# NCERT Solutions for Class 12 Maths Chapter 7 - Integrals - Exercise 7.9

Exercise 7.9

Evaluate the definite integrals.

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

$$\textbf{2.}\space\int^{\textbf{3}}_{\textbf{2}}\frac{\textbf{1}}{\textbf{x}}\textbf{dx}\\\textbf{Sol.\space}\text{Let I}=\int^{3}_{2}\frac{1}{x}dx=[log x]^{3}_{2}\\=\text{log 3- log 2}=\frac{3}{2}$$

$$\textbf{3.\space}\int^{\textbf{2}}_{\textbf{1}}\textbf{(4x}^\textbf{3}\textbf{-5x}^\textbf{2}\textbf{+6x+9)}\textbf{dx.}\\\textbf{Sol.}\space\text{Let \space I}=\int^{2}_{1}(4x^3-5x^2+6x+9)dx\\=\bigg[\frac{4x^4}{4}-\frac{5x^3}{3}+\frac{6x^2}{2}+9x\bigg]^{2}_{1}\\=\bigg[x^4-\frac{5x^3}{3}+3x^2+9x\bigg]^{2}_{1}\\=\bigg[(2)^4-\frac{5(2)^3}{3}+3(2)^2+9(2)\bigg]-\\\bigg[(1)^4-\frac{5(1)^3}{3}+3(1)^2+9(1)\bigg]$$

$$=\bigg[16-\frac{40}{3}+12+18\bigg]-\\\bigg[1-\frac{5}{3}+3+9\bigg]\\=\bigg[46-\frac{40}{3}\bigg]-\bigg[13-\frac{5}{3}\bigg]\\=\bigg[\frac{138-40}{3}\bigg]-\bigg[\frac{39-5}{3}\bigg]\\=\frac{98}{3}-\frac{34}{3}=\frac{64}{3}$$

$$\textbf{4.\space}\int^{\frac{\pi}{\textbf{4}}}_{\textbf{0}}\space\textbf{sin 2x dx.}\\\textbf{Sol.}\space\text{Let}\space\text{I}=\int^{\frac{\pi}{4}}_{0}\space\text{sin2xdx}\\=\int^{\pi/4}_{0}\text{2 sin x cosx dx}\\\lbrack\because\space\text{sin 2x = 2 sinx cosx}\rbrack\\\text{Put sin x = t}\\\Rarr\space\text{cos x}=\frac{dt}{dx}\Rarr\space dx=\frac{dt}{\text{cos x}}\\\text{Upper limit, at x=}\frac{\pi}{4}\\\Rarr\space t=\text{sin}\frac{\pi}{4}=\frac{1}{\sqrt{2}}$$

Lower limit, at x = 0

$$\Rarr\space t=\text{sin 0 = 0}\\\therefore\space\text{I}=2\int^{\frac{1}{\sqrt{2}}}_{0}\space\text{t cos x}\frac{dt}{cos x}\\=2\int^{\frac{1}{\sqrt{2}}}_{0}\space\text{t dt}\\= 2\bigg[\frac{t^2}{2}\bigg]^{\frac{1}{\sqrt{2}}}_{0}\\=\bigg(\frac{1}{\sqrt{2}}\bigg)^2=\frac{1}{2}$$

Note : If we change the integral function by considerable another variable, then remember to change the limit or after integration change into the given variable and then apply the limit.

$$\textbf{5.\space}\int^{\frac{\pi}{\textbf{2}}}_{\textbf{0}}\space\textbf{cos 2x\space dx}\\\textbf{Sol.}\space\int^{\frac{\pi}{\textbf{2}}}_{\textbf{0}}\space\text{cos 2x dx}=\bigg[\frac{sin 2x}{2}\bigg]^{\frac{\pi}{2}}_{0}\\\bigg(\because\space\int\text{cos ax dx}=\frac{sin\space ax}{a}\bigg)\\=\frac{1}{2}[\text{sin}\space 2x]^{\frac{\pi}{2}}_{0}=\frac{1}{2}\bigg[\bigg(\text{sin 2}×\frac{\pi}{2}\bigg)-\text{sin}(0)\bigg]\\=\frac{1}{2}[0-0]=0$$

$$\textbf{6.\space}\int^{\textbf{5}}_{\textbf{4}}\space \textbf{e}^\textbf{x}\space \textbf{dx}\\\textbf{Sol.}\space\int^{5}_{4} e^xdx=[e^x]^{5}_{4}\\=[e^5-e^4]=e^4[e-1]$$

$$\textbf{7.}\space\int^{\frac{\pi}{\textbf{4}}}_{\textbf{0}}\space\textbf{tan x dx}\\\textbf{Sol.\space}\int^{\frac{\pi}{4}}_{0}\space\text{tan x dx}=\bigg[\text{-log|cos x|}\bigg]^{\frac{\pi}{4}}_{0}\\=-\text{log}\begin{vmatrix}\text{cos}\frac{\pi}{4}\end{vmatrix}-(-log\space|\text{cos 0}|)\\=-\space log\bigg(\frac{1}{\sqrt{2}}\bigg)+\text{log 1}= - \text{log}\space 2^{-1/2}+0\\\lbrack\because\space \text{log 1 = 0 and log}\space m^n= nlog m\rbrack\\=\frac{1}{2}\text{log 2}$$

$$\textbf{8.}\space\int^{\frac{\pi}{\textbf{4}}}_{\frac{\pi}{\textbf{6}}}\space\textbf{cosec x}\space \textbf{dx.}\\\textbf{Sol.\space}\text{I}=\int^{\frac{\pi}{4}}_{\frac{\pi}{6}}\space\text{cosec x dx}=\\\bigg[\text{log}|\text{cosec x -cot x}|\bigg]^{\frac{\pi}{4}}_{\frac{\pi}{6}}\\=\text{log}\begin{vmatrix}\text{cosec}\frac{\pi}{4}-\text{cot}\frac{\pi}{4}\end{vmatrix}-\\\text{log}\begin{vmatrix}\text{cosec}\frac{\pi}{6}-\text{cot}\frac{\pi}{6}\end{vmatrix}\\=\text{log}|\sqrt{2}-1- \log|2-\sqrt{3}||\\=\text{log}\begin{vmatrix}\frac{\sqrt{2}-1}{2-\sqrt{3}}\end{vmatrix}\\\bigg(\because\text{log a - log b}= log\frac{a}{b}\bigg)$$

$$\textbf{9.}\space\int^{\textbf{1}}_{\textbf{0}}\space\frac{\textbf{1}}{\sqrt{\textbf{1-x}^\textbf{2}}}\textbf{dx}\\\textbf{Sol.}\space\int^{1}_{0}\space\frac{1}{\sqrt{1-x^2}}dx\\=[\text{sin}^{\normalsize-1}x]^{1}_{0}\\\bigg(\because\space\int\frac{dx}{\sqrt{1-x^2}}=\text{sin}^{\normalsize-1}x\bigg)\\=\text{sin}^{\normalsize-1}-\text{sin}^{\normalsize-1}0=\frac{\pi}{2}-0\\=\frac{\pi}{2}$$

$$\textbf{10.\space}\int^{\textbf{1}}_{\textbf{0}}\frac{\textbf{1}}{\textbf{1+x}^\textbf{2}}\textbf{dx}\\\textbf{Sol.}\space\int^{1}_{0}\frac{1}{1+x^2}dx\\=[\text{tan}^{\normalsize-1}x]^{1}_{0}\\\bigg(\because\space\int\frac{dx}{1+x^2}=\text{tan}^{\normalsize-1}x\bigg)\\=\text{tan}^{\normalsize-1}1-\text{tan}^{\normalsize-1}0=\frac{\pi}{4}$$

$$\textbf{11.}\space\int^{\textbf{3}}_{\textbf{2}}\space\frac{\textbf{1}}{\textbf{x}^\textbf{2}\textbf{-1}}\space\textbf{dx}\\\textbf{Sol.}\space\int^{3}_{2}\frac{1}{x^2-1}dx=\bigg[\frac{1}{2}\text{log} \begin{vmatrix}\frac{x-1}{x+1}\end{vmatrix}\bigg]^{3}_{2}\\\bigg[\because\space\int\frac{dx}{x^2-a^2}=\frac{1}{2a}\text{log}\begin{vmatrix}\frac{x-a}{x+a}\end{vmatrix}\bigg]\\=\frac{1}{2}\text{log} \begin{vmatrix}\frac{3-1}{3+1}\end{vmatrix}-\frac{1}{2}\text{log}\begin{vmatrix}\frac{2-1}{2+1}\end{vmatrix}\\=\frac{1}{2}\text{log}\begin{vmatrix}\frac{2}{4}\end{vmatrix}-\frac{1}{2}\text{log}\begin{vmatrix}\frac{1}{3}\end{vmatrix}\\=\frac{1}{2}\bigg[\text{log}\begin{vmatrix}\frac{1}{2}\end{vmatrix}-\text{log}\begin{vmatrix}\frac{1}{3}\end{vmatrix}\bigg]\\\frac{1}{2}[\text{log}(1)-\text{log}(2)-\text{log}(1)+\text{log}(3)]$$

$$\bigg[\because\space\text{log}\bigg(\frac{a}{b}\bigg)=\text{log a - log b}\bigg]\\=\frac{1}{2}\text{log}\bigg(\frac{3}{2}\bigg)$$

$$\textbf{12.}\space\int^{\frac{\pi}{\textbf{2}}}_{\textbf{0}}\space\textbf{cos}^\textbf{2}\textbf{x dx.}\\\textbf{Sol.}\space\int^{\frac{\pi}{\text{2}}}_{0}\space\text{cos}^2 xdx=\int^{\frac{\pi}{2}}_{0}\bigg(\frac{\text{cos 2x+1}}{2}\bigg)dx\\(\because\space \text{cos 2x = 2 cos}^2 x-1)\\=\frac{1}{2}\int^{\frac{\pi}{2}}_{0}\space\text{cos 2x dx}+\frac{1}{2}\int^{\frac{\pi}{2}}_{0}\space 1 dx\\=\frac{1}{2}\bigg[\frac{\text{sin 2x}}{2}\bigg]^{\frac{\pi}{2}}_{0}+\frac{1}{2}[x]^{\frac{\pi}{2}}_{0}\\=\frac{1}{2}\bigg[\bigg(\frac{\text{sin}\space \pi}{2}-\frac{\text{sin 0}}{2}\bigg)+\bigg(\frac{\pi}{2}-0\bigg)\bigg]\\=\frac{\pi}{4}+0-0=\frac{\pi}{4}.$$

$$\textbf{13.}\space\int^{\textbf{3}}_{\textbf{2}}\space\frac{\textbf{x}}{\textbf{x}^\textbf{2}\textbf{+1}}\\\textbf{Sol.}\space\text{Let}\space\text{I}=\int^{3}_{2}\frac{x}{x^2+1}dx\\\text{Put\space}x^2+1=t\\\Rarr\space 2x=\frac{dt}{dx}\Rarr\space dx=\frac{dt}{2x}\\\therefore\space\text{I}=\int^{3}_{2}\frac{x}{t}\frac{dt}{2x}=\frac{1}{2}\int^{3}_{2}\frac{1}{t}dt\\=\frac{1}{2}[\text{log}|t|]^{3}_{2}=\frac{1}{2}[\text{log}|x^2+1|]^{3}_{2}$$

(∵ t = x2 + 1)

$$=\frac{1}{2}\bigg[\text{log}|3^2+1|-\text{log}|2^2+1|\bigg]\\=\frac{1}{2}[\text{log}|10|-\text{log}|5|]\\=\frac{1}{2}\text{log}\begin{vmatrix}\frac{10}{5}\end{vmatrix}=\frac{1}{2}\text{log 2}\\\bigg(\because\space \text{log a- log b = log}\frac{a}{b}\bigg)$$

$$\textbf{14.}\space\int^{\textbf{1}}_{\textbf{0}}\space\frac{\textbf{2x+3}}{\textbf{(}\textbf{5x}^\textbf{2}\textbf{+1}\textbf{)}}\textbf{dx}\\\textbf{Sol.\space}\text{Let}\space\text{I}=\int^{1}_{0}\frac{2x+3}{(5x^2+1)}dx\\=\int^{1}_{0}\frac{2x}{(5x^2+1)}dx+3\int^{1}_{0}\frac{1}{(5x^2+1)}dx\\\text{Put 5x}^2+1=t\Rarr 10x=\frac{dt}{dx}\\\Rarr\space dx=\frac{dt}{10x}$$

When x = 0, t = 0, When x = 1, t = 6

$$\therefore\space\text{I}=\int^{6}_{0}\frac{2x}{t}\frac{dt}{10 x}+\frac{3}{5}\int^{1}_{0}\frac{1}{x^2+\bigg(\frac{1}{\sqrt{5}}\bigg)^2}dx\\=\frac{1}{5}\int^{6}_{0}\frac{1}{t}dt+\frac{3}{5}\int^{1}_{0}\frac{1}{x^2+\bigg(\frac{1}{\sqrt{5}}\bigg)^2}dx\\=\frac{1}{5}[\text{log}]^{6}_{0}+\frac{3}{5}×\frac{1}{\frac{1}{\sqrt{5}}}\bigg[\text{tan}^{\normalsize-1}\frac{x}{\frac{1}{\sqrt{5}}}\bigg]^{1}_{0}\\\bigg(\because\space\int\frac{dx}{a^2+x^2}=\frac{1}{a}\text{tan}{\normalsize-1}\frac{x}{a}\bigg)\\=\frac{1}{5}[\text{log}|6-log 1|]^{1}_{0}+\frac{3}{5}\sqrt{5}[\text{tan}^{\normalsize-1}\sqrt{5}x]^{1}_{0}$$

$$=\frac{1}{5}\text{log 6}+\frac{3}{\sqrt{5}}[\text{tan}^{\normalsize-1}\sqrt{5}(1)-\text{tan}^{\normalsize-1}\sqrt{5}(0)]\\=\frac{1}{5}\text{log 6}+\frac{3}{\sqrt{5}}\text{tan}^{\normalsize-1}\sqrt{5}$$

(∵ log 1 = 0 and tan–1 (0) = 0)

$$\textbf{15.\space}\int^{\textbf{1}}_{\textbf{0}}\space \textbf{xe}^{\textbf{x}^\textbf{2}}\textbf{dx.}\\\textbf{Sol.}\space\text{Let}\space\text{I}=\int^{1}_{0}xe^{x^{2}}dx.\\\text{Put}\space x^2=t\\\Rarr 2x=\frac{dt}{dx}\\\Rarr\space dx=\frac{dt}{2x}$$

When x = 0

$$\Rarr\space t=0\space\text{and\space when x=1}\Rarr t=1 \\\therefore\space\text{I}=\int^{1}_{0}xe^{t}\frac{dt}{2x}=\frac{1}{2}\int e^{t}dt=\frac{1}{2}[e^t]^{1}_{0}\\=\frac{1}{2}[e^1-e^0]=\frac{1}{2}[e-1]$$

$$\textbf{16.\space}\int^{\textbf{2}}_{\textbf{1}}\space\frac{\textbf{5x}^\textbf{2}}{\textbf{x}^\textbf{2}\textbf{+4x+3}}\textbf{dx}\\\textbf{Sol.\space}\text{let}\space\text{I}=\int^{2}_{1}\frac{5x^2}{x^2+4x+3}dx$$

Here, the degree of numerator and denominator is same, so on dividing numerator by denominator, we get

$$\frac{5x^2}{x^2+4x+3}=5-\frac{20x-15}{x^2+4x+3}\\\therefore\space\text{I}=\int^{2}_{1}\bigg[5-\frac{20x+15}{x^2+4x+3}\bigg]dx\\=\int^{2}_{1} 5dx-\int^{2}_{1}\frac{20x+15}{(x+1)(x+3)}dx$$

...(i)

$$\text{Let}\space\frac{20x+15}{(x+1)(x+3)}dx\\=\frac{A}{(X+1)}+\frac{B}{(X+2)}$$

On equating the coefficients of x and constant term on both sides, we get
20 = A + B ...(ii)
and
15 = 3A + B ...(iii)

On solving equations (ii) and (iii), w
e get

$$\Rarr\space 20x + 15 = A(x+3)+B(x+1)\\\Rarr\space 20x+15=Ax+3A+Bx+B$$

On equating the coefficients of x and constant term on both sides, we get

$$A=-\frac{5}{2},B=\frac{45}{2}\\\therefore\space \text{I}=5\int^{2}_{1}\space 1 dx+\frac{5}{2}\int^{2}_{1}\frac{1}{(x+1)}dx-\\\frac{45}{2}\int^{2}_{1}\frac{1}{(x+3)}dx\\\lbrack\text{from eq.(i)}\rbrack\\=\bigg[5x+\frac{5}{2}\text{log}|x+1|-\frac{45}{2}\text{log}|x+3|\bigg]^{2}_{1}\\= \begin{bmatrix}10+\frac{5}{2}\text{log}|3|-\frac{45}{2}\text{log}|5|-5-\\\frac{5}{2}\text{log}|2|+\frac{45}{2}\text{log}|4|\end{bmatrix}\\\Rarr\space\text{I}= 5+\frac{5}{2}\text{log}\begin{vmatrix}\frac{3}{2}\end{vmatrix}-\frac{45}{2}\text{log}\begin{vmatrix}\frac{5}{4}\end{vmatrix}\\\bigg[\because\space\text{m log a - m log b = m log}\bigg(\frac{a}{b}\bigg)\bigg]$$

$$\textbf{17.\space}\int^{\frac{\pi}{\textbf{4}}}_{\textbf{0}}\space(\textbf{2 sec}^\textbf{2}\textbf{x} \textbf{+ x}^\textbf{3}\textbf{+2})\textbf{dx.}\\\textbf{Sol.}\space\int^{\frac{\pi}{\textbf{4}}}_{\textbf{0}}(\text{2 sec}^\text{2}\text{x} + x^3+2)dx\\=2\int^{\frac{\pi}{4}}_{0}\text{sec}^2\space x dx + \int^{\frac{\pi}{4}}_{0}\space x^3 dx+\\2\int^{\frac{\pi}{\textbf{4}}}_{\textbf{0}}\space 1 dx\\=2[\text{tan x}]^{\frac{\pi}{4}}_{0}+\frac{1}{4}[x^4]^{\frac{\pi}{4}}_{0}+2[x]^{\frac{\pi}{4}}_{0}\\= 2\text{tan}\frac{\pi}{4}+\frac{1}{4}\bigg(\frac{\pi}{4}\bigg)^4+2\bigg(\frac{\pi}{4}\bigg)\\=2+\frac{\pi}{2}+\frac{\pi^4}{1024}$$

$$\textbf{18.\space}\int^{\pi}_{\textbf{0}}\bigg(\textbf{sin}^{\textbf{2}}\frac{\textbf{x}}{\textbf{2}}-\textbf{cos}^{\textbf{2}}\frac{\textbf{x}}{\textbf{2}}\bigg)\textbf{dx.}\\\textbf{Sol.}\space\int^{\pi}_{\textbf{0}}\bigg(\text{sin}^{2}\frac{x}{2}-\text{cos}^{2}\frac{x}{2}\bigg)dx-\\\int^{\pi}_{0}\bigg(\text{cos}^{2}\frac{\pi}{2}-\text{sin}^{2}\frac{x}{2}\bigg)dx\\=-\int^{\pi}_{0}\space\text{cos xdx}\\\lbrack\because\space\text{cos 2x = cos}^{2}x-\text{sin}^{2}x\rbrack\\=-\lbrack\text{sin x}\rbrack^{\pi}_{0}=-\lbrack\text{sin}\pi - \text{sin 0}\rbrack=0$$

$$\textbf{19.\space}\int^{\textbf{2}}_{\textbf{0}}\space\frac{\textbf{6x+3}}{\textbf{x}^\textbf{2}\textbf{+4}}\textbf{dx.}\\\textbf{Sol.}\space\text{Let}\space\text{I}=\int^{2}_{0}\frac{6x+3}{x^2+4}dx\\=\int^{2}_{0}\frac{6x}{x^2+4}dx+\int^{2}_{0}\frac{3}{x^2+4}dx\\\text{Put}\space x^2+4=t\\\Rarr\space 2x=\frac{dt}{dx}\\\Rarr\space dx=\frac{dt}{2x}$$

Lower limit, when x = 0, t = 0 + 4 = 4

upper limit, when x = 2, t = 4 + 4 = 8

$$\therefore\space\text{I}=\int^{8}_{4}\frac{6x}{t}\frac{dt}{2x}+\int^{2}_{0}\frac{3}{x^2+4}dx\\=3\int^{8}_{4}\frac{1}{t}dt+ 3\int^{2}_{0}\frac{1}{x^2+2^2}dx\\=3[\text{log t}]^{8}_{4}+\frac{3}{2}\bigg[\text{tan}^{\normalsize-1}\frac{x}{2}\bigg]^{2}_{0}\\\bigg[\because\space\int\frac{dx}{a^2+x^2}=\frac{1}{a}\text{tan}^{\normalsize-1}\frac{x}{a}\bigg]\\=3[\text{log}(8)-\text{log}(4)+\frac{3}{2}\bigg[\text{tan}^{\normalsize-1}\frac{2}{2}\bigg]]\\=\text{3 log}\bigg(\frac{8}{4}\bigg)+\frac{3}{2}×\frac{\pi}{4}\\\bigg[\because\space\text{log b - log a}=\text{log}\frac{b}{a}\bigg]$$

$$=\text{3 log 2}+\frac{3\pi}{8}$$

$$\textbf{20.\space}\int^{\textbf{1}}_{\textbf{0}}\bigg(\textbf{xe}^\textbf{x}\textbf{+}\textbf{sin}\frac{\pi \textbf{x}}{\textbf{4}}\bigg)\textbf{dx.}\\\textbf{Sol.\space}\int^{1}_{0}\bigg(xe^{x}+\text{sin}\frac{\pi.x}{4}\bigg)dx\\=\int^{1}_{0}xe^{x}dx+\int^{1}_{0}\text{sin}\bigg(\frac{\pi.x}{4}\bigg)dx\\=[xe^x-\int 1e^x dx]^{1}_{0}+\bigg[\frac{-\text{cos}\frac{\pi.x}{4}}{\frac{\pi}{4}}\bigg]^{1}_{0}$$

First integral solved by using the integration by parts and for second

$$\bigg(\int\text{sin xdx}=-\frac{\text{cos ax}}{a}\bigg)\\=[xe^x-e^x]^{1}_{0}-\frac{4}{\pi}\bigg[\text{cos}\bigg(\frac{\pi.x}{4}\bigg)\bigg]^{1}_{0}\\=(e-e)-(0-e^0)-\frac{4}{\pi}\bigg(\text{cos}\frac{\pi}{4}-\text{cos 0}\bigg)\\=-(\normalsize-1)-\frac{4}{\pi}\bigg(\frac{1}{\sqrt{2}}-1\bigg)=\\1-\frac{4}{\pi}\bigg(\frac{1-\sqrt{2}}{2}\bigg)×\frac{\sqrt{2}}{\sqrt{2}}\\=1+\frac{2}{\pi}(2-\sqrt{2})\\=1+\frac{4}{\pi}-\frac{2\sqrt{2}}{\pi}.$$

$$\textbf{21.}\int^{\sqrt{\textbf{3}}}_{\textbf{1}}\bigg(\frac{\textbf{1}}{\textbf{1+x}}\bigg)\textbf{dx\space equals :}\\\textbf{(a)\space}\frac{\pi}{\textbf{3}}\\\textbf{(b)\space}\frac{\textbf{2}\pi}{\textbf{3}}\\\textbf{(c)\space}\frac{\pi}{\textbf{6}}\\\textbf{(d)\space}\frac{\pi}{\textbf{12}}\\\textbf{Sol.}\space\text{(d)\space}\frac{\pi}{12}\\\int^{\sqrt{3}}_{1}\frac{1}{1+x^2}dx=\int^{\sqrt{3}}_{1}\frac{1}{x^2+1}dx\\=\bigg[\frac{1}{1}\text{tan}^{\normalsize-1}\bigg(\frac{x}{1}\bigg)\bigg]^{\sqrt{3}}_{1}$$
$$=\text{tan}^{\normalsize-1}\sqrt{3}-\text{tan}^{\normalsize-1}1\\\bigg(\because\space\int\frac{dx}{1+x^2}=\text{tan}^{\normalsize-1}x\bigg)\\=\frac{\pi}{3}-\frac{\pi}{4}\\\Rarr\space\frac{4\pi-3\pi}{12}=\frac{\pi}{12}.$$
$$\textbf{22.\space}\int^{\frac{\textbf{2}}{\textbf{3}}}_{\textbf{0}}\frac{\textbf{1}}{\textbf{4+9x}^\textbf{2}}\textbf{dx}\\\textbf{(a)\space}\frac{\pi}{\textbf{6}}\\\textbf{(b)\space}\frac{\pi}{\textbf{12}}\\\textbf{(c)\space}\frac{\pi}{\textbf{24}}\\\textbf{(d)\space}\frac{\pi}{\textbf{4}}\\\textbf{Sol.\space}\text{(c)}\space\frac{\pi}{24}\\=\int^{\frac{2}{3}}_{0}\frac{1}{4+9x^2}dx=\\\frac{1}{9}\int^{\frac{2}{3}}_{0}\frac{1}{\frac{4}{9}+x^2}dx$$
$$=\frac{1}{9}\int^{\frac{2}{3}}_{0}\frac{1}{\bigg(\frac{2}{3}\bigg)^2+x^2}dx\\=\frac{1}{9}.\frac{1}{2/3}\bigg[\text{tan}^{\normalsize-1}\bigg(\frac{x}{2/3}\bigg)\bigg]^{\frac{2}{3}}_{0}\\\bigg[\because\space\int\frac{dx}{a^2+x^2}=\frac{1}{a}\text{tan}^{\normalsize-1}\frac{x}{a}\bigg]\\=\frac{1}{6}\bigg[\text{tan}^{\normalsize-1}\bigg(\frac{3x}{2}\bigg)\bigg]^{2/3}_{0}\\=\frac{1}{6}\bigg[\text{tan}^{\normalsize-1}\bigg(\frac{3}{2}.\frac{2}{3}\bigg)-\text{tan}^{\normalsize-1}0\bigg]\\=\frac{1}{6}(\text{tan}^{\normalsize-1}1-0)=\frac{1}{6}.\frac{\pi}{4}=\frac{\pi}{24}.$$