } \right| + C\\&= \frac{{\left( {x + 2} \right)}}{2}\sqrt {{x^2} + 4x + 6} + \log \left| {\left( {x + 2} \right) + \sqrt {{x^2} + 4x + 6} } \right| + C\end{align}

## Chapter 7 Ex.7.7 Question 4

Integrate $$\sqrt {{x^2} + 4x + 1}$$

### Solution

Consider,

\begin{align}I &= \int {\sqrt {{x^2} + 4x + 1} } dx\\&= \int {\sqrt {\left( {{x^2} + 4x + 4} \right) - 3} } dx\\&= \int {\sqrt {{{\left( {x + 2} \right)}^2} - {{\left( {\sqrt 3 } \right)}^2}} } dx\\{\text{Since, }}\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\\\therefore I &= \frac{{\left( {x + 2} \right)}}{2}\sqrt {{x^2} + 4x + 1} - \frac{3}{2}\log \left| {\left( {x - 2} \right) + \sqrt {{x^2} + 4x + 1} } \right| + C\end{align}

## Chapter 7 Ex.7.7 Question 5

Integrate $$\sqrt {1 - 4x - {x^2}}$$

### Solution

\begin{align}{\text{Consider, }}I &= \int {\sqrt {1 - 4x - {x^2}} } dx\\&= \int {\sqrt {1 - \left( {{x^2} + 4x + 4 - 4} \right)} } dx\\&= \int {\sqrt {1 + 4 - {{\left( {x + 2} \right)}^2}} } dx\\&= \int {\sqrt {{{\left( {\sqrt 5 } \right)}^2} - {{\left( {x + 2} \right)}^2}} } dx\\{\text{Since, }}\sqrt {{a^2} - {x^2}} dx& = \frac{x}{2}\sqrt {{a^2} - {x^2}} + \frac{{{a^2}}}{2}{\sin ^{ - 1}}\frac{x}{a} + C\\\therefore I &= \frac{{\left( {x + 2} \right)}}{2}\sqrt {1 - 4x - {x^2}} + \frac{5}{2}{\sin ^{ - 1}}\left( {\frac{{x + 2}}{{\sqrt 5 }}} \right) + C\end{align}

## Chapter 7 Ex.7.7 Question 6

Integrate $$\sqrt {{x^2} + 4x - 5}$$

### Solution

\begin{align}{\text{Let}}\;I &= \int {\sqrt {{x^2} + 4x - 5} } dx\\&= \int {\sqrt {\left( {{x^2} + 4x + 4} \right) - 9} } dx = \int {\sqrt {{{\left( {x + 2} \right)}^2} - {{\left( 3 \right)}^2}} dx} \\{\text{Since, }}\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\\\therefore I &= \frac{{\left( {x + 2} \right)}}{2}\sqrt {{x^2} + 4x - 5} - \frac{9}{2}\log \left| {\left( {x + 2} \right) + \sqrt {{x^2} + 4x - 5} } \right| + C\end{align}

## Chapter 7 Ex.7.7 Question 7

Integrate $$\sqrt {1 + 3x - {x^2}}$$

### Solution

Put, $$I = \int {\sqrt {1 + 3x - {x^2}} } dx$$

\begin{align}&= \int {\sqrt {1 - \left( {{x^2} - 3x + \frac{9}{4} - \frac{9}{4}} \right)} } dx\\&= \int {\sqrt {\left( {1 + \frac{9}{4}} \right) - {{\left( {x - \frac{3}{2}} \right)}^2}} } dx = \int {\sqrt {{{\left( {\frac{{\sqrt {13} }}{2}} \right)}^2} - {{\left( {x - \frac{3}{2}} \right)}^2}} } dx\end{align}

\begin{align}{\text{Since, }}\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\\\therefore I &= \frac{{x - \frac{3}{2}}}{2}\sqrt {1 + 3x - {x^2}} + \frac{{13}}{{4 \times 2}}{\sin ^{ - 1}}\left( {\frac{{x - \frac{3}{2}}}{{\frac{{\sqrt {13} }}{2}}}} \right) + C\\&= \frac{{2x - 3}}{4}\sqrt {1 + 3x - {x^2}} + \frac{{13}}{8}{\sin ^{ - 1}}\left( {\frac{{2x - 3}}{{\sqrt {13} }}} \right) + C\end{align}

## Chapter 7 Ex.7.7 Question 8

Integrate $$\sqrt {{x^2} + 3x}$$

### Solution

\begin{align}{\text{Let}}\;I &= \int {\sqrt {{x^2} + 3x} } dx\\&= \int {\sqrt {{x^2} + 3x + \frac{9}{4} - \frac{9}{4}} } dx\\&= \int {\sqrt {{{\left( {x + \frac{3}{2}} \right)}^2} - {{\left( {\frac{3}{2}} \right)}^2}} } dx\\{\text{Since, }}\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\\\therefore I &= \frac{{\left( {x + \frac{3}{2}} \right)}}{2}\sqrt {{x^2} + 3x} - \frac{{\frac{9}{4}}}{2}\log \left| {\left( {x + \frac{3}{2}} \right) + \sqrt {{x^2} + 3x} } \right| + C\\&= \frac{{\left( {2x + 3} \right)}}{4}\sqrt {{x^2} + 3x} - \frac{9}{8}\log \left| {\left( {x + \frac{3}{2}} \right) + \sqrt {{x^2} + 3x} } \right| + C\end{align}

## Chapter 7 Ex.7.7 Question 9

Integrate $$\sqrt {1 + \frac{{{x^2}}}{9}}$$

### Solution

\begin{align}{\text{Let}}\;I &= \int {\sqrt {1 + \frac{{{x^2}}}{9}} } dx = \frac{1}{3}\int {\sqrt {9 + {x^2}} } dx = \frac{1}{3}\int {\sqrt {{{\left( 3 \right)}^2} + {x^2}dx} } \\{\text{Since, }}\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\\\therefore I &= \frac{1}{3}\left[ {\frac{x}{2}\sqrt {{x^2} + 9} + \frac{9}{2}\log \left| {x + \sqrt {{x^2} + 9} } \right|} \right] + C\\&= \frac{x}{6}\sqrt {{x^2} + 9} + \frac{3}{2}\log \left| {x + \sqrt {{x^2} + 9} } \right| + C\end{align}

## Chapter 7 Ex.7.7 Question 10

$$\int {\sqrt {1 + {x^2}} }$$is equal to

\begin{align}&A.\,\,\,\frac{x}{2}\sqrt {1 + {x^2}} + \frac{1}{2}\log \left| {x + \sqrt {1 + {x^2}} } \right| + C\\&B.\,\,\,\frac{2}{3}{\left( {1 + {x^2}} \right)^{\frac{2}{3}}} + C\\&C.\,\,\,\frac{2}{3}x{\left( {1 + {x^2}} \right)^{\frac{2}{3}}} + C\\&D.\,\,\,\frac{{{x^3}}}{2}\sqrt {1 + {x^2}} + \frac{1}{2}{x^2}\log \left| {x + \sqrt {1 + {x^2}} } \right| + C\end{align}

### Solution

\begin{align}&{\text{Since}}{\text{. }}\sqrt {{a^2} + {x^2}} dx = \frac{x}{2}\sqrt {{a^2} + {x^2}} + \frac{{{a^2}}}{2}\log \left| {x + \sqrt {{x^2} + {a^2}} } \right| + C\\&\therefore \int {\sqrt {1 + {x^2}} dx = \frac{x}{2}} \sqrt {1 + {x^2}} + \frac{1}{2}\log \left| {x + \sqrt {1 + {x^2}} } \right| + C\end{align}

Thus, the correct option is A.

## Chapter 7 Ex.7.7 Question 11

$$\int {\sqrt {{x^2} - 8x + 7} } dx$$ is equal to

\begin{align}&A.\,\,\,\frac{1}{2}\left( {x - 4} \right)\sqrt {{x^2} - 8x + 7} + 9\log \left| {x - 4 + \sqrt {{x^2} - 8x + 7} } \right| + C\\&B.\,\,\,\frac{1}{2}\left( {x + 4} \right)\sqrt {{x^2} - 8x + 7} + 9\log \left| {x + 4 + \sqrt {{x^2} - 8x + 7} } \right| + C\\&C.\,\,\,\frac{1}{2}\left( {x - 4} \right)\sqrt {{x^2} - 8x + 7} - 3\sqrt 2 \log \left| {x - 4 + \sqrt {{x^2} - 8x + 7} } \right| + C\\&D.\,\,\,\frac{1}{2}\left( {x - 4} \right)\sqrt {{x^2} - 8x + 7} - \frac{9}{2}\log \left| {x - 4 + \sqrt {{x^2} - 8x + 7} } \right| + C\end{align}

### Solution

\begin{align}{\text{Let}}\;I &= \int {\sqrt {{x^2} - 8x + 7} } dx\\&= \int {\sqrt {\left( {{x^2} - 8x + 16} \right) - 9} } dx\\&= \int {\sqrt {{{\left( {x - 4} \right)}^2} - {{\left( 3 \right)}^2}} dx} \\{\text{Since, }}\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\\\therefore I &= \frac{{\left( {x - 4} \right)}}{2}\sqrt {{x^2} - 8x + 7} - \frac{9}{2}\log \left| {\left( {x - 4} \right) + \int {\sqrt {{x^2} - 8x + 7} } } \right| + C\end{align}

Thus, the correct option is D.

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