Class 12 Maths NCERT Solutions Chapter 7 Ex 7.1 – Anti Derivatives & Indefinite Integrals

\dfrac{d}{dx}(\cos 2x) = -2\sin 2x

\Rightarrow \sin 2x = \dfrac{d}{dx}\left(-\dfrac12\cos 2x\right)

\dfrac{d}{dx}(\sin 3x) = 3\cos 3x

\Rightarrow \cos 3x = \dfrac{d}{dx}\left(\dfrac13\sin 3x\right)

\dfrac{d}{dx}(e^{2x}) = 2e^{2x}

\Rightarrow e^{2x} = \dfrac{d}{dx}\left(\dfrac12 e^{2x}\right)

By the chain rule, \dfrac{d}{dx}\left[\dfrac{(ax+b)^3}{3a}\right] = \dfrac{3(ax+b)^2\cdot a}{3a} = (ax+b)^2

From Q1, anti derivative of \sin 2x is -\dfrac12\cos 2x.

\dfrac{d}{dx}(e^{3x}) = 3e^{3x} \Rightarrow anti derivative of e^{3x} is \dfrac13 e^{3x}, so anti derivative of -4e^{3x} is -\dfrac43 e^{3x}.

Let I = \displaystyle\int (4e^{3x}+1)\,dx = 4\displaystyle\int e^{3x}dx + \int dx

Using \int e^{ax}dx = \dfrac{e^{ax}}{a} and \int dx = x: I = 4\cdot\dfrac{e^{3x}}{3} + x

x^2\left(1-\dfrac{1}{x^2}\right) = x^2 - 1

Let I = \displaystyle\int (x^2-1)\,dx. Using \int x^n dx = \dfrac{x^{n+1}}{n+1}: I = \dfrac{x^3}{3} - x

Let I = \displaystyle\int (ax^2+bx+c)\,dx = a\displaystyle\int x^2 dx + b\displaystyle\int x\,dx + c\displaystyle\int dx

Using \int x^n dx = \dfrac{x^{n+1}}{n+1}: I = a\cdot\dfrac{x^3}{3} + b\cdot\dfrac{x^2}{2} + cx

Let I = \displaystyle\int (2x^2+e^x)\,dx = 2\displaystyle\int x^2 dx + \displaystyle\int e^x dx = \dfrac{2x^3}{3} + e^x

\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)^2 = x - 2 + \dfrac{1}{x}

Let I = \displaystyle\int \left(x - 2 + \dfrac{1}{x}\right)dx. Using \int x\,dx = \dfrac{x^2}{2} and \int \dfrac{1}{x}\,dx = \log|x|: I = \dfrac{x^2}{2} - 2x + \log|x|

\dfrac{x^3+5x^2-4}{x^2} = x + 5 - \dfrac{4}{x^2}

Let I = \displaystyle\int \left(x + 5 - \dfrac{4}{x^2}\right)dx = \dfrac{x^2}{2} + 5x - 4\left(-\dfrac1x\right) = \dfrac{x^2}{2} + 5x + \dfrac{4}{x}

\dfrac{x^3+3x+4}{\sqrt{x}} = x^{5/2} + 3x^{1/2} + 4x^{-1/2}

Let I = \displaystyle\int \left(x^{5/2} + 3x^{1/2} + 4x^{-1/2}\right)dx. Using \int x^n dx = \dfrac{x^{n+1}}{n+1}: I = \dfrac27 x^{7/2} + 2x^{3/2} + 8x^{1/2}

x^3-x^2+x-1 = x^2(x-1)+1(x-1) = (x-1)(x^2+1)

\therefore \dfrac{x^3-x^2+x-1}{x-1} = x^2+1   (for x\neq1)

Let I = \displaystyle\int (x^2+1)\,dx = \dfrac{x^3}{3} + x

(1-x)\sqrt{x} = x^{1/2} - x^{3/2}

Let I = \displaystyle\int \left(x^{1/2} - x^{3/2}\right)dx = \dfrac23 x^{3/2} - \dfrac25 x^{5/2}

\sqrt{x}\,(3x^2+2x+3) = 3x^{5/2}+2x^{3/2}+3x^{1/2}

Let I = \displaystyle\int \left(3x^{5/2}+2x^{3/2}+3x^{1/2}\right)dx = \dfrac67 x^{7/2} + \dfrac45 x^{5/2} + 2x^{3/2}

Let I = \displaystyle\int (2x-3\cos x+e^x)\,dx = \displaystyle\int 2x\,dx - 3\displaystyle\int \cos x\,dx + \displaystyle\int e^x\,dx

I = x^2 - 3\sin x + e^x

Let I = \displaystyle\int (2x^2-3\sin x+5\sqrt{x})\,dx = \displaystyle\int 2x^2\,dx - 3\displaystyle\int \sin x\,dx + 5\displaystyle\int x^{1/2}dx

I = \dfrac23 x^3 - 3(-\cos x) + 5\cdot\dfrac23 x^{3/2} = \dfrac23 x^3 + 3\cos x + \dfrac{10}{3}x^{3/2}

\sec x(\sec x+\tan x) = \sec^2 x + \sec x\tan x

Let I = \displaystyle\int (\sec^2 x + \sec x\tan x)\,dx. Using \int \sec^2 x\,dx = \tan x and \int \sec x\tan x\,dx = \sec x: I = \tan x + \sec x

\dfrac{\sec^2 x}{\text{cosec}^2 x} = \dfrac{\sin^2 x}{\cos^2 x} = \tan^2 x

Let I = \displaystyle\int \tan^2 x\,dx. Using the identity \tan^2 x = \sec^2 x - 1: I = \displaystyle\int(\sec^2 x - 1)\,dx = \tan x - x

\dfrac{2-3\sin x}{\cos^2 x} = \dfrac{2}{\cos^2 x} - 3\cdot\dfrac{\sin x}{\cos x}\cdot\dfrac{1}{\cos x} = 2\sec^2 x - 3\sec x\tan x

Let I = \displaystyle\int (2\sec^2 x - 3\sec x\tan x)\,dx. Using \int \sec^2 x\,dx = \tan x and \int \sec x\tan x\,dx = \sec x: I = 2\tan x - 3\sec x

\sqrt{x}+\dfrac{1}{\sqrt{x}} = x^{1/2}+x^{-1/2}

Let I = \displaystyle\int \left(x^{1/2}+x^{-1/2}\right)dx. Using \int x^n dx = \dfrac{x^{n+1}}{n+1}: I = \dfrac23 x^{3/2} + 2x^{1/2}

Given f'(x)=4x^3-\dfrac{3}{x^4} = 4x^3-3x^{-4}

\therefore f(x) = \displaystyle\int (4x^3-3x^{-4})\,dx = x^4 - 3\cdot\left(-\dfrac{1}{3}x^{-3}\right) = x^4+\dfrac{1}{x^3}+C  ...(1)

Given f(2)=0. Putting x=2 in (1): 2^4+\dfrac{1}{2^3}+C=0 \Rightarrow 16+\dfrac18+C=0 \Rightarrow C=-\dfrac{129}{8}