# NCERT Solutions for Class 12 Maths Chapter 7 - Integrals - Exercise 7.1

Exercise 7.1

1. sin 2x.

$$\textbf{Sol.}\space\int\space sin 2x\space dx=-\frac{1}{2}\text{cos 2x}\\\begin{bmatrix}\because\space\frac{d}{dx}\bigg(-\frac{1}{2}\text{cos 2x}\bigg)=\\-\frac{1}{2}(-\text{sin 2x})2=\text{sin2x}\end{bmatrix}$$

2. cos 3x.

$$\textbf{Sol.}\space\int\space\text{cos 3x\space dx}=\frac{1}{3}\space\text{sin 3x}\\\begin{pmatrix}\because\space\frac{d}{dx}\bigg(\frac{1}{3}\text{sin 3x}\bigg)=\\\frac{1}{3}\text{(cos 3x)3}=\text{cos 3x}\end{pmatrix}$$

3. e2x.

$$\textbf{Sol.}\space\int e^{2x}dx=\frac{1}{2}e^{2x}\\\bigg(\because\space\frac{d}{dx}\bigg(\frac{1}{2}e^{2x}\bigg)=\frac{1}{2}e^{2x}2=e^{2x}\bigg)$$

4. (ax + b)2.

$$\textbf{Sol.}\space\int(ax+b)^2 dx=\frac{1}{3a}(ax+b)^3\\\begin{pmatrix}\because\space\frac{d}{dx}\bigg(\frac{1}{3a}(ax+b)^3\bigg)=\\\frac{1}{3a}.3(ax+b)^2a=(ax+b)^2\end{pmatrix}$$

5. sin 2x – 4e3x.

$$\textbf{Sol.}\space\int\text{(sin 2x - 4 e}^{3x})dx\\=\int\text{sin 2xdx - 4}\int e^{3x}dx\\=\frac{-\text{cos 2x}}{2}-4\frac{e^{3x}}{3}\\=-\frac{1}{2}\text{cos 2x}-\frac{4}{3}e^{3x}\\\begin{pmatrix}\because\space\frac{d}{dx}\begin{Bmatrix}-\frac{1}{2}\text{cos 2x}=\end{Bmatrix}\\-\frac{1}{2}(-\text{sin 2x})2= sin\space2x\end{pmatrix}\\=\text{sin 2x and}\frac{d}{dx}\bigg(\frac{4}{3}e^{3x}\bigg)\\=\frac{4}{3}(e^{3x}3)=4e^{3x}$$

Direction (Q.6 to 20) : Find the following integrals.
$$\textbf{6.}\space\int(\textbf{4 e}^{\textbf{3x}}\textbf{+1})\textbf{dx.}\\\textbf{Sol.\space}\int (4e^{3x}+1)dx=\\4\int e^{3x}dx+\int\space 1dx\\=\frac{4e^{3x}}{3}+x+C\\\bigg[\because\space\int e^{nx}dx=\frac{e^{nx}}{n}\bigg]$$

$$\textbf{7.\space}\int \textbf{x}^2\bigg(\textbf{1}-\frac{\textbf{1}}{\textbf{x}^\textbf{2}}\bigg)\space \textbf{dx.}\\\textbf{Sol.}\space\int x^2\bigg(1-\frac{1}{x^2}\bigg)dx=\\\int\bigg(x^2-\frac{x^2}{x^2}\bigg)dx\\=\int(x^2-1)dx=\int x^2dx-\int 1 dx\\=\frac{x^3}{3}x+C$$

8. ∫ (ax2 + bx + c)dx.

Sol. ∫ (ax2 + bx + c) dx = ∫ ax2 dx + ∫ bx dx + c ∫ 1 dx

$$=\space\frac{ax^3}{3}+\frac{bx^2}{2}+cx+k$$

(where, K is a constant of integration)

9. ∫ (2x2 + ex) dx.

Sol. ∫ (2x2 + ex) dx = 2 ∫ x2dx + ∫ ex dx

$$=\frac{2x^3}{3}+e^x+C$$

$$\textbf{10.}\space\int\bigg(\sqrt{\textbf{x}}-\frac{\textbf{1}}{\sqrt{\textbf{x}}}\bigg)^\textbf{2}\textbf{ dx.}\\\textbf{Sol.}\space\int\bigg(\sqrt{x}-\frac{1}{\sqrt{x}}\bigg)^2 dx\\=\int\bigg[(\sqrt{x})^2+\bigg(\frac{1}{x}\bigg)^2-2\sqrt{x}×\frac{1}{\sqrt{x}}\bigg]dx$$

[∵ (a – b)2 = a2 + b2 – 2ab]

$$=\int\space\bigg(x+\frac{1}{x}-2\bigg)dx=\\\int xdx+\int\frac{1}{x}dx-2\int 1 dx\\=\frac{x^2}{2}+\text{log}|x|-2x+C.$$

$$\textbf{11.}\space\frac{\textbf{x}^\textbf{3}\textbf{+5x}^\textbf{2}\textbf{-4}}{\textbf{x}^\textbf{2}}\textbf{dx.}\\\textbf{Sol.\space}\int\frac{x^3+5x^2-4}{x^2}dx=\\\int\frac{x^3}{x^2}dx+5\int\frac{x^2}{x^2}dx-4\int\frac{1}{x^2}dx\\=\int x\space dx + 5\int 1.dx -4\int x^{\normalsize-2}dx\\=\frac{x^2}{2}+5x-4\bigg(\frac{x^{-2+1}}{-2+1}\bigg)+C\\=\frac{x^2}{2}+5x-4\bigg(\frac{x^{\normalsize-1}}{-1}\bigg)+\text{C}\\=\frac{x^2}{2}+5x+\frac{4}{x}+\text{C}.$$

$$\textbf{12}.\space\int\frac{\textbf{x}^\textbf{3}\textbf{+3x+4}}{\sqrt{\textbf{x}}}\textbf{dx}.\\\textbf{Sol.}\space\int\frac{x^3+3x+4}{\sqrt{x}}dx\\=\int\frac{x^3}{\sqrt{x}}dx+3\int\frac{x}{\sqrt{x}}dx+4\int\frac{1}{\sqrt{x}}dx$$

= ∫ x3 × x– 1/2 dx + 3 ∫ x1/2 dx + 4 ∫ x– 1/2 dx

= ∫ x(6 – 1)/2 dx + 3 ∫ x1/2 dx + 4 ∫ x–1/2 dx

= ∫ x5/2 dx + 3 ∫ x1/2 dx + 4 ∫ x–1/2 dx

$$=\frac{x^{(5/2)+1}}{(5/2)+1}+\frac{3x^{(1/2)+1}}{(1/2)+1}+\frac{4x^{(-1/2)+1}}{(-1/2)+1}+\text{C}\\\bigg(\because\space \int x^ndx=\frac{x^{n+1}}{n+1}\bigg)\\=\frac{x^{7/2}}{7/2}+\frac{3x^{3/2}}{3/2}+\frac{4x^{1/2}}{1/2}+\text{C}\\=\frac{2}{7}x^{7/2}+\frac{2}{3}x^{3/2}+8x^{1/2}+\text{C}$$

$$\textbf{13.}\space\int\frac{\textbf{x}^\textbf{3}\textbf{-x}^\textbf{2+x}\textbf{-1}}{\textbf{(x-1)}}\textbf{dx}\\\textbf{Sol.}\space\int\frac{x^3-x^2+x-1}{(x-1)}dx\\=\int\frac{x^2(x-1)+1(x-1)}{(x-1)}dx\\=\int\frac{(x^2+1)(x-1)}{(x-1)}dx\\=\int(x^2+1)dx=\frac{x^3}{3}+x+C$$

$$\textbf{14.}\space\int\textbf{(1-x)}\sqrt{\textbf{x}}\space\textbf{dx}\\\textbf{Sol.}\space\int(1-x)\sqrt{x}dx=\\\int\bigg(\sqrt{x}-x^{\frac{3}{2}}\bigg)dx\\=\frac{x^{(1/2)+1}}{(1/2)+1}-\frac{x^{(3/2)+1}}{(3/2)+1}+\text{C}\\=\frac{x^{(3/2)}}{(3/2)}-\frac{x^{(5/2)}}{(5/2)}+\text{C}\\\bigg(\because\space \int x^ndx=\frac{x^{n+1}}{n+1}\bigg)\\=\frac{2}{3}x^{3/2}-\frac{2}{5}x^{5/2}+\text{C}$$

$$\textbf{15.\space}\int\sqrt{\textbf{x}}(\textbf{3x}^\textbf{2}\textbf{+2x+3)}\textbf{dx}\\\textbf{Sol.}\space\int\sqrt{x}(3x^2+2x+3)dx\\=3\int x^{5/2}dx+2\int x^{3/2}dx+3\int x^{1/2}dx\\=\frac{3.x^{(5/2)+1}}{(5/2)+1}+\frac{2.x^{(3/2)+1}}{(3/2)+1}+\frac{3.x^{(1/2)+1}}{(1/2)+1}+\text{C}\\=\frac{3x^{7/2}}{7/2}+\frac{2x^{5/2}}{5/2}+\frac{3x^{3/2}}{3/2}+\text{C}\\=\frac{6}{7}x^{7/2}+\frac{4}{5}x^{5/2}+2x^{3/2}+\text{C}$$

16. ∫ (2x – 3 cos x + ex) dx.

Sol. ∫ (2x – 3 cos x + ex) dx

= 2 ∫x dx – 3 ∫ cos x dx + ∫ ex dx

$$=\frac{2x^2}{2}-3\space\text{sin x + e}^x+\text{C}$$

= x2 – 3 sin x + ex + C

$$\textbf{17.}\space\int\textbf{(2x}^\textbf{2}\textbf{- 3 sin x + 5}\sqrt{\textbf{x}})\textbf{dx.}\\\textbf{Sol.}\space\int(2x^2-3 sin x+5\sqrt{x})dx\\=2\int x^2dx-3\int\text{sin xdx}+5\int x^{1/2}dx\\=\frac{2x^3}{3}-3(-\text{cos x})+5\frac{x^{(1/2)+1}}{(1/2)+1}+\text{C}\\=\frac{2}{3}x^3+3 cos x+\frac{5 ×2}{3}x^{3/2}+\text{C}\\=\frac{2}{3}x^3+3\space\text{cos x}+\frac{10}{3}x^{3/2}+C$$

18. ∫ sec x (sec x + tan x) dx.

Sol. ∫ sec x (sec x + tan x) dx

= ∫ (sec2 x + sec x. tan x) dx

= ∫ sec2 x dx + ∫ sec x.tan x dx

= tan x + sec x + C

$$\textbf{19.}\space\int\frac{\textbf{sec}^\textbf{2}\textbf{x}}{\textbf{cosec}^\textbf{2}\textbf{x}}\textbf{dx.}\\\textbf{Sol.}\space\int\frac{\text{sec}^2x}{\text{cosec}^2x}dx=\\\int\frac{\text{sin}^2dx}{\text{cos}^2dx}dx$$

= ∫ tan2 x dx

= ∫ (sec2 x – 1) dx

$$\begin{pmatrix}\because\space\frac{1}{\text{cosec}^2x}=\text{sin}^2x\space\text{and sec}^2x=\frac{1}{\text{cos}^2x}\end{pmatrix}$$

= ∫ sec2 x dx – ∫ 1dx

= tan x – x + C

$$\textbf{20.}\space\int\frac{\textbf{2-3 sin x}}{\textbf{cos}^\textbf{2}\textbf{x}}\textbf{dx.}$$

$$\textbf{Sol.}\space\int\frac{2-3\space\text{sin x}}{\text{cos}^2x}dx\\=\int\frac{2}{\text{cos}^2x}dx-3\int\frac{\text{sin x}}{\text{cos}^2x}dx$$

= 2 ∫ sec2 x dx – 3 ∫ tan x.sec x dx

= 2 tan x – 3 sec x + C

Choose the correct answer in Q 21 and 22 :

$$\textbf{21. The anti-derivative of}\\\bigg(\sqrt{\textbf{x}}+\frac{\textbf{1}}{\sqrt{\textbf{x}}}\bigg)\space\textbf{equals}:$$

$$\textbf{(a)}\space\frac{\textbf{1}}{{3}}\textbf{x}^{\textbf{1/3}}\textbf{+}\textbf{2x}^{\textbf{1/2}}\textbf{+ C}\\\textbf{(b)}\space\frac{\textbf{2}}{\textbf{3}}x^\textbf{2/3}+\frac{\textbf{1}}{\textbf{2}}\textbf{x}^\textbf{2}+\textbf{C}\\\textbf{(c)}\space\frac{\textbf{2}}{\textbf{3}}\textbf{x}^{\textbf{3/2}}+\textbf{2x}^{\textbf{1/2}}+\textbf{C}\\\textbf{(d)}\space\frac{\textbf{3}}{\textbf{2}}\textbf{x}^{\textbf{3/2}}+\frac{\textbf{1}}{\textbf{2}}\textbf{x}^{\textbf{1/2}}+\textbf{C}\\\textbf{Sol.}\space(c)\space\frac{2}{3}x^{3/2}+2x^{1/2}+\text{C}\\\int\bigg(\sqrt{x}+\frac{1}{\sqrt{x}}\bigg)dx\\=\int\sqrt{x}dx+\int\frac{1}{\sqrt{x}}dx$$

$$=\int x^{1/2}dx+\int x^{-1/2}dx\\=\frac{x^{3/2}}{3/2}+\frac{x^{1/2}}{1/2}+\text{C}\\=\frac{2}{3}x^{3/2}+2x^{1/2}+\text{C}$$

$$\textbf{22. If}\space\frac{\textbf{d}}{\textbf{dx}} \textbf{f(x)}\textbf{= 4x}^\textbf{3}\textbf{-}\frac{\textbf{3}}{\textbf{x}^\textbf{4}}\space\\\textbf{such that f(2) = 0. Then, f(x) is :}\\\textbf{(a)}\space \textbf{x}^\textbf{4}+\frac{\textbf{1}}{\textbf{3}}-\frac{\textbf{129}}{\textbf{8}}\\\textbf{(b)\space}\textbf{x}^\textbf{3}+\frac{\textbf{1}}{\textbf{x}^\textbf{4}}+\frac{\textbf{129}}{\textbf{8}}\\\textbf{(c)\space}\textbf{x}^\textbf{4}+\frac{1}{\textbf{x}^\textbf{3}}+\frac{\textbf{129}}{\textbf{8}}\\\textbf{(d)\space}\textbf{x}^\textbf{3}+\frac{\textbf{1}}{\textbf{x}^\textbf{4}}-\frac{\textbf{129}}{\textbf{8}}\\\textbf{Sol.}\space(a)\space x^4+\frac{1}{x^3}-\frac{129}{8}\\\text{Given,}\space\frac{d}{dx}\text{f(x)}=4x^3-\frac{3}{x^4}$$

$$\Rarr\space\text{Anti-derivative of}\space\bigg(4x^3-\frac{3}{x^4}\bigg)=\text{f(x)}\\\therefore\space\text{f(x)}=\int\bigg(4x^3-\frac{3}{x^4}\bigg)dx\\=4\int x^3 dx-3\int x^{\normalsize-4} dx\\\Rarr\space\text{f(x)}=4\bigg(\frac{x^4}{4}\bigg)-3\bigg(\frac{x^{\normalsize-3}}{-3}\bigg)+\text{C}\\= x^4+\frac{1}{x^3}+\text{C}\\\text{Also, f(2) = 0}\\\therefore\space f(2)=(2)^4+\frac{1}{(2)^3}+\text{C}$$

$$\Rarr\space 16+\frac{1}{8}+\text{C}=0\\\Rarr\space \text{C}=-\bigg(16+\frac{1}{8}\bigg)\\\Rarr\space \text{C}=-\frac{129}{8}\\\therefore\space\text{f(x)}=x^4+\frac{1}{x^3}-\frac{129}{8}.$$