# NCERT Solutions for Class 12 Maths Chapter 7 - Integrals - Exercise 7.7

Exercise 7.7

Direction (Q. 1 to 9) : Integrate the function.

$$\textbf{1.}\space\sqrt{\textbf{4-x}^\textbf{2}}\\\textbf{Sol.}\space\text{Let}\space\text{I}=\int\sqrt{4-x^2}dx\\=\int\sqrt{2^2-x^2}dx\\=\frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}\text{sin}^{\normalsize-1}\frac{x}{2}+\text{C}\\\begin{bmatrix}\because\space\int\sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\\\frac{a^2}{2}\text{sin}^{\normalsize-1}\frac{x}{2}+\text{C}\end{bmatrix}\\\Rarr\space\text{I}=\frac{x}{2}\sqrt{4-x^2}+ 2\text{sin}^{\normalsize-1}\frac{x}{2}+\text{C}$$

$$\textbf{2.\space}\sqrt{\textbf{1-4x}^\textbf{2}}\\\textbf{Sol.}\space\text{Let}\space\text{I}=\int\sqrt{1-4x^2}dx\\=\int\sqrt{4\bigg(\frac{1}{4}-x^2\bigg)dx}\\=2\int\sqrt{\bigg(\bigg(\frac{1}{2}\bigg)^2-x^2\bigg)}dx\\\begin{bmatrix}\because\space \int\sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2} + \\\frac{a^2}{2}\text{sin}^{\normalsize-1}\frac{x}{2}+\text{C}\end{bmatrix}\\=2.\frac{x}{2}\sqrt{\frac{1}{4}-x^2}+\frac{1}{4}.\frac{2}{2}\text{sin}^{\normalsize-1}\frac{x}{1/2}+\text{C}\\\Rarr\space\text{I}=\frac{x}{2}\sqrt{1-4x^2}+\frac{1}{4}\text{sin}^{\normalsize-1}(2x)+\text{C}$$

$$\textbf{3.}\space\sqrt{\textbf{x}^\textbf{2}\textbf{+4x+6}}\\\textbf{Sol.}\space\text{Let I}=\int\sqrt{x^2+4x+6}dx\\=\int\sqrt{x^2+4x+2^2+6-4}dx\\=\int\sqrt{(x+2)^2 + (\sqrt{2})^2}dx\\\begin{bmatrix}\because\space \int\sqrt{x^2+a^2}dx=\frac{x}{2}\sqrt{x^2+a^2}+\\\frac{a^2}{2}\text{log}|x + \sqrt{x^2+a^2}|\end{bmatrix}\\=\frac{x+2}{2}\sqrt{x^2+4x+6}+\frac{2}{2}\text{log}|(x+2)+\\\sqrt{x^2+4x+6}|+\text{C}\\\Rarr\space\text{I}=\frac{x+2}{2}\sqrt{x^2+4x+6}+\\\text{log}|(x+2)|+ \sqrt{x^2+4x+6}+\text{C}$$

$$\textbf{4.\space}\sqrt{\textbf{x}^\textbf{2}\textbf{+4x+1}}\\\textbf{Sol.}\space \text{Let}\space\text{I}=\int\sqrt{x^2+4x+1}dx\\=\int\sqrt{x^2+4x+1-2^2+2^2}dx\\=\int\sqrt{(x+2)^2-(\sqrt{3})^2}dx\\\begin{bmatrix}\because\space \int\sqrt{x^2-a^2}dx=\frac{x}{2}\sqrt{x^2-a^2}-\\\frac{a^2}{2}\text{log}|x+\sqrt{x^2-a^2}|\end{bmatrix}\\\Rarr\space \text{I}=\frac{x+2}{2}\sqrt{x^2+4x+1}-\\\frac{3}{2}\text{log}|(x+2) + \sqrt{x^2+4x+1}|+\text{C}$$

$$\textbf{5.\space}\sqrt{\textbf{1-4x-x}^2}\\\textbf{Sol.}\space\text{Let}\space\text{I}=\int\sqrt{1-4x-x^2}dx\\=\int\sqrt{-(x^2 + 4x-1-2^2+2^2)}dx\\=\int\sqrt{-[(x+2)^2-(\sqrt{5})^2]}dx\\=\int\sqrt{(\sqrt{5})^2-(x+2)^2}dx\\\begin{bmatrix}\because\space \int\sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\\\frac{a^2}{2}\text{sin}^{\normalsize-1}\frac{x}{a}+\text{C}\end{bmatrix}\\\Rarr\space\text{I}=\frac{x+2}{2}\sqrt{1-4x-x^2}+\\\frac{5}{2}\text{sin}^{\normalsize-1}\frac{(x+2)}{\sqrt{2}}+\text{C}$$

$$\textbf{6.\space}\sqrt{\textbf{x}^\textbf{2}\textbf{+4x-5}}\\\textbf{Sol.}\space\text{Let}\space\text{I}=\int\sqrt{x^2+4x-5-2^2+2^2}dx\\=\int\sqrt{(x+2)^2-5-4}dx\\=\int\sqrt{(x+2)^2-(3)^2}dx\\\Rarr\space\text{I}=\frac{x+2}{2}\sqrt{x^2+4x-5}-\\\frac{9}{2}\text{log}|(x+2)+\sqrt{x^2+4x-5}|+\text{C}\\\begin{bmatrix}\because\space\int\sqrt{x^2-a^2}dx=\frac{x}{2}\sqrt{x^2-a^2}-\\\frac{a^2}{2}\text{log}|x+\sqrt{x^2-a^2}|\end{bmatrix}$$

$$\textbf{7.\space}\sqrt{\textbf{1+3x-x}^\textbf{2}}\\\textbf{Sol.}\space\text{Let}\space\text{I}=\int\sqrt{1+3x-x^2}dx\\=\int\sqrt{-(x^2-3x-1)}dx\\=\int\sqrt{-\bigg[x^2-3x+\bigg(\frac{3}{2}\bigg)^2-\bigg(\frac{3}{2}\bigg)^2-1\bigg]}dx\\=\int\sqrt{-\bigg[\bigg(x-\frac{3}{2}\bigg)^2-\frac{9}{4}-1\bigg]}dx\\=\int\sqrt{-\bigg[\bigg(x-\frac{3}{2}\bigg)^2-\bigg(\frac{9+4}{4}\bigg)\bigg]}dx\\=\int\sqrt{-\bigg[\bigg(x-\frac{3}{2}\bigg)^2-\frac{13}{4}\bigg]}dx$$

$$=\int\sqrt{\bigg(\frac{\sqrt{13}}{2}\bigg)^2-\bigg(x-\frac{3}{2}\bigg)^2}dx\\\therefore\space \text{I}=\frac{x-\frac{3}{2}}{2}\sqrt{1+3x-x^2}+\\\frac{13}{4×2}\text{sin}^{\normalsize-1}\bigg(\frac{x-\frac{3}{2}}{\frac{\sqrt{13}}{2}}\bigg)$$

$$\begin{bmatrix}\because\space\int\sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\\\frac{a^2}{2}\text{sin}^{\normalsize-1}\frac{x}{a}+\text{C}\end{bmatrix}\\\Rarr\space \text{I}=\frac{2x-3}{4}\sqrt{1+3x-x^2}+\\\frac{13}{8}\text{sin}^{\normalsize-1}\bigg(\frac{2x-13}{\sqrt{13}}\bigg)+\text{C}$$

$$\textbf{8.\space}\sqrt{\textbf{x}^\textbf{2}\textbf{+3x}}\\\textbf{Sol.}\space\textbf{Let}\space\text{I}=\int\sqrt{x^2+3x}dx\\=\int\sqrt{x^2+3x-\bigg(\frac{3}{2}\bigg)^2+\bigg(\frac{3}{2}\bigg)^2}dx\\=\int\sqrt{\bigg(x+\frac{3}{2}\bigg)^2-\bigg(\frac{3}{2}\bigg)^2}dx\\\Rarr\space \text{I}=\frac{x+\frac{3}{2}}{2}\sqrt{\bigg(x+\frac{3}{2}\bigg)^2-\bigg(\frac{3}{2}\bigg)^2}\\-\frac{9}{4×2}\\\text{log}\begin{vmatrix}\bigg(x+\frac{3}{2}\bigg)+\sqrt{x^2+3x}\end{vmatrix}+\text{C}$$

$$\begin{bmatrix}\because\space \int\sqrt{x^2-a^2}dx=\frac{x}{2}\sqrt{x^2-a^2}-\\\frac{a^2}{2}\text{log}|x+\sqrt{x^2-a^2}|\end{bmatrix}\\\Rarr\space \text{I}=\frac{2x+3}{4}\sqrt{x^2+3x}-\frac{9}{8}\text{log}\\\begin{vmatrix}\bigg(\frac{2x+3}{2}\bigg)+\sqrt{x^2+3x}\end{vmatrix}+\text{C}$$

$$\textbf{9.\space}\sqrt{\textbf{1+}\frac{\textbf{x}^\textbf{2}}{\textbf{9}}}\\\textbf{Sol.}\space\text{Let}\space\text{I}=\int\sqrt{1+\frac{x^2}{9}}dx\\=\int\sqrt{\frac{9+x^2}{9}}dx\\=\frac{1}{3}\int\sqrt{(3)^2+x^2}dx\\=\frac{1}{3}.\frac{x}{2}\sqrt{(3)^2+x^2}+\\\frac{9}{2×3}\text{log}|x+\sqrt{9+x^2}|+\text{C}\\\begin{bmatrix}\because\space\int\sqrt{a^2+x^2}\space dx=\frac{1}{2}x\sqrt{a^2-x^2}\\+\frac{1}{2}a^2\space\text{log}|x+\sqrt{a^2+x^2}|+\text{C}\end{bmatrix}\\=\frac{x}{6}\sqrt{x^2+9}+\frac{3}{2}\text{log}|x+\sqrt{9+x^2}|+\text{C}$$

Choose the correct answer in the given question.

$$\textbf{10.\space}\int \sqrt{\textbf{1+x}^\textbf{2}}\space\textbf{dx is equal to :}\\\textbf{(a)\space}\frac{\textbf{x}}{\textbf{2}}\sqrt{\textbf{1+x}^\textbf{2}}+\frac{\textbf{1}}{\textbf{2}}\textbf{log}|x+\sqrt{\textbf{1+x}^\textbf{2}}|+\textbf{C}.\\\textbf{(b)\space}\frac{\textbf{2}}{\textbf{3}}(\textbf{1+x}^\textbf{2})^{\frac{\textbf{3}}{\textbf{2}}}+\textbf{C}\\\textbf{(c)\space}\frac{\textbf{2}}{\textbf{3}}\textbf{x(1+x}^\textbf{2})^{\frac{\textbf{3}}{\textbf{2}}}+\textbf{C}\\\textbf{(d)\space}\frac{\textbf{x}^\textbf{2}}{\textbf{2}}+\frac{\textbf{1}}{\textbf{2}}\textbf{x}^\textbf{2}\textbf{log}\textbf{|x+}\sqrt{\textbf{1+x}^\textbf{2}}|+\textbf{C}$$

$$\textbf{Sol.\space}(a)\frac{x}{2}\sqrt{1+x^2}+\frac{1}{2}\text{log}|x+\sqrt{1+x^2}|+\text{C}$$

$$\text{Let}\space\text{I}=\int\sqrt{1+x^2}\space dx\\=\sqrt{1^2+x^2}dx\\\Rarr\space\text{I}=\frac{x}{2}\sqrt{1+x^2}+\frac{1}{2}\text{log}|x+\sqrt{1+x^2}|+\text{C}\\\begin{bmatrix}\because\space\int\sqrt{x^2+a^2}dx=\frac{x}{2}\sqrt{x^2+a^2}+\\\frac{a^2}{2}\text{log}|x+\sqrt{x^2+a^2}|\end{bmatrix}$$

$$\textbf{11.\space}\sqrt{\textbf{x}^\textbf{2}\textbf{-8x+7}}\space\textbf{dx is equal to :}\\\textbf{(a)\space}\frac{\textbf{1}}{\textbf{2}}(\textbf{x-4})\sqrt{\textbf{x}^\textbf{2}\textbf{-8x+7}}+\\\textbf{9 log}|\textbf{x-4 +} \sqrt{\textbf{x}^\textbf{2}\textbf{-8x+7}}|\textbf{+ C}\\\textbf{(b)\space}\frac{\textbf{1}}{\textbf{2}}(\textbf{x+4})\sqrt{\textbf{x}^\textbf{2}\textbf{-8x+7}}\textbf{+ 9}\space \textbf{log}|\textbf{x+4} +\\\sqrt{\textbf{x}^\textbf{2}\textbf{-8x+7}}|+\textbf{C}\\\textbf{(c)\space}\frac{\textbf{1}}{\textbf{2}}\textbf{(x-4)}\sqrt{\textbf{x}^\textbf{2}\textbf{-8x+7}}-\textbf{3}\sqrt{\textbf{2}}\space\textbf{log}\textbf{|x-4|+}\\\sqrt{\textbf{x}^\textbf{2}\textbf{-8x+7}}+\textbf{C}\\\textbf{(d)\space}\frac{\textbf{1}}{\textbf{2}}\textbf{(x-4)}\sqrt{\textbf{x}^\textbf{2}\textbf{-8x+7}}\textbf{-}\frac{\textbf{9}}{\textbf{2}}\space\textbf{log}|\textbf{x-4}+\\\sqrt{\textbf{x}^\textbf{2}\textbf{-8x+7}}|+\textbf{C}$$

$$\textbf{Sol.\space}\text{(d)}\frac{1}{2}(x-4)\sqrt{x^2-8x+7}-\\\frac{9}{2}\text{log}|x-4+\sqrt{x^2-8x+7}|+\text{C}\\\text{Let}\space\text{I}=\int\sqrt{x^2-8x+7}dx\\\Rarr\space\text{I}=\int\sqrt{x^2-8x+7+(4)^2-(4)^2}dx\\=\int\sqrt{(x-4)^2+7-16}\space dx\\=\int\sqrt{(x-4)^2-(3)^2}dx\\\begin{bmatrix}\because\space\int\sqrt{x^2-a^2}\space dx=\frac{x}{2}\sqrt{x^2-a^2}\\-\frac{a^2}{2}\text{log}|x+\sqrt{x^2-a^2}|\end{bmatrix}\\\Rarr\space\text{I}=\frac{x-4}{2}\sqrt{x^2-8x+7}-\\\frac{9}{2}\text{log}|x-4 + \sqrt{x^2-8x+7}|+\text{C}$$