NCERT Solutions for Class 12 Maths Chapter 7 - Integrals - Exercise 7.10
Access Exercises of 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
Exercise 7.10
Direction (Q. 1 to 8) : Evaluate the integrals using substitution.
$$\textbf{1.\space}\int^{\textbf{1}}_{\textbf{0}}\frac{\textbf{x}}{\textbf{x}^\textbf{2}\textbf{+1}}\textbf{dx}\\\textbf{Sol.\space}\text{Let}\space\text{I}=\int^{1}_{0}\frac{x}{x^2+1}dx\\\text{Put}\space x^2+1=t\\\Rarr 2x=\frac{dt}{dx}\\\Rarr\space dx=\frac{dt}{2x}$$
For limit, when x = 0
$$\Rarr\space t=1\space\text{and}\\\text{when}\space x=1\Rarr\space t=2\space \\(\because t=x^2+1)\\\therefore\space\text{I}=\int^{2}_{1}\frac{x}{t}\frac{dt}{2x}\\=\frac{1}{2}\int^{2}_{1}\frac{1}{t}dt\\=\frac{1}{2}[\text{log}|t|]^{2}_{1}\\=\frac{1}{2}[\text{log}|2|-\text{log}|1|]\\=\frac{1}{2}\text{log 2}$$
(∵ log 1 = 0)
$$\textbf{2.\space}\int^{\frac{\pi}{\textbf{2}}}_{\textbf{0}}\sqrt{\textbf{sin}\space\phi\space\textbf{cos}^{\textbf{5}}\phi}d\phi\\\textbf{Sol.\space}\int^{\frac{\pi}{2}}_{0}\sqrt{\text{sin}\space\phi}\space\text{cos}^{5}\phi d\phi\\=\int^{\frac{\pi}{2}}_{0}\sqrt{\text{sin}\phi}(1-\text{sin}^{2}\phi)^{2}\text{cos}\phi d\phi\\(\because\space \text{cos}^{2}\theta+\text{sin}^{2}\theta=1)\\\text{Put\space}\text{sin}\phi=t\Rarr\space\text{cos}\phi=\frac{dt}{d\phi}\\\Rarr\space d\phi=\frac{dt}{\text{cos}\phi}\\\text{For limit when}\phi=0\\\Rarr\space t=1\space\text{and when}\phi=\frac{\pi}{12}$$
$$\Rarr\space t=1\\\therefore\space\text{I}=\int^{1}_{0}\sqrt{t}(1-t^2)dt=\\\int^{1}_{0}\sqrt{t}(t^{4}+1-2t^2)dt\\=\int^{1}_{0}(t^{9/2}+t^{1/2}-2t^{5/2})dt\\=\bigg[\frac{t^{\frac{9}{2}+1}}{\bigg(\frac{9}{2}+1\bigg)}+\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}-2\bigg(\frac{t^{\frac{5}{2}+1}}{\frac{5}{2}+1}\bigg)\bigg]^{1}_{0}\\=\bigg[\frac{2}{11}+\frac{2}{3}-\frac{4}{7}\bigg]-0\\=\frac{42+154-132}{11×3×7}=\frac{64}{231}$$
$$\textbf{3.\space}\int^{1}_{0}\textbf{sin}^{\normalsize-1}\bigg(\frac{\textbf{2x}}{\textbf{1+x}^\textbf{2}}\bigg)\textbf{dx}\\\textbf{Sol.\space}\text{Let I}=\int^{1}_{0}\text{sin}^{\normalsize-1}\bigg(\frac{2x}{1+x^2}\bigg)dx$$
$$\text{Put \space}\text{tan}\space\theta=x\\\Rarr\space\text{sec}^{2}\theta d\theta= dx or \theta=\text{tan}{\normalsize-1}x\\\text{For limit when x = 0}\\\Rarr\space \theta=0\space\text{and when x=1}\\\Rarr\space\theta=\frac{\pi}{4}(\because\space\theta=\text{tan}^{\normalsize-1}x)\\\therefore\space\text{I}=\int^{\frac{\pi}{4}}_{0}\text{sin}^{\normalsize-1}\bigg(\frac{2 tan\theta}{1+\text{tan}^2\theta}\bigg)\text{sec}^{2}\theta.d\theta\\=\int^{\frac{\pi}{4}}_{0}\text{sin}^{\normalsize-1}(\text{sin}\space2\theta)\text{sec}^{2}\theta d\theta\\\bigg(\because\space\text{sin} \space2\theta=\frac{\text{2 tan}\theta}{1+\text{tan}^2\theta}\bigg)\\=2\underset{\text{I}}{ \int^{\frac{\pi}{4}}_{0}}\underset{\text{II}}{\theta\text{sec}^{2}\theta d\theta}$$
Now, applying rule of integration by parts taking θ as the first function and sec2 θ as second function, we get
$$\text{I}=2\begin{bmatrix}\theta\int\text{sec}^{2}d\theta-\\\int\begin{Bmatrix}\bigg(\frac{d}{d\theta}\bigg)\end{Bmatrix}\int\text{sec}^{2}\theta d\theta\end{bmatrix}^{\frac{\pi}{4}}_{0}$$
$$= 2\bigg[\theta\space\text{tan}\space\theta-\int 1\text{tan}\space\theta\space d\theta\bigg]^{\frac{\pi}{4}}_{0}\\(\because\space\int\text{sec}^{2}\theta d\theta=\text{tan}\space\theta)\\=2\lbrack\theta\space\text{tan}\space\theta+\text{log}|\text{cos}\space\theta|\rbrack^{\frac{\pi}{4}}_{0}\\(\because\space\int\text{tan}\theta d\theta=-\text{log cos}\space\theta)\\=2\begin{Bmatrix}\bigg(\frac{\pi}{4}\text{tan}\frac{\pi}{4}+\text{log}\begin{vmatrix}\text{cos}\frac{\pi}{4}\end{vmatrix}\bigg)-\\(10 + \text{log}|\text{cos 0}|)\end{Bmatrix}\\=\frac{\pi}{2}+2\text{log}\frac{1}{\sqrt{2}}-0\space(\because\space\text{log 1=0})\\=\frac{\pi}{2}+2\text{log}2^{\frac{-1}{2}}$$
$$\frac{\pi}{2}-2×\frac{1}{2}\text{log 2}\\(\because\space \text{log m}^n= n\space log\space m)\\=\frac{\pi}{2}-\text{log 2}$$
$$\textbf{4.\space}\int^{\textbf{2}}_{\textbf{0}}\textbf{x}\sqrt{\textbf{x+2}}\space\textbf{dx}\qquad[\text{Put x+2=t}^2]\\\textbf{Sol.\space}\int^{2}_{0}x\sqrt{x+2}dx=\\\int^{2}_{0}\lbrace(x+2)-2\rbrace\sqrt{x+2}\space\text{dx}$$
(add and subtract 2)
$$=\int^{2}_{0}\lbrace(x+2)^{3/2}-2(x+2)^{\frac{1}{2}}\rbrace dx\\=\bigg[\frac{(x+2)^{(\frac{3}{2})+1}}{(\frac{3}{2}+1)}-\frac{2(x+2)^{(\frac{1}{2})+1}}{(\frac{1}{2}+1)}\bigg]^{2}_{0}\\=\frac{2}{5}[(x+2)^{5/2}]^{2}_{0}-\frac{4}{3}[(x+2)^{\frac{3}{2}}]^{2}_{0}\\=\frac{2}{5}[4^{\frac{5}{2}}-2^{\frac{5}{2}}]-\frac{4}{3}[4^{\frac{3}{2}}-2^{\frac{3}{2}}]\\=\frac{2}{5}[2^5-2^{\frac{5}{2}}]-\frac{4}{3}[2^3-2^{\frac{3}{2}}]\\=\frac{2}{5}[32-4\sqrt{2}]-\frac{4}{3}[8-2\sqrt{2}]\\=\frac{8}{5}[8-\sqrt{2}]-\frac{8}{3}[4-\sqrt{2}]$$
$$=\bigg(\frac{64}{5}-\frac{32}{3}\bigg)+\bigg(\frac{8\sqrt{2}}{3}-\frac{8\sqrt{2}}{5}\bigg)\\=\frac{32}{15}+\frac{16\sqrt{2}}{15}$$
$$\textbf{5.\space}\int^{\frac{\pi}{\textbf{2}}}_{0}\frac{\textbf{sin x}}{\textbf{1 + cos}^\textbf{2}\textbf{x}}\textbf{dx}\\\textbf{Sol.\space}\text{let}\space\text{I}=\int^{\frac{\pi}{2}}_{0}\frac{\text{sin x}}{\text{1 + cos}^2x}dx\\\text{Put \space}\text{cos x}=t\\\Rarr\space -\text{sin x}\space dx=dt\\\Rarr\space dx=\frac{dt}{-\text{sin x}}\\\text{For limit when x=0}\\\Rarr\space -\text{sin x dx = dt}\Rarr\space dx=\frac{dt}{\text{- sin x}}\\\text{For limit when x = 0}\\\Rarr\space t=\text{cos 0 = 1}(\because\space t = cos x)\\\text{and when}\space x=\frac{\pi}{2}$$
$$\Rarr\space t=\text{cos}\frac{\pi}{2}=0\\\therefore\space\text{I}=\int^{0}_{1}\frac{\text{sin x}}{1+t^2}\frac{dt}{(-\text{sinx})}dx\\=-\int^{0}_{1}\frac{1}{1+t^2}dx\\\bigg(\because\space\int\frac{1}{a^2+x^2}dx=\frac{1}{a}\text{tan}^{\normalsize-1}\frac{x}{a}\bigg)\\=\bigg[\frac{1}{1}\text{tan}^{\normalsize-1}\bigg(\frac{t}{1}\bigg)\bigg]^{0}_{1}\\=[\text{tan}^{\normalsize-1}]^{0}_{1}\\=[\text{tan}^{\normalsize-1}(0)-\text{tan}^{\normalsize-1}(1)]\\=-\bigg(0-\frac{\pi}{4}\bigg)=\frac{\pi}{4}.$$
$$\textbf{6.\space}\int^{\textbf{2}}_{\textbf{0}}\frac{\textbf{1}}{\textbf{x+4-x}^\textbf{2}}\textbf{dx}\\\textbf{Sol.\space}\int^{2}_{0}\frac{1}{x+4-x^2}dx\\=\int^{2}_{0}\frac{1}{x+4-x^2-\bigg(\frac{1}{2}\bigg)^2+\bigg(\frac{1}{2}\bigg)^2}dx\\=\int^{2}_{0}\frac{1}{4+\frac{1}{4}-\bigg[\bigg(x^2-x+\frac{1}{4}\bigg)\bigg]}dx\\=\int^{2}_{0}\frac{1}{4+\frac{1}{4}-\bigg(x-\frac{1}{2}\bigg)^2}dx$$
$$=\int^{2}_{0}\frac{1}{\bigg(\frac{\sqrt{17}}{2}\bigg)^2-\bigg(x-\frac{1}{2}\bigg)^2}dx\\=\frac{1}{2.\frac{\sqrt{17}}{2}}\bigg[\text{log}\begin{vmatrix}\frac{\frac{\sqrt{17}}{2}+x-\frac{1}{2}}{\frac{\sqrt{17}}{2}-x+\frac{1}{2}}\end{vmatrix}\bigg]^{2}_{0}\\\bigg(\because\space\int\frac{dx}{a^2-x^2}=\frac{1}{2a}\text{log}\begin{vmatrix}\frac{a+x}{a-x}\end{vmatrix}\bigg)\\=\frac{1}{\sqrt{17}}\bigg[\text{log}\begin{vmatrix}\frac{\sqrt{17}+2x-1}{\sqrt{17}-2x+1}\end{vmatrix}\bigg]^{2}_{0}\\=\frac{1}{\sqrt{17}}\bigg[\text{log}\begin{vmatrix}\frac{\sqrt{17}+3}{\sqrt{17-3}}\end{vmatrix}-\text{log}\begin{vmatrix}\frac{\sqrt{17}-1}{\sqrt{17}+1}\end{vmatrix}\bigg]\\=\frac{1}{\sqrt{17}}\text{log}\begin{vmatrix}\frac{\sqrt{17}+3}{\sqrt{17}-3}÷\frac{\sqrt{17}-1}{\sqrt{17}-1}\end{vmatrix}$$
$$\bigg(\because\space \text{log m - log n}=\text{log}\frac{m}{n}\bigg)\\=\frac{1}{\sqrt{17}}\text{log}\begin{vmatrix}\frac{(\sqrt{17}+3)(\sqrt{17}+1)}{(\sqrt{17}-3)(\sqrt{17}-1)}\end{vmatrix}\\=\frac{1}{\sqrt{17}}\space\text{log}\begin{vmatrix}\frac{20+4\sqrt{17}}{20-4\sqrt{17}}\end{vmatrix}\\=\frac{1}{\sqrt{17}}\text{log}\begin{vmatrix}\frac{5+\sqrt{17}}{5-\sqrt{17}}×\frac{5+\sqrt{17}}{5+\sqrt{17}}\end{vmatrix}$$
To remove square root term from denominator, we multiply by its conjugate i.e.,$$5+\sqrt{17}\space\text{in both numerator and denominator.}$$
$$=\frac{1}{\sqrt{17}}\space\text{log}\begin{vmatrix}\frac{(5+\sqrt{17})^2}{(5)^2-)(\sqrt{17})^2}\end{vmatrix}\\\lbrack\because (a-b)(a+b)=a^2-b^2\rbrack\\=\frac{1}{\sqrt{17}}\text{log}\begin{vmatrix}\frac{25+17+10\sqrt{17}}{25-17}\end{vmatrix}\\=\frac{1}{\sqrt{17}}\text{log}\begin{vmatrix}\frac{42+10\sqrt{17}}{8}\end{vmatrix}\\=\frac{1}{\sqrt{17}}\text{log}\begin{vmatrix}\frac{21+5\sqrt{17}}{4}\end{vmatrix}$$
$$\textbf{7.\space}\int^{\textbf{1}}_{\textbf{\normalsize-1}}\frac{\textbf{1}}{\textbf{x}^\textbf{2}\textbf{+2x+5}}\textbf{dx}\\\textbf{Sol.\space}\int^{1}_{-1}\frac{1}{x^2+2x+5}dx=\\\int^{1}_{-1}\frac{1}{(x^2+2x+1)+4}dx\\=\int^{1}_{\normalsize-1}\frac{1}{(x+1)^2+2^2}dx\\=\frac{1}{2}\bigg[\text{tan}^{\normalsize-1}\frac{x+1}{2}\bigg]^{1}_{\normalsize-1}\\\bigg[\because\int\frac{dx}{a^2+x^2}=\frac{1}{a}\text{tan}^{\normalsize-1}\frac{x}{a}\bigg]$$
$$=\frac{1}{2}\bigg[\text{tan}^{\normalsize-1}\bigg(\frac{2}{2}\bigg)-\text{tan}^{\normalsize-1}\bigg(\frac{0}{2}\bigg)\bigg]\\=\frac{1}{2}[\text{tan}^{\normalsize-1}]=\frac{1}{2}×\frac{\pi}{4}=\frac{\pi}{8}.$$
$$\textbf{8.\space}\int^{\textbf{2}}_{\textbf{1}}\bigg(\frac{\textbf{1}}{\textbf{x}}-\frac{\textbf{1}}{\textbf{2x}^\textbf{2}}\bigg)\textbf{e}^{\textbf{2x}}\textbf{dx}\\\textbf{Sol.\space}\text{Let}\space\text{I}=\int^{2}_{1}\bigg(\frac{1}{x}-\frac{1}{2x^2}\bigg)e^{2x}dx\\=\int^{2}_{1}\frac{1}{x}e^{2x}dx-\int^{2}_{1}\frac{1}{2x^2}e^{2x}dx$$
Evaluate the first integral by rule of integration by parts taking
$$\frac{1}{x}\space\text{as the first function, we get}$$
$$\text{I = }\bigg[\frac{1}{x}.\frac{e^{2x}}{2}\bigg]^{2}_{1}+\int^{2}_{1}\frac{1}{x^2}.\frac{e^{2x}}{2}dx-\\\int^{2}_{1}\frac{1}{2x^2}e^{2x}dx\\=\bigg[\frac{1}{x}.\frac{e^{2x}}{2}\bigg]^{2}_{1}=\frac{e^{4}}{4}-\frac{e^{2}}{2}\\=\frac{e^{2}}{2}\bigg(\frac{e^{2}}{2}-1\bigg)$$
Choose the correct answer.
9. The value of the integral
$$\int^{\textbf{1}}_{\textbf{1/3}}\frac{\textbf{(x-x}^\textbf{3}\textbf{)}^{\textbf{1/3}}}{\textbf{x}^\textbf{4}}\textbf{dx}\space\textbf{is}:$$
(a) 6
(b) 0
(c) 3
(d) 4
Sol. (a) 6
$$\text{Let\space I}=\int^{1}_{\frac{1}{3}}\frac{(x-x^3)^{\frac{1}{3}}}{x^4}dx\\=\int^{1}_{\frac{1}{3}}\frac{(x^3)^{\frac{1}{3}}\bigg(\frac{x}{x^3}-\frac{x^3}{x^3}\bigg)^{\frac{1}{3}}}{x^4}dx$$
[Multiply and divide the numerator by (x3)1/3]
$$=\int^{1}_{\frac{1}{3}}\frac{\bigg(\frac{1}{x^2}-1\bigg)^{1/3}}{x^3}dx\\\text{Put}\space\frac{1}{x^2}=t\Rarr\space\frac{-2}{x^3}dx=dt\\\Rarr\space\frac{dx}{x^3}=\frac{dt}{(\normalsize-2)}$$
For limit when x = 1
$$\Rarr\space t=1\space\text{and when x =}\frac{1}{3}\\\Rarr\space t-3^2=9\space\bigg(\because t = \frac{1}{x^2}\bigg)\\\therefore\space\text{I}=\int^{1}_{9}(t-1)^{1/3}\frac{dt}{(\normalsize-2)}\\=-\frac{1}{2}\begin{bmatrix}\frac{(t-1)^{\frac{1}{3}+1}}{\frac{1}{3}+1}\end{bmatrix}^{1}_{9}\\=-\frac{1}{2}×\frac{3}{4}\bigg[(t-1)^{4/3}\bigg]^{1}_{9}\\=-\frac{3}{8}\bigg[(1-1)^{\frac{4}{3}}-(9-1)^{\frac{4}{3}}\bigg]\\=-\frac{3}{8}\bigg[0-(2^3)^{\frac{4}{3}}\bigg]=\\-\frac{3}{8}×(-16)=6$$
$$\textbf{10.\space}\textbf{If f(x)}=\int^{\textbf{x}}_{\textbf{0}}\textbf{t sintdt},\\\textbf{then f'(x) is}$$
(a) cos x + x sin x
(b) x sin x
(c) x cos x
(d) sin x + x cos x
Sol. (b) x sin x
$$\text{Given,\space}f(x)=\int^{x}_{0}\text{t sint dt}$$
On integrating by parts, we get
$$\text{f(x) = t}\int^{x}_{0}\text{sin t dt}-\\\int^{x}_{0}\bigg[\bigg(\frac{d}{dt}t\bigg)\int\text{sin t dt}\bigg]dt\\=[t(\text{- cos t})]^{x}_{0}-\int^{x}_{0}(-\text{cos t})dt\\\lbrack-\text{t\space cos t + sin t}\rbrack^{x}_{0}\\\Rarr\space\text{f(x)= -x\space cos x + sin x}$$
Now, differentiate w.r.t. x, we get
f'(x) = – [{x(– sin x)} + cos x] + cos x
= x sin x – cos x + cos x = x sin x
$$\text{Using the product rule}\space\\\frac{d}{dx}(u.v)=u\frac{d}{dx}+v\frac{du}{dx}$$
Share page on
