# NCERT Solutions for Class 12 Maths Chapter 7 - Integrals - Exercise 7.10

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

$$\text{I}=2\begin{bmatrix}\theta\int\text{sec}^{2}d\theta-\\\int\begin{Bmatrix}\bigg(\frac{d}{d\theta}\bigg)\end{Bmatrix}\int\text{sec}^{2}\theta d\theta\end{bmatrix}^{\frac{\pi}{4}}_{0}$$

$$= 2\bigg[\theta\space\text{tan}\space\theta-\int 1\text{tan}\space\theta\space d\theta\bigg]^{\frac{\pi}{4}}_{0}\\(\because\space\int\text{sec}^{2}\theta d\theta=\text{tan}\space\theta)\\=2\lbrack\theta\space\text{tan}\space\theta+\text{log}|\text{cos}\space\theta|\rbrack^{\frac{\pi}{4}}_{0}\\(\because\space\int\text{tan}\theta d\theta=-\text{log cos}\space\theta)\\=2\begin{Bmatrix}\bigg(\frac{\pi}{4}\text{tan}\frac{\pi}{4}+\text{log}\begin{vmatrix}\text{cos}\frac{\pi}{4}\end{vmatrix}\bigg)-\\(10 + \text{log}|\text{cos 0}|)\end{Bmatrix}\\=\frac{\pi}{2}+2\text{log}\frac{1}{\sqrt{2}}-0\space(\because\space\text{log 1=0})\\=\frac{\pi}{2}+2\text{log}2^{\frac{-1}{2}}$$

$$\frac{\pi}{2}-2×\frac{1}{2}\text{log 2}\\(\because\space \text{log m}^n= n\space log\space m)\\=\frac{\pi}{2}-\text{log 2}$$

$$\textbf{4.\space}\int^{\textbf{2}}_{\textbf{0}}\textbf{x}\sqrt{\textbf{x+2}}\space\textbf{dx}\qquad[\text{Put x+2=t}^2]\\\textbf{Sol.\space}\int^{2}_{0}x\sqrt{x+2}dx=\\\int^{2}_{0}\lbrace(x+2)-2\rbrace\sqrt{x+2}\space\text{dx}$$

$$=\int^{2}_{0}\lbrace(x+2)^{3/2}-2(x+2)^{\frac{1}{2}}\rbrace dx\\=\bigg[\frac{(x+2)^{(\frac{3}{2})+1}}{(\frac{3}{2}+1)}-\frac{2(x+2)^{(\frac{1}{2})+1}}{(\frac{1}{2}+1)}\bigg]^{2}_{0}\\=\frac{2}{5}[(x+2)^{5/2}]^{2}_{0}-\frac{4}{3}[(x+2)^{\frac{3}{2}}]^{2}_{0}\\=\frac{2}{5}[4^{\frac{5}{2}}-2^{\frac{5}{2}}]-\frac{4}{3}[4^{\frac{3}{2}}-2^{\frac{3}{2}}]\\=\frac{2}{5}[2^5-2^{\frac{5}{2}}]-\frac{4}{3}[2^3-2^{\frac{3}{2}}]\\=\frac{2}{5}[32-4\sqrt{2}]-\frac{4}{3}[8-2\sqrt{2}]\\=\frac{8}{5}[8-\sqrt{2}]-\frac{8}{3}[4-\sqrt{2}]$$

$$=\bigg(\frac{64}{5}-\frac{32}{3}\bigg)+\bigg(\frac{8\sqrt{2}}{3}-\frac{8\sqrt{2}}{5}\bigg)\\=\frac{32}{15}+\frac{16\sqrt{2}}{15}$$

$$\textbf{5.\space}\int^{\frac{\pi}{\textbf{2}}}_{0}\frac{\textbf{sin x}}{\textbf{1 + cos}^\textbf{2}\textbf{x}}\textbf{dx}\\\textbf{Sol.\space}\text{let}\space\text{I}=\int^{\frac{\pi}{2}}_{0}\frac{\text{sin x}}{\text{1 + cos}^2x}dx\\\text{Put \space}\text{cos x}=t\\\Rarr\space -\text{sin x}\space dx=dt\\\Rarr\space dx=\frac{dt}{-\text{sin x}}\\\text{For limit when x=0}\\\Rarr\space -\text{sin x dx = dt}\Rarr\space dx=\frac{dt}{\text{- sin x}}\\\text{For limit when x = 0}\\\Rarr\space t=\text{cos 0 = 1}(\because\space t = cos x)\\\text{and when}\space x=\frac{\pi}{2}$$

$$\Rarr\space t=\text{cos}\frac{\pi}{2}=0\\\therefore\space\text{I}=\int^{0}_{1}\frac{\text{sin x}}{1+t^2}\frac{dt}{(-\text{sinx})}dx\\=-\int^{0}_{1}\frac{1}{1+t^2}dx\\\bigg(\because\space\int\frac{1}{a^2+x^2}dx=\frac{1}{a}\text{tan}^{\normalsize-1}\frac{x}{a}\bigg)\\=\bigg[\frac{1}{1}\text{tan}^{\normalsize-1}\bigg(\frac{t}{1}\bigg)\bigg]^{0}_{1}\\=[\text{tan}^{\normalsize-1}]^{0}_{1}\\=[\text{tan}^{\normalsize-1}(0)-\text{tan}^{\normalsize-1}(1)]\\=-\bigg(0-\frac{\pi}{4}\bigg)=\frac{\pi}{4}.$$

$$\textbf{6.\space}\int^{\textbf{2}}_{\textbf{0}}\frac{\textbf{1}}{\textbf{x+4-x}^\textbf{2}}\textbf{dx}\\\textbf{Sol.\space}\int^{2}_{0}\frac{1}{x+4-x^2}dx\\=\int^{2}_{0}\frac{1}{x+4-x^2-\bigg(\frac{1}{2}\bigg)^2+\bigg(\frac{1}{2}\bigg)^2}dx\\=\int^{2}_{0}\frac{1}{4+\frac{1}{4}-\bigg[\bigg(x^2-x+\frac{1}{4}\bigg)\bigg]}dx\\=\int^{2}_{0}\frac{1}{4+\frac{1}{4}-\bigg(x-\frac{1}{2}\bigg)^2}dx$$

$$=\int^{2}_{0}\frac{1}{\bigg(\frac{\sqrt{17}}{2}\bigg)^2-\bigg(x-\frac{1}{2}\bigg)^2}dx\\=\frac{1}{2.\frac{\sqrt{17}}{2}}\bigg[\text{log}\begin{vmatrix}\frac{\frac{\sqrt{17}}{2}+x-\frac{1}{2}}{\frac{\sqrt{17}}{2}-x+\frac{1}{2}}\end{vmatrix}\bigg]^{2}_{0}\\\bigg(\because\space\int\frac{dx}{a^2-x^2}=\frac{1}{2a}\text{log}\begin{vmatrix}\frac{a+x}{a-x}\end{vmatrix}\bigg)\\=\frac{1}{\sqrt{17}}\bigg[\text{log}\begin{vmatrix}\frac{\sqrt{17}+2x-1}{\sqrt{17}-2x+1}\end{vmatrix}\bigg]^{2}_{0}\\=\frac{1}{\sqrt{17}}\bigg[\text{log}\begin{vmatrix}\frac{\sqrt{17}+3}{\sqrt{17-3}}\end{vmatrix}-\text{log}\begin{vmatrix}\frac{\sqrt{17}-1}{\sqrt{17}+1}\end{vmatrix}\bigg]\\=\frac{1}{\sqrt{17}}\text{log}\begin{vmatrix}\frac{\sqrt{17}+3}{\sqrt{17}-3}÷\frac{\sqrt{17}-1}{\sqrt{17}-1}\end{vmatrix}$$

$$\bigg(\because\space \text{log m - log n}=\text{log}\frac{m}{n}\bigg)\\=\frac{1}{\sqrt{17}}\text{log}\begin{vmatrix}\frac{(\sqrt{17}+3)(\sqrt{17}+1)}{(\sqrt{17}-3)(\sqrt{17}-1)}\end{vmatrix}\\=\frac{1}{\sqrt{17}}\space\text{log}\begin{vmatrix}\frac{20+4\sqrt{17}}{20-4\sqrt{17}}\end{vmatrix}\\=\frac{1}{\sqrt{17}}\text{log}\begin{vmatrix}\frac{5+\sqrt{17}}{5-\sqrt{17}}×\frac{5+\sqrt{17}}{5+\sqrt{17}}\end{vmatrix}$$

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

$$\frac{1}{x}\space\text{as the first function, we get}$$

$$\text{I = }\bigg[\frac{1}{x}.\frac{e^{2x}}{2}\bigg]^{2}_{1}+\int^{2}_{1}\frac{1}{x^2}.\frac{e^{2x}}{2}dx-\\\int^{2}_{1}\frac{1}{2x^2}e^{2x}dx\\=\bigg[\frac{1}{x}.\frac{e^{2x}}{2}\bigg]^{2}_{1}=\frac{e^{4}}{4}-\frac{e^{2}}{2}\\=\frac{e^{2}}{2}\bigg(\frac{e^{2}}{2}-1\bigg)$$

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

$$\text{Using the product rule}\space\\\frac{d}{dx}(u.v)=u\frac{d}{dx}+v\frac{du}{dx}$$