NCERT Class 12 Mathematics — Integrals, Exercise 7.1. All 22 questions solved.

Exercise 7.1 is the first exercise on integration, and every question in it is solved by the same idea: integration is differentiation run backwards. Nothing here needs substitution, partial fractions or integration by parts — those come later in the chapter. What the exercise actually tests is whether you can rewrite an awkward-looking integrand into a sum of terms you already know how to integrate.

Almost every difficulty in this exercise is algebraic rather than calculus. The integrals that look hardest — question 10, question 13, question 19 — become one-line problems the moment the integrand is tidied up.

Key insight. Do the algebra first, integrate second. Expand brackets, divide out fractions, factorise, and convert every root into a fractional power, until the integrand is a sum of terms of the form $x^n$ and standard trigonometric or exponential functions. Only then apply $\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$ term by term.

Questions 1 to 5 — anti derivatives by inspection.

Find an anti derivative (or integral) of the following functions by the method of inspection.

“By inspection” means: think of a function whose derivative is the one given, then fix the constant by trial. Because $\frac{d}{dx}\sin kx = k\cos kx$, a factor of $\frac{1}{k}$ always appears when the angle is a multiple of $x$. These five answers are anti derivatives, so the constant $C$ is conventionally omitted here — NCERT asks for an anti derivative, not the general one.

Question 1

$$\sin 2x$$

Solution. Differentiating $\cos 2x$ gives $-2\sin 2x$, which is $-2$ times what we want. So divide by $-2$.

$$\frac{d}{dx}\left(-\frac{1}{2}\cos 2x\right) = \sin 2x$$

Question 2

$$\cos 3x$$

Solution. Differentiating $\sin 3x$ gives $3\cos 3x$, so scale by $\frac{1}{3}$.

$$\frac{d}{dx}\left(\frac{1}{3}\sin 3x\right) = \cos 3x$$

Question 3

$$e^{2x}$$

Solution. The exponential reproduces itself on differentiation, but the chain rule brings down the factor $2$ from the exponent, so we divide by $2$.

$$\frac{d}{dx}\left(\frac{1}{2}e^{2x}\right) = e^{2x}$$

Question 4

$$(ax + b)^2$$

Solution. Raising the power by one gives $(ax+b)^3$, whose derivative is $3a(ax+b)^2$ — the extra $a$ comes from differentiating the inside. Divide by $3a$ to cancel both factors.

$$\frac{d}{dx}\left(\frac{1}{3a}(ax+b)^3\right) = (ax+b)^2$$

Question 5

$$\sin 2x – 4e^{3x}$$

Solution. An anti derivative of a difference is the difference of anti derivatives, so treat the two terms separately. The first is question 1. For the second, $\frac{d}{dx}e^{3x} = 3e^{3x}$, so an anti derivative of $4e^{3x}$ is $\frac{4}{3}e^{3x}$.

$$-\frac{1}{2}\cos 2x – \frac{4}{3}e^{3x}$$

Questions 6 to 20 — find the integrals.

Find the following integrals in Exercises 6 to 20.

From here the answers are indefinite integrals, so the constant of integration $C$ must be written every time.

Question 6

$$\int \left(4e^{3x} + 1\right)dx$$

Solution. Integrate term by term. The $1$ integrates to $x$, since $\frac{d}{dx}x = 1$.

$$\int 4e^{3x}\,dx + \int 1\,dx = \frac{4}{3}e^{3x} + x + C$$

Question 7

$$\int x^2\left(1 – \frac{1}{x^2}\right)dx$$

Solution. There is no product rule for integration, so the bracket must be multiplied out before anything else. Doing so collapses the integrand to a polynomial.

$$x^2\left(1 – \frac{1}{x^2}\right) = x^2 – 1$$

$$\int (x^2 – 1)\,dx = \frac{x^3}{3} – x + C$$

Question 8

$$\int \left(ax^2 + bx + c\right)dx$$

Solution. Here $a$, $b$ and $c$ are constants, so each pulls straight out of its integral and only the powers of $x$ are affected.

$$a\cdot\frac{x^3}{3} + b\cdot\frac{x^2}{2} + cx + C$$

Question 9

$$\int \left(2x^2 + e^x\right)dx$$

Solution. Term by term, using $\int e^x\,dx = e^x + C$.

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

Question 10

$$\int \left(\sqrt{x} – \frac{1}{\sqrt{x}}\right)^{2}dx$$

Solution. Square the bracket first. The cross term is where the interest lies: $2 \cdot \sqrt{x} \cdot \frac{1}{\sqrt{x}} = 2$, a constant.

$$\left(\sqrt{x} – \frac{1}{\sqrt{x}}\right)^{2} = x – 2 + \frac{1}{x}$$

The $\frac{1}{x}$ term is the one exception to the power rule, because $\frac{x^{n+1}}{n+1}$ is undefined when $n = -1$. It integrates to $\log|x|$.

$$\int \left(x – 2 + \frac{1}{x}\right)dx = \frac{x^2}{2} – 2x + \log|x| + C$$

Question 11

$$\int \frac{x^3 + 5x^2 – 4}{x^2}\,dx$$

Solution. Divide each term of the numerator by $x^2$ separately. There is no quotient rule for integration either, so a fraction like this must be split before integrating.

$$\frac{x^3 + 5x^2 – 4}{x^2} = x + 5 – 4x^{-2}$$

$$\int \left(x + 5 – 4x^{-2}\right)dx = \frac{x^2}{2} + 5x – 4\cdot\frac{x^{-1}}{-1} + C$$

Note the sign: dividing by $-1$ turns $-4x^{-2}$ into $+\frac{4}{x}$.

Question 12

$$\int \frac{x^3 + 3x + 4}{\sqrt{x}}\,dx$$

Solution. Write $\sqrt{x}$ as $x^{1/2}$ and divide term by term, subtracting $\tfrac{1}{2}$ from each exponent.

$$\frac{x^3 + 3x + 4}{x^{1/2}} = x^{5/2} + 3x^{1/2} + 4x^{-1/2}$$

Now the power rule applies to each term:

$$\frac{x^{7/2}}{7/2} + 3\cdot\frac{x^{3/2}}{3/2} + 4\cdot\frac{x^{1/2}}{1/2} + C$$

Question 13

$$\int \frac{x^3 – x^2 + x – 1}{x – 1}\,dx$$

Solution. This one rewards a look before any calculus. The numerator factorises by grouping, and the factor of $x – 1$ cancels.

$$x^3 – x^2 + x – 1 = x^2(x – 1) + 1(x – 1) = (x – 1)(x^2 + 1)$$

$$\int \frac{(x-1)(x^2+1)}{x-1}\,dx = \int (x^2 + 1)\,dx$$

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

Question 14

$$\int (1 – x)\sqrt{x}\,dx$$

Solution. Multiply through by $x^{1/2}$, adding $\tfrac{1}{2}$ to each exponent.

$$(1 – x)x^{1/2} = x^{1/2} – x^{3/2}$$

$$\int \left(x^{1/2} – x^{3/2}\right)dx = \frac{x^{3/2}}{3/2} – \frac{x^{5/2}}{5/2} + C$$

Question 15

$$\int \sqrt{x}\left(3x^2 + 2x + 3\right)dx$$

Solution. The same expansion as question 14, with three terms.

$$x^{1/2}\left(3x^2 + 2x + 3\right) = 3x^{5/2} + 2x^{3/2} + 3x^{1/2}$$

$$3\cdot\frac{x^{7/2}}{7/2} + 2\cdot\frac{x^{5/2}}{5/2} + 3\cdot\frac{x^{3/2}}{3/2} + C$$

Question 16

$$\int \left(2x – 3\cos x + e^x\right)dx$$

Solution. Three standard integrals. The sign on the cosine term is worth care: $\int \cos x\,dx = \sin x$, so $-3\cos x$ integrates to $-3\sin x$.

$$x^2 – 3\sin x + e^x + C$$

Question 17

$$\int \left(2x^2 – 3\sin x + 5\sqrt{x}\right)dx$$

Solution. Here the sine term flips sign, because $\int \sin x\,dx = -\cos x$. So $-3\sin x$ integrates to $+3\cos x$.

$$2\cdot\frac{x^3}{3} + 3\cos x + 5\cdot\frac{x^{3/2}}{3/2} + C$$

Question 18

$$\int \sec x\,(\sec x + \tan x)\,dx$$

Solution. Expand the bracket, and both terms become standard derivatives you already know from Chapter 5: $\frac{d}{dx}\tan x = \sec^2 x$ and $\frac{d}{dx}\sec x = \sec x\tan x$.

$$\int \left(\sec^2 x + \sec x\tan x\right)dx = \tan x + \sec x + C$$

Question 19

$$\int \frac{\sec^2 x}{\operatorname{cosec}^2 x}\,dx$$

Solution. Neither function integrates usefully as a quotient, so convert both to sines and cosines. Since $\operatorname{cosec}^2 x = \frac{1}{\sin^2 x}$, dividing by it means multiplying by $\sin^2 x$.

$$\frac{\sec^2 x}{\operatorname{cosec}^2 x} = \frac{\sin^2 x}{\cos^2 x} = \tan^2 x$$

$\tan^2 x$ has no elementary anti derivative on sight, but the identity $\tan^2 x = \sec^2 x – 1$ turns it into one that does.

$$\int \left(\sec^2 x – 1\right)dx = \tan x – x + C$$

Question 20

$$\int \frac{2 – 3\sin x}{\cos^2 x}\,dx$$

Solution. Split the fraction over the two terms in the numerator, then read each piece as a standard form.

$$\frac{2}{\cos^2 x} – \frac{3\sin x}{\cos^2 x} = 2\sec^2 x – 3\sec x\tan x$$

The second rearrangement uses $\frac{\sin x}{\cos^2 x} = \frac{1}{\cos x}\cdot \frac{\sin x}{\cos x} = \sec x\tan x$.

$$\int \left(2\sec^2 x – 3\sec x\tan x\right)dx = 2\tan x – 3\sec x + C$$

Questions 21 and 22 — multiple choice.

Choose the correct answer in Exercises 21 and 22.

Question 21

The anti derivative of $\left(\sqrt{x} + \dfrac{1}{\sqrt{x}}\right)$ equals

Solution. Write both terms as powers and integrate.

$$\int \left(x^{1/2} + x^{-1/2}\right)dx = \frac{x^{3/2}}{3/2} + \frac{x^{1/2}}{1/2} + C$$

$$= \frac{2}{3}x^{3/2} + 2x^{1/2} + C$$

Option (A) raises the power the wrong way, and (D) has the reciprocal of the correct coefficient — both are what you get by dividing by the old exponent instead of the new one.

Question 22

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

Solution. The derivative of $f$ is given, so $f$ itself is the integral of it. Write the second term as a power first.

$$f(x) = \int \left(4x^3 – 3x^{-4}\right)dx = x^4 – 3\cdot\frac{x^{-3}}{-3} + C = x^4 + \frac{1}{x^3} + C$$

The condition $f(2) = 0$ is what pins down $C$ — without it, any value of $C$ would satisfy the derivative, which is why the question supplies one.

$$0 = 2^4 + \frac{1}{2^3} + C = 16 + \frac{1}{8} + C$$

$$C = -\frac{129}{8}$$

Common mistakes

• Integrating a product or a quotient term by term without expanding it first. There is no product rule or quotient rule for integration. In question 7, integrating $x^2$ and $\left(1 – \frac{1}{x^2}\right)$ separately and multiplying the results gives a completely wrong answer. Expand first, every time.
• Forgetting that $\int \frac{1}{x}\,dx = \log|x| + C$, not $\frac{x^0}{0}$. The power rule genuinely fails at $n = -1$. Question 10 is the only place in this exercise where it bites.
• Losing the sign when integrating $\sin x$. $\int \sin x\,dx = -\cos x$ but $\int \cos x\,dx = +\sin x$. Questions 16 and 17 sit next to each other partly to catch this: the sine term in question 17 flips sign, the cosine term in question 16 does not.
• Dividing by the old exponent rather than the new one. $\int x^{1/2}dx$ is $\frac{x^{3/2}}{3/2} = \frac{2}{3}x^{3/2}$, not $2x^{3/2}$. This is exactly the error that distractor (D) in question 21 is built from.
• Dropping the constant of integration. Questions 1 to 5 ask for an anti derivative, so $C$ is optional there — but from question 6 onwards these are indefinite integrals and $C$ is part of the answer.
• Missing the factorisation in question 13. Attempting polynomial long division works, but grouping the numerator as $x^2(x-1) + (x-1)$ is faster and far less error-prone.

Practise next

• Exercise 7.2 — integration by substitution, where the rewriting stops being algebraic and starts being a change of variable.
• Chapter 5, Exercise 5.2 — if the derivatives used here felt unfamiliar, the standard derivative list is worth revising before going further into this chapter.