NCERT Solutions for Class 12 Maths Chapter 7 - Integrals
Exercise 7.1 Solutions 22 Questions
Exercise 7.2 Solutions 39 Questions
Exercise 7.3 Solutions 24 Questions
Exercise 7.4 Solutions 25 Questions
Exercise 7.5 Solutions 23 Questions
Exercise 7.6 Solutions 24 Questions
Exercise 7.7 Solutions 11 Questions
Exercise 7.8 Solutions 6 Questions
Exercise 7.9 Solutions 22 Questions
Exercise 7.10 Solutions 10 Questions
Exercise 7.11 Solutions 21 Questions
Miscellaneous Exercise on Chapter 7 Solutions 44 Questions
Miscellaneous Exercise
Integrate the following functions w.r.t.x.
$$\textbf{1.\space}\frac{\textbf{1}}{\textbf{x-x}^\textbf{3}}\\\textbf{Sol.}\int\frac{1}{x-x^3}dx=\int\frac{1}{x(1-x^2)}dx\\=\int\frac{1}{x(1-x)(1+x)}dx\\\text{Let}\space\frac{1}{x(1-x)(1+x)}=\\\frac{A}{x}+\frac{B}{(1-x)}+\frac{\text{C}}{(1 + x)}$$
$$\Rarr\space 1 = A(1-x)(1+x)+B(x)(1+x)+\\\text{C(x)}(1-x)\\\Rarr\space 1=\text{A}(1-x^2)+ B(x+x^2) + \text{C}(x-x^2)\\\Rarr\space 1 = x^2(\text{B-A-C}) + x(\text{B+C})+\text{A}$$
On equating the coefficients of x2, x and constant term on both sides, we get
– A + B – C = 0, B + C = 0 and A = 1
On solving these equations, we get
$$\text{A = 1, B =}\frac{1}{2}\text{and C}=-\frac{1}{2}\\\text{From eq. (i), we get}\\\int\frac{1}{x(1-x^2)}dx=\int\frac{1}{x}dx + \\\frac{1}{2}\int\frac{1}{1-x}dx-\frac{1}{2}\int\frac{1}{1+x}dx\\=\text{log}|x|-\frac{1}{2}\text{log}|1-x|-\\\frac{1}{2}\text{log}|1+x|+\text{C}\\=\text{log}|x|-\frac{1}{2}\text{log}\lbrace(1-x)(1+x)\rbrace+\text{C}\\=\frac{1}{2}\text{log}|x|^2-\frac{1}{2}\text{log}|1-x^2|+\text{C}\\=\frac{1}{2}\text{log}\begin{vmatrix}\frac{x^2}{1-x^2}\end{vmatrix}+\text{C}$$
$$\textbf{2.\space}\frac{\textbf{1}}{\sqrt{\textbf{x+a}} + \sqrt{\textbf{x+b}}}\\\textbf{Sol.\space}\int\frac{1}{\sqrt{x+a}\sqrt{x+b}}dx\\=\int\frac{1}{\sqrt{x+a} + \sqrt{x+b}}×\\\frac{\sqrt{x+a}-\sqrt{x+b}}{\sqrt{x+a}-\sqrt{x+b}}dx\\=\int\frac{\sqrt{x+a} - \sqrt{x+b}}{(x+a)-(x+b)}dx\\=\frac{1}{a-b}\int\lbrace(x+a)^{\frac{1}{2}}- (x+b)^{\frac{1}{2}}\rbrace dx$$
$$=\frac{1}{a-b}\begin{Bmatrix}\frac{(x+a)^{(\frac{1}{2})+1}}{\frac{1}{2}+1}-\frac{(x+b)^{\frac{1}{2}+1}}{\frac{1}{2}+1}\end{Bmatrix}+\text{C}\\=\frac{1}{a-b}\begin{Bmatrix}\frac{(x+a)^{\frac{3}{2}}}{\frac{3}{2}}- \frac{(x+b)^{\frac{3}{2}}}{\frac{3}{2}}\end{Bmatrix}+\text{C}\\=\frac{2}{3(a-b)}[(x+a)^{\frac{3}{2}}-(x+b)^{\frac{3}{2}}]+\text{C}$$
$$\textbf{3.\space}\frac{\textbf{1}}{\textbf{x}\sqrt{\textbf{ax-x}^\textbf{2}}}\\\textbf{Sol.\space}\text{Let I = }\int\frac{1}{x\sqrt{ax-x^2}}dx\space\text{Put x =}\frac{a}{t}\\\Rarr\space dx=\frac{-a dt}{t^2}\\\therefore\space\text{I = }\int\frac{1}{\frac{a}{t}\sqrt{a\bigg(\frac{a}{t}\bigg)-\frac{a^2}{t^2}}}\bigg(\frac{-a}{t^2}\bigg)dt\\=\int\frac{-a}{a.a\sqrt{t-1}}dt\\=-\frac{1}{a}\int(t-1)^{\normalsize-\frac{1}{2}}dt$$
$$=-\frac{1}{a}.\frac{(t-1)^{\frac{-1}{2}}+1}{-(\frac{1}{2})+1}+\text{C}\\=-\frac{2}{a}\sqrt{t-1}+\text{C}\\=-\frac{2}{a}\sqrt{\frac{a}{x}-1}+\text{C}\\\bigg(\because x=\frac{a}{t}\Rarr\space t=\frac{a}{x}\bigg)\\=-\frac{2}{a}\sqrt{\frac{a-x}{x}}+\text{C}$$
$$\textbf{4.\space}\frac{\textbf{1}}{\textbf{x}^\textbf{2}\textbf{(x}^\textbf{4}\textbf{+1)}^{\frac{\textbf{3}}{\textbf{4}}}}\\\textbf{Sol.\space}\text{Let \space I =}\int\frac{1}{x^2(x^4+1)^{\frac{3}{4}}}dx\\=\int\frac{1}{x^2\begin{Bmatrix}x^4\bigg(1 + \frac{1}{x^4}\bigg)\end{Bmatrix}^{\frac{3}{4}}}dx\\=\int\frac{1}{x^2.x^3\bigg(1 + \frac{1}{x^4}\bigg)^{\frac{3}{4}}}dx\\=\int\frac{1}{x^5\bigg(1 + \frac{1}{x^4}\bigg)^{\frac{3}{4}}}dx$$
$$\text{Put}\space 1+\frac{1}{x^4}=t\\\Rarr\space-\frac{4}{x^5}dx=dt\\\Rarr\space \frac{1}{x^5}dx=-\frac{dt}{4}\\\therefore\space\text{I} = \frac{1}{-4}\int\frac{1}{t^{3/4}}dt\\=-\frac{1}{4}\bigg[\frac{t^{\frac{1}{4}}}{\frac{1}{4}}\bigg]+\text{C}$$
= – (1 + x–4)1/4 + C
$$\textbf{5.\space}\frac{\textbf{1}}{\textbf{x}^{\frac{\textbf{1}}{\textbf{2}}}\textbf{+ x}^{\frac{\textbf{1}}{\textbf{3}}}}\\\textbf{Sol. \space}\text{Let}\space \text{I = }\int\frac{1}{x^{\frac{1}{2}} + x^{\frac{1}{3}}}dx\\\text{Put \space} x^{\frac{1}{6}}=t\space\Rarr\space x = t^{6}\\\Rarr\space dx=6 t^{5}dt\\\therefore\space\text{I = }\int\frac{6t^{5}}{(t^{6})^{\frac{1}{2}} + (t^{6})^{\frac{1}{3}}}dt\\ = 6\int\frac{t^5}{t^3+t^2}dt$$
$$= 6\int\frac{t^5}{t^2(t+1)}dt=6\int\frac{t^3}{(t+1)}dt\\=6\int\bigg(\frac{t^3+1-1}{t-1}\bigg)dt\\=6\int\bigg[\frac{(t+1)(t^2-1+1)}{t+1}-\frac{1}{t+1}\bigg]dt\\=6\int\begin{Bmatrix}t^2-t+1+\frac{\normalsize-1}{t+1}\end{Bmatrix}dt\\6\begin{Bmatrix}\frac{t^3}{3}-\frac{t^2}{2}+t-\text{log}|t+1|\end{Bmatrix}+\text{C}\\=6\begin{Bmatrix}\frac{(x^{\frac{1}{6}})^3}{3}-\frac{(x^{\frac{1}{6}})^2}{2} + (x^{\frac{1}{6}})-\text{log}|x^{\frac{1}{6}}+1|\end{Bmatrix}+\text{C}$$
= 2x1/2 – 3x1/3 + 6x1/6 – 6 log|x1/6 + 1| + C
$$\textbf{6.\space}\frac{\textbf{5x}}{\textbf{(x+1)(x}^\textbf{2}\textbf{+9)}}\\\textbf{Sol.\space}\int\frac{5x}{(x+1)(x^2+9)}dx\\\text{Let}\space\frac{5x}{(x+1)(x^2+9)}=\frac{A}{x+1} + \frac{\text{Bx+C}}{x^2+9}\\\Rarr\space 5x = A(x^2+9)+(Bx+C)(x+1)\\\Rarr\space 5x= Ax^2+9A+Bx^2+Bx+Cx+C $$
On equating the coefficients of x2, x and constant term on both sides, we get
A + B = 0, B + C = 5, 9A + C = 0
On solving these equations, we get
$$\text{A}=-\frac{1}{2},\text{B}=\frac{1}{2}\text{and C}=\frac{9}{2}\\\therefore\space\int\frac{5x}{(x+1)(x^2+9)}dx=\\\int\frac{\bigg(-\frac{1}{2}\bigg)}{(x+1)}dx+\int\frac{\frac{1}{2}x+\frac{9}{2}}{x^2+9}dx\\=-\frac{1}{2}\int\frac{1}{x+1}dx + \frac{1}{2}\int\bigg(\frac{x+9}{x^2+9}\bigg)dx\\=-\frac{1}{2}\text{log}|x+1|+\frac{1}{2}.\frac{1}{2}\int\frac{2x}{x^2+9}dx+\\\frac{9}{2}\int\frac{1}{x^2+9}dx\\=-\frac{1}{2}\text{log}|x+1|+\frac{1}{4}\text{log}|x^2+9|+\\\frac{9}{2}.\frac{1}{3}\text{tan}^{\normalsize-1}\text{tan}^{\normalsize-1}\bigg(\frac{x}{3}\bigg)+\text{C}$$
$$\bigg[\because\space\int\frac{dx}{a^2+x^2}=\frac{1}{a}\text{tan}^{\normalsize-1}\bigg(\frac{x}{a}\bigg)\bigg]\\=-\frac{1}{2}\text{log}|x+1|+\frac{1}{4}\text{log}|x^2+9|+\\\frac{3}{2}\text{tan}^{\normalsize-1}\bigg(\frac{x}{3}\bigg)+\text{C}$$
$$\textbf{7.\space}\frac{\textbf{sin x}}{\textbf{sin}\textbf{(x-a)}}\\\textbf{Sol.\space}\int\frac{\text{sin x}}{\text{sin}(x-a)}dx=\\\int\frac{\text{sin}(x+a)+a}{\text{sin}(x-a)}dx\\=\int\frac{\text{sin(x-a)cos a + cos (x-a) sin a}}{\text{sin}(x-a)}dx\\\lbrack\because\space\text{sin}(A+B) = \text{sin A cos B + cos A sin B}\rbrack\\=\int\frac{\text{sin}(x-a)\text{cos a}}{\text{sin}(x-a)}dx+\int\frac{\text{cos}(x-a) sina }{\text{sin}(x-a)}dx$$
= cos a ∫ 1 dx + sin a ∫ cot (x – a) dx
= (cos a) x + sin a.log|sin(x – a)| + C
$$\textbf{8.\space}\frac{\textbf{e}^{\textbf{5 log x}}\textbf{- e}^{\textbf{4 log x}}}{\textbf{e}^{\textbf{3 log x}}\textbf{- e}^{\textbf{2 log x}}}\\\textbf{Sol.\space}\int\frac{e^{\text{5 log x}}-e^{\text{4 log x}}}{e^{\text{3 log x}}-e^{\text{2 log x}}}=\\\int\frac{e^{\text{log x}^{5}}-e^{\text{log x}^{4}}}{e^{\text{log x}^{3}}-e^{\text{log x}^{2}}}dx\\=\int\frac{x^5-x^4}{x^3-x^2}dx\space (\because\space e^{log x}=x)\\=\int\frac{x^{4}(x-1)}{x^{2}(x-1)}dx\\=\int x^2 dx=\frac{1}{3}x^{3}+\text{C}$$
$$\textbf{9.\space}\frac{\textbf{cos x}}{\sqrt{ \textbf{4 - sin}^{2}\textbf{x}}}\\\textbf{Sol.\space}\text{Let}\space\text{I = }\int\frac{\text{cos x}}{\sqrt{ \text{4 - sin}^{2}\text{x}}}dx\\\text{Put sin x = t}$$
$$\Rarr\space\text{cos x dx = dt}\\\therefore\space\text{I} = \int\frac{\text{cos x}}{\sqrt{4-t^2}}\frac{dt}{\text{cos x}}\\=\int\frac{t}{\sqrt{4-t^{2}}}dt\\=\text{sin}^{\normalsize-1}\bigg(\frac{t}{2}\bigg)+\text{C}\\\bigg[\because\space \int\frac{dx}{\sqrt{a^2-x^2}} = \text{sin}^{\normalsize-1}\bigg(\frac{x}{a}\bigg)\bigg]\\=\text{sin}^{\normalsize-1}\bigg(\frac{\text{sin x}}{2}\bigg)+\text{C}$$
$$\textbf{10.\space}\frac{\textbf{sin}^{\textbf{8}}\textbf{x}\textbf{ - cos}^{\textbf{8}}\textbf{x}}{\textbf{1 - 2 sin}^{\textbf{2}}\textbf{x}\textbf{cos}^{\textbf{2}}\textbf{x}}\\\textbf{Sol.\space}\int\frac{\text{sin}^{8}x - \text{cos}^{8}x}{\text{1 - 2 sin}^{2}\text{x cos}^{2}x}dx=\\\int\frac{\lbrace(\text{sin}^{4}x)^{2} - (\text{cos}^{4} x)^{2}\rbrace}{\text{1 - 2 sin}^{2} \text{x cos}^{2}x}dx\\=\int\frac{(\text{sin}^{4}x - \text{cos}^{4}x)(\text{sin}^{4}x+\text{cos}^{4}x)}{\text{1 - 2 sin}^{2}\text{x cos}^{2}x}dx\\\lbrack\because\space a^{2}-b^{2}= (a-b)(a+b)\rbrack\\\text{(sin}^{2}x - \text{cos}^{2}x)(\text{sin}^{2}x + \text{cos}^{2}x)\\=\int\frac{\lbrack(\text{sin}^{2}x)^{2} + (\text{cos}^{2}x)^{2}\rbrack}{\text{1 - 2 sin}^{2}x\text{cos}^{2}x}dx$$
$$=\\\int\frac{(\text{sin}^{2}x \text{- cos}^{2}x)1\lbrace(\text{sin}^{2}x + \text{cos}^{2}x)^{2}-2\space\text{sin}^{2}\text{x. cos}^{2}x\rbrace}{\text{1 - 2 sin}^{2}\text{x cos}^{2}\text{x}}$$
[∵ sin2 x + cos 2x = 1 and a2 + b2= (a + b)2 – 2ab]
$$=\int\frac{\text{- cos 2x}\lbrace 1 - 2 \text{sin}^{2}x.\text{cos}^{2}x\rbrace}{\text{1 - 2 sin}^{2}\text{x cos}^{2}x}dx\\(\because\space\text{cos 2x = cos}^{2}x \text{- sin}^{2}x)\\=-\int\text{cos 2x dx}=-\frac{\text{sin 2x}}{2}+\text{C}$$
$$\textbf{11.\space}\frac{\textbf{1}}{\textbf{cos}\textbf{(x+a) cos (x+b)}}\\\textbf{Sol.\space}\int\frac{1}{\text{cos}(x+a)\text{cos}(x+b)}dx$$
On multiplying and dividing by sin(a – b), we get
$$\text{I = }\frac{1}{\text{sin}(a-b)}\int\frac{\text{sin}(a-b)}{\text{cos}(x+a)\text{cos}(x+b)}dx\\\Rarr\space\text{I =}\frac{1}{\text{sin}(a-b)}\int\frac{\text{sin}(x+a) - (x+b)}{\text{cos}(x+a) \text{cos}(x+b)}dx$$
sin(x+a)cos(x+b)
$$=\frac{1}{\text{sin}(a-b)}\int\frac{\text{-cos}(x+a)\text{sin}(x+b)}{\text{cos}(x+a)\text{cos}(x+b)}dx\\\lbrack\because\space \text{sin}(A-B) = \text{sin A cos B - sin B cos A}\rbrack\\=\frac{1}{\text{sin}(a-b)}\int\bigg[\frac{\text{sin}(x+a)}{\text{cos}(x+a)}-\frac{\text{sin}(x+b)}{\text{cos}(x+b)}\bigg]dx\\=\frac{1}{\text{sin}(a-b)}\int[\text{tan}(x+a) - \text{tan}(x+b)]dx\\=\frac{1}{\text{sin}(a-b)} \begin{bmatrix}-\text{log}|\text{cos}(x+a)| + \\\text{log}|\text{cos}(x+b)|\end{bmatrix}+\text{C}\\=\frac{1}{\text{sin}(a-b)}\text{log}\begin{vmatrix}\frac{\text{cos}(x+b)}{\text{cos}(x+a)}\end{vmatrix}+\text{C}\\\bigg(\because\space\text{log m - log n = log}\frac{m}{n}\bigg)$$
$$\textbf{12.\space}\frac{\textbf{x}^\textbf{3}}{\sqrt{\textbf{1-x}^{\textbf{8}}}}.\\\textbf{Sol.\space}\text{Let}\space\text{I} = \int\frac{x^3}{\sqrt{1-x^{8}}}dx\\\text{Put\space x}^{4}=t\Rarr\space 4x^{3}dx=dt\\\Rarr dx = \frac{dt}{4x^{3}}\\\therefore\space\text{I = }\int\frac{x^{3}}{\sqrt{1-t^{2}}}\frac{dt}{4x^{3}}\\=\frac{1}{4}\int\frac{1}{\sqrt{1-t^{2}}}dt\\=\frac{1}{4}\text{sin}^{\normalsize-1}t+\text{C}\\\bigg[\because\space\int\frac{dx}{\sqrt{a^{2}-x^{2}}}=\text{sin}^{\normalsize-1}\bigg(\frac{x}{a}\bigg)\bigg]$$
$$=\frac{1}{4}\text{sin}^{\normalsize-1}(x^{4})+\text{C}$$
$$\textbf{13.\space}\frac{\textbf{e}^\textbf{x}}{\textbf{(1+e}^{\textbf{x}})\textbf{(2+e}^{\textbf{x}})}\\\textbf{Sol.\space}\text{Let}\space\text{I = }\int\frac{e^{x}}{(1+e^{x})(2+e^{x})}dx$$
Put ex = t
$$\Rarr\space e^{x}dx=dt\\\therefore\space\text{I = }\int\frac{e^{x}}{(1+t)(2+t)}\frac{dt}{e^{x}}\\=\int\frac{1}{(1+t)(2+t)}dt\\\text{Let}\space\frac{1}{(1+t)(2+t)}=\\\frac{A}{(1+t)}+\frac{B}{(2+t)}$$
$$\Rarr\space 1 = A(2 + t)+B(1+t)$$
= (2A + B) + t(A + B)
On equating the coefficients of t and constant term on both sides, we get A + B = 0 and 2A + B = 1.
On solving both equations, we get A = 1 and B = – 1
$$\therefore\space\text{I} = \int\frac{A}{(1+t)}dt - \int\frac{1}{(2+t)}dt\\=\text{log}|1+t| - \text{log}|2 + t|+\text{C}\\=\text{log}\begin{vmatrix}\frac{1+t}{2+t}\end{vmatrix}+\text{C}\\=\text{log} \begin{vmatrix}\frac{1+e^{x}}{2+e^{x}}\end{vmatrix}+\text{C}$$
$$\textbf{14.\space}\frac{\textbf{1}}{\textbf{(x}^\textbf{2}\textbf{+1)(x}^\textbf{2}\textbf{+4)}}\\\textbf{Sol.\space}\int\frac{1}{(x^2+1)(x^2+4)}dx\\\text{Let}\frac{1}{(x^2+1)(x^2+4)}=\\\frac{\text{Ax+B}}{x^2+1} + \frac{\text{(Cx+D)}}{x^2+4} $$
$$\Rarr 1 = (Ax + B) (x^2 + 4) + (Cx + D) (x^2 + 1)$$
On comparing the coefficients of x3, x2, x and constant term on both sides, we get
A + C = 0, B + D = 0, 4A + C = 0 and 4B + D = 1
On solving these equations, we get
$$\text{A = 0, C = 0, B=}\frac{1}{3}\space\text{and D}=-\frac{1}{3}\\\therefore\space\int\frac{1}{(x^2+1)(x^2+4)}dx=\\\frac{1}{3}\int\bigg(\frac{1}{(x^2+1)}-\frac{1}{(x^2+4)}\bigg)dx\\=\frac{1}{3}\begin{Bmatrix}\text{tan}^{\normalsize -1}x-\frac{1}{2}\text{tan}^{\normalsize-1}\bigg(\frac{x}{2}\bigg)\end{Bmatrix}+\text{C}$$
$$\textbf{15. cos}^{\textbf{3}}\textbf{xe}^{\textbf{log sinx }}\\\textbf{Sol.\space}\text{Let}\space\text{I = }\int\text{cos}^{3}xe^{log \space sinx}dx = \\\int\text{(cos x)}^{3}\text{sin x dx}\\(\because\space e^{log x}=x)\\=-\int(\text{cos x})^3(-\text{sin x})dx\\\text{Put cos x = t}\\\Rarr\space -\text{sin x dx=dt}\\\therefore\space\text{I} =-\int t^{3}dt =-\frac{t^{4}}{4}+\text{C}\\=-\frac{\text{cos}^{4}x}{4}+\text{C}$$
$$\textbf{16.\space}\textbf{e}^{\textbf{3} \space \textbf{log x}}(\textbf{x}^{\textbf{4}}\textbf{+1})^{\normalsize\textbf{-1}}\\\textbf{Sol.\space}\text{Let}\space\text{I} = \int e^{3 log x}(x^{4}+1)^{\normalsize-1}dx\\=\int\frac{e^{log x^{3}}}{(x^{4}+1)}dx\\=\int\frac{x^{3}}{(x^4+1)}dx\space(\because\space e^{log x}=x)\\\text{Put x}^{4}+1=t\Rarr\space 4x^{3}dx=dt\\\Rarr\space dx=\frac{dt}{4x^{3}}\\\therefore\space\int\frac{x^{3}}{t}\frac{dt}{4x^{3}}=\frac{1}{4}\int\frac{1}{t}dt\\=\frac{1}{4}\text{log}|t|+\text{C}\\=\frac{1}{4}\text{log}|x^{4}+1|\text{C}$$
$$\textbf{17.\space}\textbf{f'(ax+b)}[\textbf{f(ax+b)}]^{\textbf{n}}\\\textbf{Sol.\space}\text{Let\space I}=\int f'(ax+b)[f(ax+b)]^n dx\\\text{Put f(ax+b) = t}\\\Rarr\space f'(ax+b) a\space dx=dt\\\Rarr\space dx=\frac{dt}{af'(ax+b)}\\\therefore\space\text{I} = \int f'(ax+b)t^n\frac{dt}{f'(ax+b)a}\\=\frac{1}{a}\int t^{n}dt=\frac{1}{a}\bigg(\frac{t^{n+1}}{n+1}\bigg)+\text{C}\\=\frac{1}{a}.\frac{[f(ax+b)]^{n+1}}{n+1}+\text{C}$$
