# Integrals

## Class 12 NCERT

### NCERT

1   Find an anti-derivative (or integral) of the function $\sin 2x$ by the method of inspection.

The anti-derivative of $\sin 2x$ is a function of $x$ whose derivative is $\sin 2x$. It is known that, $\\$ $\dfrac{d}{dx}(\cos2x) =-2\sin2x\\ \implies \sin 2x=-\dfrac{1}{2}\dfrac{d}{dx}(cos 2x)\\ \therefore \sin2x =\dfrac{d}{dx}(-\dfrac{1}{2}\cos 2x)$$\\ Therefore, the anti-derivative of \sin 2x is -\dfrac{1}{2}\cos2x. 2 Find an anti-derivative (or integral) of the function \cos 3x by the method of inspection. ##### Solution : The anti-derivative of \cos 3x is a function of x whose derivative is \cos 3x.$$\\$ It is known that,$\\$ $\dfrac{d}{dx}(\sin3x)=3\cos3x\\ \implies \cos3x =\dfrac{1}{3}\dfrac{d}{dx}(\sin3x)\\ \therefore \cos3x=\dfrac{d}{dx}(\dfrac{1}{3}\sin3x)$$\\ Therefore, the anti-derivative of \cos3x is \dfrac{1}{3}\sin3x 3 Find an anti-derivative (or integral) of the function e^{ 2x} by the method of inspection. ##### Solution : The anti–derivative of e^{ 2x} is the function of x whose derivative is e^{ 2x} It is known that,\\ \dfrac{d}{dx}(e^{2x}) =2e^{2x}\\ \implies e^{2x} = \dfrac{1}{2}\dfrac{d}{dx}(e^{2x})\\ \therefore e^{2x} =\dfrac{d}{dx}(\dfrac{1}{2}e^{2x}).$$\\$ Therefore, the anti-derivative of $e^{2x}$ is $\dfrac{1}{2}e^{2x}$

4   Find an anti-derivative (or integral) of the function $(ax + b)^ 2$ by the method of inspection.

The anti-derivative of $(ax + b)^ 2$ is the function of $x$ whose derivative is $(ax + b)^ 2$.$\\$ It is known that,$\\$ $\dfrac{d}{dx}(ax+b)^3=3a(ax+b)^2\\ \implies (ax+b)^2=\dfrac{1}{3a}\dfrac{d}{dx}(ax+b)^3\\ \therefore (ax+b)^2=\dfrac{d}{dx}(\dfrac{1}{3a}(ax+b)^3)$$\\ Therefore, the anti-derivative of (ax+b)^2 is \dfrac{1}{3a}(ax+b)^3 5 Find an anti-derivative (or integral) of the function \sin 2x - 4e^{ 3x} by the method of inspection. ##### Solution : The anti-derivative of \sin 2x - 4e^{ 3x} is the function of x whose derivative is \sin2x - 4e^{ 3x} It is known that,\\ \dfrac{d}{dx}(-\dfrac{1}{2}\cos2x-\dfrac{4}{3}e^{3x}) =\\ \sin2x-4e^{3x}$$\\$ Therefore, the anti derivative of $(\sin2x -4e^{3x})$ is $(-\dfrac{1}{2}\cos2x-\dfrac{4}{3}e^{3x})$

6   $\int (4e^{3x}+1)\mathrm dx$

##### Solution :

$\int (4e^{3x}+1)\mathrm dx $$\\ =4 \int e^{3x} \mathrm dx+\int 1 \mathrm dx\\ 4(\dfrac{e^{3x}}{3})+x+C\\ \dfrac{4}{3}e^{3x}+x+C$$\\$ where $C$ is an arbitrary constant.

7   $\int x^2(1-\dfrac{1}{x^2}) \mathrm dx$

##### Solution :

$\int x^2(1-\dfrac{1}{x^2}) \mathrm dx$$\\ =\int(x^2-1)\mathrm dx\\ \int x^2 \mathrm dx -\int 1 \mathrm dx \\ \dfrac{x^3}{3}-x+C$$\\$ where $C$ is an arbitrary constant.

8   $\int (ax^2+ bx+c)\mathrm dx$

##### Solution :

$\int (ax^2+ bx+c)\mathrm dx$$\\ =a \int x^2 \mathrm dx + b\int x \mathrm dx +c \int 1. \mathrm dx\\ a(\dfrac{x^3}{3})+b(\dfrac{x^2}{2})+cx + C\\ =\dfrac{ax^3}{3}+\dfrac{bx^2}{2} + cx + C$$\\$ where $C$ is an arbitrary constant.

9   $\int (2x^2+e^x) \mathrm dx$

##### Solution :

$\int (2x^2+e^x) \mathrm dx$$\\ =2 \int x^2 \mathrm dx+ \int e^x \mathrm dx\\ =2(\dfrac{x^3}{3})+e^x+C\\ =\dfrac{2}{3}x^3+ e^x+C$$\\$ where $C$ is an arbitrary constant.

10   $\int (\sqrt{x}-\dfrac{1}{\sqrt{x}})^2 \mathrm dx$

##### Solution :

$\int (\sqrt{x}-\dfrac{1}{\sqrt{x}})^2 \mathrm dx$$\\ =\int (x+\dfrac{1}{x}-2)\mathrm dx\\ =\int x \mathrm dx+\int \dfrac{1}{x} \mathrm dx -2\int 1.\mathrm dx \\ =\dfrac{x^2}{2}+\log|x|-2x+C$$\\$ where $C$ is an arbitrary constant.

11   $\int \dfrac{x^3+ 5x^2-4}{x^2} \mathrm dx$

##### Solution :

$\int \dfrac{x^3+ 5x^2-4}{x^2} \mathrm dx$$\\ =\int (x+5-4x^{-2})\mathrm dx\\ =\int x \mathrm dx + 5\int 1.\mathrm dx -4\int x^{-2}\mathrm dx\\ \dfrac{x^2}{2}+5x-4(\dfrac{x^{-1}}{-1})+C\\ =\dfrac{x^2}{2}+5x+\dfrac{4}{x}+C$$\\$ where $C$ is an arbitrary constant.

12   $\int \dfrac{x^3+3x+4}{\sqrt{x}}\mathrm dx$

##### Solution :

$\int \dfrac{x^3+3x+4}{\sqrt{x}}\mathrm dx$$\\ =\int(x^{\frac{5}{2}}+3x^{\frac{1}{2}}+4x^{-\frac{1}{2}})\mathrm dx\\ \dfrac{x^{\frac{7}{2}}}{\frac{7}{2}}\\ +\dfrac{3(x^{\frac{3}{2}})}{\frac{3}{2}}\\ +\dfrac{4(x^{\frac{1}{2}})}{\dfrac{1}{2}}+C\\ =\dfrac{2}{7}x^{\frac{7}{2}}+2x^{3}{2}+8x^{\frac{1}{2}}+C\\ =\dfrac{2}{7}x^(\frac{7}{2})+2x^{\frac{3}{2}}+8\sqrt{x}+C$$\\$ where $C$ is an arbitrary constant.

13   $\int \dfrac{x^3-x^2+x-1}{x-1} \mathrm dx$

##### Solution :

$\int \dfrac{x^3-x^2+x-1}{x-1} \mathrm dx $$\\ On factorising, we obtain\\ \int \dfrac{(x^2+1)(x-1)}{x-1} \mathrm dx\\ =\int (x^2+1) \mathrm dx\\ =\int x^2 \mathrm dx + \int 1 \mathrm dx \\ \dfrac{x^3}{3}+x+C$$\\$ where $C$ is an arbitrary constant.

14   $\int (1-x)\sqrt{x} \mathrm dx$

##### Solution :

$\int (1-x)\sqrt{x} \mathrm dx $$\\ =\int(\sqrt{x}-x^{\frac{3}{2}}) \mathrm dx\\ =\int x^{\frac{1}{2}}\mathrm dx -\int x^{\frac{3}{2}} \mathrm dx\\ =\dfrac{x^{\frac{3}{2}}}{\dfrac{3}{2}}-\dfrac{x^{\frac{5}{2}}}{\dfrac{5}{2}}+C\\ =\dfrac{2}{3}x^{\frac{3}{2}}-\dfrac{2}{5}x^{5}{2}+C$$\\$ where $C$ is an arbitrary constant.

15   $\int \sqrt{x}(3x^2+2x+3)\mathrm dx$

##### Solution :

$\int \sqrt{x}(3x^2+2x+3)\mathrm dx $$\\ =3\int x^{\frac{5}{2}}\mathrm dx+2\int x^{\frac{3}{2}}\mathrm dx+3\int x ^{\frac{1}{2}} \mathrm dx\\ =3(\dfrac{x^{\frac{7}{2}}}{\dfrac{7}{2}})+2(\dfrac{x^{\frac{5}{2}}}{\dfrac{5}{2}})\\ +3 \dfrac{(x^{\frac{3}{2}})}{\dfrac{3}{2}}+C\\ =\dfrac{6}{7}x^{\frac{7}{2}}+\dfrac{4}{5}x^{\frac{5}{2}}+2x^{\frac{3}{2}}+C$$\\$ where $C$ is an arbitrary constant.

16   $\int (2x-3 \cos x + e^x) \mathrm dx$

##### Solution :

$\int (2x-3 \cos x + e^x) \mathrm dx$$\\ =2\int x \mathrm dx -3 \int \cos x \mathrm dx +\int e^x \mathrm dx\\ =\dfrac{2 x^2}{2}-3(\sin x)+e^x+C\\ =x^2-3 \sin x +e^x+C$$\\$ where $C$is an arbitrary constant.

17   $\int (2x^2-3 \sin x +5\sqrt{x})\mathrm dx$

##### Solution :

$\int (2x^2-3 \sin x +5\sqrt{x})\mathrm dx$$\\ =2\int x^2 \mathrm dx -3 \int \sin x \mathrm dx\\ +5\int x^{\frac{1}{2}}\mathrm dx \\ =\dfrac{2 x^3}{3}-3(-\cos x)+5(\dfrac{x^{\frac{3}{2}}}{\dfrac{3}{2}})+C\\ =\dfrac{2}{3}x^3+3 \cos x+\dfrac{10}{3}x^{\frac{3}{2}}+C$$\\$ where $C$ is an arbitrary constant.

18   $\int \sec x(\sec x+ \tan x)\mathrm dx$

##### Solution :

$\int \sec x(\sec x+ \tan x)\mathrm dx$$\\ =\int (\sec^2 x+\sec x \tan x)\mathrm dx\\ =\int \sec^2 x \mathrm dx +\int \sec x \tan x \mathrm dx\\ =\tan x + \sec x + C$$\\$ where $C$ is an arbitrary constant.

19   $\int \dfrac{\sec^2 x}{\csc^2 x}\mathrm dx$

##### Solution :

$\int \dfrac{\sec^2 x}{\csc^2 x}\mathrm dx$$\\ =\int \dfrac{\dfrac{1}{\cos ^2 x}}{\dfrac{1}{\sin^2 x}}\mathrm dx\\ =\int \dfrac{\sin^2 x}{\cos ^2 x}\mathrm dx\\ =\int \tan^2 x \mathrm dx\\ =\int (\sec^2 x-1) \mathrm dx\\ =\int \sec^2 x \mathrm dx -\int 1 \mathrm dx \\ =\tan x- x+C$$\\$ where $C$ is an arbitrary constant.

20   $\int \dfrac{2-3 \sin x}{\cos^2 x}\mathrm dx$

$\int \dfrac{2-3 \sin x}{\cos^2 x}\mathrm dx\\ =\int (\dfrac{2}{\cos^2 x}-\dfrac{3 \sin x}{\cos^2 x})\mathrm dx\\ =\int 2 \sec^2 x \mathrm dx -3 \int \tan x \sec x \mathrm dx\\ =2 \tan x -3\sec x +C$$\\ where C is an arbitrary constant. 21 The anti-derivative of (\sqrt{x}+\dfrac{1}{\sqrt{x}}) equals ##### Solution : \int \sqrt{x}+\dfrac{1}{\sqrt{x}}\mathrm dx \\ =\int x^{\frac{1}{2}} \mathrm dx + \int x^{\frac{1}{2}} \mathrm dx\\ =\dfrac{x^{\frac{3}{2}}}{\dfrac{3}{2}}\\ +\dfrac{x^{\frac{1}{2}}}{\dfrac{1}{2}}+C\\ =\dfrac{2}{3}x^{\frac{3}{2}}+2x^{\frac{1}{2}}+C,$$\\$ where $C$ is an arbitrary constant.$\\$ Hence, the correct Answer is C.

22   If $\dfrac{\mathrm d}{\mathrm dx} f(x)=4x^3-\dfrac{3}{x^4}$such that $f(2)=0,$ then $f(x)$ is

It is given that,$\dfrac{\mathrm d}{\mathrm d} f(x)=4x^3-\dfrac{3}{x^4}$$\\ Anti-derivative of 4 x^3-\dfrac{3}{x^4}=f(x)\\ \therefore f(x)=\int 4 x^3-\dfrac{3}{x^4}=f(x)\\ f(x)=4\int x^3 \mathrm dx -3 \int (x^{-4}) \mathrm dx\\ f(x)=4(\dfrac{x^4}{4})-3 (\dfrac{x^{-3}}{-3})+C\\ f(x) x^4 + \dfrac{1}{x^3}+C$$\\$ Also,$\\$ $f(2)=0\\ \therefore f(2) =(2)^4+\dfrac{1}{(2)^3}+C=0\\ \implies 16+\dfrac{1}{8}+C=0\\ \implies C=-(16+\dfrac{1}{8})\\ \implies C=\dfrac{-129}{8}\\ \therefore f(x)=x^4+ \dfrac{1}{x^3}-\dfrac{129}{8}$$\\ Hence, the correct Answer is A. 23 Integrate \dfrac{2x}{1+x^2} ##### Solution : Let 1+x^2 =t\\ \therefore 2x \mathrm dx =\mathrm dt\\ \implies \int \dfrac{2x}{1+x^2}\mathrm dx\\ =\int \dfrac{1}{t} \mathrm dt\\ =\log|t|+C\\ =\log|1+x^2|+C\\ \log(1+x^2)+C$$\\$ where C is an arbitrary constant.

24   Integrate$\dfrac{(\log x)^2}{x}$

Let $\log x=t\\ \therefore \dfrac{1}{x} \mathrm dx =\mathrm dt\\ \implies \int \dfrac{(\log|x|)^2}{x}\mathrm dx \\ =\int t^2 \mathrm dt \\ =\dfrac{t^3}{3} + C\\ =\dfrac{(\log|x|)^3}{3}+C$$\\where C is an arbitrary constant. 25 Integrate \dfrac{1}{x+ x \log x} ##### Solution : The given function can be rewritten as\\ \dfrac{1}{x+x \log x}=\dfrac{1}{x(1+ \log x)}$$\\$ Let $1+\log x =t \\ \therefore \dfrac{1}{x}\mathrm dx=\mathrm dt\\ \implies \int \dfrac{1}{x(1+\log x)}\mathrm dx\\ =\int \dfrac{1}{t}\mathrm dt\\ =\log|t|+C\\ =\log |1+\log x|+C$$\\where C is an arbitrary constant. 26 Integrate \sin x . \sin (\cos x) ##### Solution : Let \cos x = t\\ \therefore -\sin x \mathrm dx = dt\\ \implies \int \sin x .\sin ( \cos x ) \mathrm dx =-\int \sin t \mathrm dt\\ =-[- \cos t ]+ C\\ =\cos t + C\\ = \cos ( \cos x )+ C$$\\$ where C is an arbitrary constant.

27   Integrate $\sin (ax + b) \cos (ax + b)$

##### Solution :

The given function can be rewritten as$\\$ $\sin (ax + b) \cos (ax + b) =\\ \dfrac{2 \sin(ax + b)\cos (ax +b)}{2}\\ =\dfrac{\sin 2(ax+b)}{2}$$\\ Let 2(ax+b)=t\\ \therefore 2 a \mathrm dx = \mathrm dt\\ \implies \int \dfrac{\sin 2 (ax+b)}{2}\mathrm dx\\ =\dfrac{1}{2}\int \dfrac{\sin t \mathrm dt}{2a}\\ =\dfrac{1}{4a}[-\cos t]+C\\ =\dfrac{-1}{4a}\cos 2(ax+b)+C$$\\$ where C is an arbitrary constant.

28   Integrate $\sqrt{ax+b}$

Let $ax + b = t\\ \implies a \mathrm dx=\mathrm dt\\ \therefore \mathrm dx=\dfrac{1}{a}\mathrm dt\\ \implies \int(ax+b)^{\frac{1}{2}} \mathrm dx\\ =\dfrac{1}{a}\int t^{\frac{1}{2}} \mathrm dt\\ =\dfrac{1}{a}(\dfrac{t^{\frac{1}{2}}}{\dfrac{3}{2}})+C\\ =\dfrac{2}{3a}(ax+b^{\frac{3}{2}})+C$$\\ where C is an arbitrary constant. 29 Integrate x\sqrt{x+2} ##### Solution : Let x + 2 = t\\ \therefore \mathrm dx = \mathrm dt\\ \implies \int x \sqrt{x+2}\mathrm dx\\ =\int (t-2)\sqrt{t}\mathrm dt\\ =\int (t^{\frac{3}{2}}-2t^{\frac{1}{2}}) \mathrm dt\\ =\int t^{\frac{3}{2}} \mathrm dt - 2 \int t^{\frac{1}{2}} \mathrm dt\\ =\dfrac{t^{\frac{5}{2}}}{\dfrac{5}{2}}\\ -2 (\dfrac{t^{\frac{3}{2}}}{\dfrac{3}{2}})+C\\ =\dfrac{2}{5}t^{\frac{5}{2}}-\dfrac{4}{3}t^{\frac{3}{2}}+C\\ =\dfrac{2}{5}(x+2)^{\frac{5}{2}}\\ -\dfrac{4}{3}(x+2)^{\frac{3}{2}}+C$$\\$where C is an arbitrary constant.

30   $x\sqrt{1+2x^2}$

Let $1+2x^2=t\\ \therefore 4x \mathrm dx =\mathrm dt\\ \implies \int x\sqrt{1+2x^2} \mathrm dx\\ =\int \dfrac{\sqrt{t}}{4}\mathrm dt\\ =\dfrac{1}{4}\int t^{\frac{1}{2}} \mathrm dt\\ =\dfrac{1}{4}(\dfrac{t^{\frac{3}{2}}}{\dfrac{3}{4}})+C\\ =\dfrac{1}{6}(1+2x^2)^{\frac{3}{2}}+C$$\\ where C is an arbitrary constant. 31 Integrate(4x+2)\sqrt{x^2+x+1} ##### Solution : Let x^2+x+1=t\\ \therefore (2x+1)\mathrm dx=\mathrm dt\\ \int (4x+2)\sqrt{x^2+x+1 }\mathrm dx\\ =\int 2 \sqrt{t}\mathrm dt\\ =2\int \sqrt{t}\mathrm dt\\ =2(\dfrac{t^{\frac{3}{2}}}{\dfrac{3}{2}})+C\\ =\dfrac{4}{3}(x^2+x+1)^{\frac{3}{2}}+C$$\\$ where C is an arbitrary constant.

32   Integrate $\dfrac{1}{x-\sqrt{x}}$

##### Solution :

The given function can be rewritten as$\\$ $\dfrac{1}{x-\sqrt{x}}=\dfrac{1}{\sqrt{x}(\sqrt{x}-1)}$$\\ Let (\sqrt{x}-1)=t\\ \therefore \dfrac{1}{2\sqrt{x}}\mathrm dx =\mathrm dt\\ \implies \int\dfrac{1}{\sqrt{x}(\sqrt{x-1})}\mathrm dx\\ =\int \dfrac{2}{t}\mathrm dt\\ =2 \log|t|+C\\ =2 \log|\sqrt{x}-1|+C$$\\$where C is an arbitrary constant.

33   Integrate $\dfrac{x}{\sqrt{x+4}},x> 0$

##### Solution :

Let $x+4=t$$\\ \therefore \mathrm dx=\mathrm dt\\ \int \dfrac{x}{\sqrt{x+4} } \mathrm dx\\ =\int \dfrac{(t-4)}{\sqrt{t}} \mathrm dt\\ =\int(\sqrt{t}-\dfrac{4}{\sqrt{t}})\mathrm dt$$\\$ $=\dfrac{t^{\frac{3}{2}}}{\dfrac{3}{2}}-4(\dfrac{t^{\frac{1}{2}}}{\dfrac{1}{2}})+C\\ =\dfrac{2}{3}(t)^{\frac{3}{2}}-8(t)^{\frac{1}{2}}+C\\ =\dfrac{2}{3}t.(t)^{\frac{1}{2}}-8(t)^{\frac{1}{2}}+C\\ =\dfrac{2}{3}t(t)^{\frac{1}{2}}(t-12)+C\\ =\dfrac{2}{3}(x+4)^{\frac{1}{2}}(x+4-12)+C\\ =\dfrac{2}{3}\sqrt{x+4}(x-8)+C$ $\\$ where C is an arbitrary constant.

34   Integrate $(x^3-1)^{\frac{1}{3}}x^5$

##### Solution :

Let $x^3 -1=t\\ \therefore 3 x^2 \mathrm dx =\mathrm dt\\ \implies \int(x^3-1)^{\frac{1}{3}} x^5 \mathrm dx\\ =\int(x^3-1)^{\frac{1}{3}} x^3.x^2 \mathrm dx\\ =\int t^{\frac{1}{3}}(t+1) \dfrac{\mathrm dt}{3}\\ =\dfrac{1}{3}\int(t^{\frac{4}{3}}+t^{\frac{1}{3}})\mathrm dt\\ =\dfrac{1}{3}[\dfrac{t^{\frac{7}{3}}}{\dfrac{7}{3}}+\dfrac{t^{\frac{4}{3}}}{\dfrac{4}{3}}] +C\\ =\dfrac{1}{3}[\dfrac{3}{7}t^{\frac{7}{3}}+\dfrac{3}{4}t^{\frac{4}{3}}]+C\\ =\dfrac{1}{7}(x^3-1)^{\frac{7}{3}}+\dfrac{1}{4}(x^3-1)^{\frac{4}{3}}+C$$\\$ where C is an arbitrary constant.