Where Mindful Learning Begins
Where Mindful Learning Begins
These Class 12 Maths NCERT Solutions Chapter 7 Ex 7.1 cover all 22 questions — finding anti derivatives by the method of inspection, and basic indefinite integrals using the standard formulae for integration.
This is the opening exercise of the Integrals chapter, so it eases you in gently before the heavier techniques later on. Questions 1–5 are solved purely by inspection — recognising which function's derivative matches the given expression, without applying any formula mechanically. From Question 6 onward, you'll apply the standard integration results directly to powers of x, exponential functions, and trigonometric expressions, including a couple of MCQs at the end that test whether you can work backward from a derivative condition to recover the original function. Getting comfortable here matters: every substitution, by-parts, and partial-fraction technique later in the chapter still rests on these basic formulae.
We need a function whose derivative is \sin 2x. Recall \dfrac{d}{dx}(\cos 2x) = -2\sin 2x, so \sin 2x = \dfrac{d}{dx}\left(-\dfrac12\cos 2x\right).
We need a function whose derivative is \cos 3x. Recall \dfrac{d}{dx}(\sin 3x) = 3\cos 3x, so \cos 3x = \dfrac{d}{dx}\left(\dfrac13\sin 3x\right).
Since \dfrac{d}{dx}(e^{2x}) = 2e^{2x}, we get e^{2x} = \dfrac{d}{dx}\left(\dfrac12 e^{2x}\right).
Try \dfrac{(ax+b)^3}{3a} — 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.
As found in Q1, an anti derivative of \sin 2x is -\dfrac12\cos 2x.
Since \dfrac{d}{dx}(e^{3x}) = 3e^{3x}, an anti derivative of e^{3x} is \dfrac13 e^{3x}.
So an anti derivative of -4e^{3x} is -\dfrac43 e^{3x}.
Splitting the integral using the sum rule and applying \int e^{ax}dx = \dfrac{e^{ax}}{a}, \int dx = x: 4\cdot\dfrac{e^{3x}}{3} + x = \dfrac43 e^{3x} + x.
Expanding the integrand first: x^2\left(1-\dfrac{1}{x^2}\right) = x^2 - 1. Integrating term by term using \int x^n dx = \dfrac{x^{n+1}}{n+1} gives \displaystyle\int (x^2-1)\,dx = \dfrac{x^3}{3} - x.
Integrating each term separately using \int x^n dx = \dfrac{x^{n+1}}{n+1}:
\displaystyle\int ax^2\,dx = a\cdot\dfrac{x^3}{3}, \displaystyle\int bx\,dx = b\cdot\dfrac{x^2}{2}, \displaystyle\int c\,dx = cx.
Splitting and integrating term by term: \displaystyle\int 2x^2\,dx = \dfrac{2x^3}{3}, and \displaystyle\int e^x\,dx = e^x.
Expanding the square first: \left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)^2 = x - 2 + \dfrac{1}{x}.
Integrating term by term: \displaystyle\int x\,dx = \dfrac{x^2}{2}, \displaystyle\int 2\,dx = 2x, and \displaystyle\int \dfrac{1}{x}\,dx = \log|x|.
Dividing each term in the numerator by x^2: \dfrac{x^3+5x^2-4}{x^2} = x + 5 - \dfrac{4}{x^2}.
Integrating term by term gives \dfrac{x^2}{2} + 5x, and since \int 4x^{-2}\,dx = -\dfrac{4}{x}, subtracting this term contributes +\dfrac{4}{x}.
Dividing each term by x^{1/2}: \dfrac{x^3+3x+4}{\sqrt{x}} = x^{5/2} + 3x^{1/2} + 4x^{-1/2}.
Integrating term by term using \int x^n dx = \dfrac{x^{n+1}}{n+1}: \displaystyle\int x^{5/2}dx = \dfrac{2}{7}x^{7/2}, \displaystyle\int 3x^{1/2}dx = 2x^{3/2}, \displaystyle\int 4x^{-1/2}dx = 8x^{1/2}.
Factoring the numerator by grouping: x^3-x^2+x-1 = x^2(x-1)+1(x-1) = (x-1)(x^2+1). Cancelling the common factor (x-1) leaves x^2+1.
Integrating: \displaystyle\int (x^2+1)\,dx = \dfrac{x^3}{3} + x.
Expanding the integrand: (1-x)\sqrt{x} = x^{1/2} - x^{3/2}. Integrating term by term: \displaystyle\int x^{1/2}dx = \dfrac23 x^{3/2}, and \displaystyle\int x^{3/2}dx = \dfrac25 x^{5/2}.
Multiplying through by x^{1/2}: \sqrt{x}\,(3x^2+2x+3) = 3x^{5/2}+2x^{3/2}+3x^{1/2}.
Integrating term by term: \displaystyle\int 3x^{5/2}dx = \dfrac67 x^{7/2}, \displaystyle\int 2x^{3/2}dx = \dfrac45 x^{5/2}, \displaystyle\int 3x^{1/2}dx = 2x^{3/2}.
Integrating term by term: \displaystyle\int 2x\,dx = x^2, \displaystyle\int 3\cos x\,dx = 3\sin x (so subtracting gives -3\sin x), and \displaystyle\int e^x\,dx = e^x.
Integrating term by term: \displaystyle\int 2x^2\,dx = \dfrac23 x^3.
\displaystyle\int 3\sin x\,dx = -3\cos x (so subtracting gives +3\cos x), and \displaystyle\int 5\sqrt{x}\,dx = \dfrac{10}{3}x^{3/2}.
Expanding the product: \sec x(\sec x+\tan x) = \sec^2 x + \sec x\tan x. Using the standard results \int \sec^2 x\,dx = \tan x and \int \sec x\tan x\,dx = \sec x gives the answer directly.
Rewriting in terms of sine and cosine: \dfrac{\sec^2 x}{\text{cosec}^2 x} = \dfrac{\sin^2 x}{\cos^2 x} = \tan^2 x.
Using the identity \tan^2 x = \sec^2 x - 1, \displaystyle\int \tan^2 x\,dx = \int(\sec^2 x - 1)\,dx = \tan x - x.
Splitting the fraction: \dfrac{2-3\sin x}{\cos^2 x} = 2\sec^2 x - 3\sec x\tan x (since \dfrac{\sin x}{\cos x}\cdot\dfrac{1}{\cos x} = \tan x \sec x).
Integrating term by term using \int \sec^2 x\,dx = \tan x and \int \sec x\tan x\,dx = \sec x gives the answer.
Writing the integrand as powers of x: \sqrt{x}+\dfrac{1}{\sqrt{x}} = x^{1/2}+x^{-1/2}.
Integrating term by term using \int x^n dx = \dfrac{x^{n+1}}{n+1}: \displaystyle\int x^{1/2}dx = \dfrac23 x^{3/2}, and \displaystyle\int x^{-1/2}dx = 2x^{1/2}.
Integrating f'(x)=4x^3-3x^{-4} term by term: \displaystyle\int 4x^3\,dx = x^4, and since \int 3x^{-4}\,dx = -x^{-3}, subtracting this term gives +x^{-3}.
So f(x) = x^4+\dfrac{1}{x^3}+C for some constant C. ...(1)
Using the given condition f(2)=0 in (1): 2^4+\dfrac{1}{2^3}+C=0 \Rightarrow 16+\dfrac18+C=0 \Rightarrow C=-\dfrac{129}{8}.
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