NCERT Class 12 Mathematics — Integrals, Miscellaneous Exercise on Chapter 7. All 40 questions solved.

Forty questions, and — unlike Exercises 7.1 to 7.10 — no heading tells you which technique to use. That is the point of the exercise, and it is what makes it the best preparation in the chapter for a board paper or JEE.

Before starting any question here, it is worth spending fifteen seconds on the following checklist, because almost every question in this exercise is decided by one of them:

1. Does an algebraic simplification collapse the problem? Questions 8, 10, 15 and 16 look hard and are trivial once rewritten. $e^{5\log x}$ is just $x^5$.
2. Is the numerator (a multiple of) the derivative of the denominator? If so it is a logarithm — questions 12 and 16.
3. Is the denominator factorable? Then partial fractions — questions 1, 6, 13, 14, 21, 32.
4. Does a substitution rationalise a root? Questions 2, 5, 19, 23, 28.
5. Is it $\int e^x[f(x) + f'(x)]\,dx$? Then the answer is $e^x f(x)$ — questions 20 and 24.
6. For a definite integral, does a property finish it? Questions 25, 26, 29, 34, 40.

Key insight. The single most valuable habit this exercise teaches is checking whether the integrand can be rewritten before deciding it needs a technique. Question 8 is $\dfrac{e^{5\log x} – e^{4\log x}}{e^{3\log x} – e^{2\log x}}$, which looks like it needs substitution and is in fact $x^2$. Question 10 has an eighth power in it and reduces to $-\cos 2x$. Students who reach for a method first lose most of their time on questions that never needed one.

Integrate the functions in Exercises 1 to 23.

Question 1

$\dfrac{1}{x – x^3}$

Solution. Factor the denominator completely: $x – x^3 = x(1-x)(1+x)$. Three distinct linear factors means three partial fractions:

$$\frac{1}{x(1-x)(1+x)} = \frac{1}{x} + \frac{1}{2(1-x)} – \frac{1}{2(1+x)}$$

Integrating gives $\log|x| – \tfrac12\log|1-x| – \tfrac12\log|1+x|$, and the last two combine into $-\tfrac12\log|1-x^2|$.

Question 2

$\dfrac{1}{\sqrt{x+a} + \sqrt{x+b}}$

Solution. Rationalise the denominator. Multiplying above and below by $\sqrt{x+a} – \sqrt{x+b}$ turns the denominator into $(x+a) – (x+b) = a – b$, a constant:

$$\frac{1}{a-b}\left(\sqrt{x+a} – \sqrt{x+b}\right)$$

Each term is now a simple power.

Question 3

$\dfrac{1}{x\sqrt{ax – x^2}}$  [Hint: Put $x = \dfrac{a}{t}$]

Solution. With $x = \dfrac{a}{t}$ we get $dx = -\dfrac{a}{t^2}\,dt$, and

$$ax – x^2 = \frac{a^2}{t} – \frac{a^2}{t^2} = \frac{a^2}{t^2}(t-1), \qquad \sqrt{ax-x^2} = \frac{a}{t}\sqrt{t-1}$$

The whole denominator becomes $\dfrac{a^2}{t^2}\sqrt{t-1}$, and the $t^2$ cancels against the one in $dx$:

$$-\frac{1}{a}\int\frac{dt}{\sqrt{t-1}} = -\frac{2}{a}\sqrt{t-1}$$

Resubstituting $t = \dfrac{a}{x}$ gives $t – 1 = \dfrac{a-x}{x}$.

Question 4

$\dfrac{1}{x^2\left(x^4 + 1\right)^{3/4}}$

Solution. The trick is to pull $x^4$ out of the bracket so that the substitution appears:

$$\left(x^4+1\right)^{3/4} = x^3\left(1 + \frac{1}{x^4}\right)^{3/4}$$

The denominator is then $x^5\left(1+x^{-4}\right)^{3/4}$. Put $t = 1 + x^{-4}$, so $dt = -4x^{-5}dx$:

$$-\frac14\int t^{-3/4}\,dt = -t^{1/4}$$

Question 5

$\dfrac{1}{x^{1/2} + x^{1/3}}$

Solution. Two fractional powers with denominators $2$ and $3$, so put $x = t^6$ — the lowest common multiple clears both at once. Then $dx = 6t^5\,dt$ and the denominator is $t^3 + t^2 = t^2(t+1)$:

$$\int\frac{6t^5}{t^2(t+1)}\,dt = 6\int\frac{t^3}{t+1}\,dt$$

Divide: $\dfrac{t^3}{t+1} = t^2 – t + 1 – \dfrac{1}{t+1}$, giving $2t^3 – 3t^2 + 6t – 6\log|1+t|$ with $t = x^{1/6}$.

Question 6

$\dfrac{5x}{(x+1)\left(x^2+9\right)}$

Solution. A linear factor and an irreducible quadratic, so the numerators are a constant and a linear expression:

$$\frac{5x}{(x+1)(x^2+9)} = \frac{A}{x+1} + \frac{Bx + C}{x^2+9}$$

Clearing denominators and putting $x = -1$ gives $A = -\tfrac12$; matching the $x^2$ coefficients gives $B = \tfrac12$, and the constant term gives $C = \tfrac92$. The quadratic piece splits again into a logarithm and an arctangent.

Question 7

$\dfrac{\sin x}{\sin(x – a)}$

Solution. The obstacle is that numerator and denominator have different arguments. Put $t = x – a$, so $x = t + a$, and expand:

$$\frac{\sin(t+a)}{\sin t} = \frac{\sin t\cos a + \cos t\sin a}{\sin t} = \cos a + \sin a\cot t$$

Both terms are now standard, with $\cos a$ and $\sin a$ as constants.

Question 8

$\dfrac{e^{5\log x} – e^{4\log x}}{e^{3\log x} – e^{2\log x}}$

Solution. No integration technique is needed at all. Since $e^{n\log x} = x^n$, the expression is

$$\frac{x^5 – x^4}{x^3 – x^2} = \frac{x^4(x-1)}{x^2(x-1)} = x^2$$

Question 9

$\dfrac{\cos x}{\sqrt{4 – \sin^2 x}}$

Solution. The numerator is the derivative of $\sin x$, which is what makes $t = \sin x$ work. Then $dt = \cos x\,dx$ and

$$\int\frac{dt}{\sqrt{2^2 – t^2}} = \sin^{-1}\frac{t}{2}$$

Question 10

$\dfrac{\sin^8 x – \cos^8 x}{1 – 2\sin^2 x\cos^2 x}$

Solution. Factor the numerator as a difference of squares twice:

$$\sin^8 x – \cos^8 x = \left(\sin^4 x – \cos^4 x\right)\left(\sin^4 x + \cos^4 x\right) = \left(\sin^2 x – \cos^2 x\right)\left(\sin^4 x + \cos^4 x\right)$$

using $\sin^2 x + \cos^2 x = 1$ on the second factor. Now note that the denominator is exactly $\sin^4 x + \cos^4 x$, since $\left(\sin^2 x + \cos^2 x\right)^2 = \sin^4 x + \cos^4 x + 2\sin^2 x\cos^2 x$. Everything cancels, leaving

$$\sin^2 x – \cos^2 x = -\cos 2x$$

Question 11

$\dfrac{1}{\cos(x+a)\cos(x+b)}$

Solution. Multiply and divide by $\sin(a-b)$, and write that as $\sin\big[(x+a) – (x+b)\big]$ expanded:

$$\frac{\sin(x+a)\cos(x+b) – \cos(x+a)\sin(x+b)}{\sin(a-b)\cos(x+a)\cos(x+b)} = \frac{\tan(x+a) – \tan(x+b)}{\sin(a-b)}$$

Introducing a factor of $1$ in this disguised form is the whole idea — it turns a product in the denominator into a difference of tangents.

Question 12

$\dfrac{x^3}{\sqrt{1 – x^8}}$

Solution. Read $x^8$ as $\left(x^4\right)^2$. Then $t = x^4$ gives $dt = 4x^3dx$, and the $x^3$ in the numerator is exactly what is needed:

$$\frac14\int\frac{dt}{\sqrt{1-t^2}} = \frac14\sin^{-1}t$$

Question 13

$\dfrac{e^x}{\left(1 + e^x\right)\left(2 + e^x\right)}$

Solution. Put $t = e^x$, so $dt = e^x dx$ and the numerator is absorbed:

$$\int\frac{dt}{(1+t)(2+t)} = \int\left(\frac{1}{1+t} – \frac{1}{2+t}\right) dt = \log\left|\frac{1+t}{2+t}\right|$$

Question 14

$\dfrac{1}{\left(x^2+1\right)\left(x^2+4\right)}$

Solution. Treat $x^2$ as the variable for the partial fractions:

$$\frac{1}{(x^2+1)(x^2+4)} = \frac13\left(\frac{1}{x^2+1} – \frac{1}{x^2+4}\right)$$

which follows from $\dfrac{1}{u+1} – \dfrac{1}{u+4} = \dfrac{3}{(u+1)(u+4)}$. Each piece is an arctangent, the second with $a = 2$.

Question 15

$\cos^3 x\,e^{\log\sin x}$

Solution. As in question 8, simplify first: $e^{\log\sin x} = \sin x$. The integrand is $\cos^3 x\sin x$, and $t = \cos x$ gives $dt = -\sin x\,dx$:

$$-\int t^3\,dt = -\frac{t^4}{4}$$

Question 16

$e^{3\log x}\left(x^4 + 1\right)^{-1}$

Solution. Again $e^{3\log x} = x^3$, so the integrand is $\dfrac{x^3}{x^4+1}$. The numerator is a quarter of the derivative of the denominator.

Question 17

$f'(ax + b)\left[f(ax + b)\right]^n$

Solution. This is the general power rule read backwards. Put $t = f(ax+b)$; then $dt = a\,f'(ax+b)\,dx$, so $f'(ax+b)\,dx = \dfrac{dt}{a}$ and the integral is

$$\frac1a\int t^n\,dt = \frac{t^{n+1}}{a(n+1)}$$

The result holds for every $n \ne -1$; at $n = -1$ it becomes a logarithm instead.

Question 18

$\dfrac{1}{\sqrt{\sin^3 x\,\sin(x + \alpha)}}$

Solution. Expand $\sin(x+\alpha) = \sin x\cos\alpha + \cos x\sin\alpha$ and pull $\sin x$ out:

$$\sin^3 x\,\sin(x+\alpha) = \sin^4 x\left(\cos\alpha + \cot x\sin\alpha\right)$$

Taking the square root gives $\sin^2 x\sqrt{\cos\alpha + \sin\alpha\cot x}$, so the integrand is

$$\frac{\operatorname{cosec}^2 x}{\sqrt{\cos\alpha + \sin\alpha\cot x}}$$

Now $t = \cos\alpha + \sin\alpha\cot x$ has $dt = -\sin\alpha\operatorname{cosec}^2x\,dx$, and the integral is $-\dfrac{2}{\sin\alpha}\sqrt{t}$.

Question 19

$\sqrt{\dfrac{1 – \sqrt{x}}{1 + \sqrt{x}}}$

Solution. Put $x = \cos^2\theta$, so $\sqrt{x} = \cos\theta$ and $dx = -2\cos\theta\sin\theta\,d\theta$. The half-angle identity

$$\frac{1 – \cos\theta}{1 + \cos\theta} = \tan^2\frac{\theta}{2}$$

removes the outer root. Since $\tan\tfrac{\theta}{2}\sin\theta = 2\sin^2\tfrac{\theta}{2} = 1 – \cos\theta$, the integrand simplifies to

$$-2(1 – \cos\theta)\cos\theta\,d\theta = \left(-2\cos\theta + 1 + \cos 2\theta\right) d\theta$$

Integrating gives $-2\sin\theta + \theta + \tfrac12\sin 2\theta$. Resubstituting $\theta = \cos^{-1}\sqrt{x}$, with $\sin\theta = \sqrt{1-x}$ and $\tfrac12\sin2\theta = \sqrt{x}\sqrt{1-x}$:

Question 20

$\dfrac{2 + \sin 2x}{1 + \cos 2x}\,e^x$

Solution. Rewrite the fraction using $\sin 2x = 2\sin x\cos x$ and $1 + \cos 2x = 2\cos^2 x$:

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

This is $f(x) + f'(x)$ with $f(x) = \tan x$, so the standard result $\int e^x\left[f(x) + f'(x)\right] dx = e^x f(x)$ applies directly.

Question 21

$\dfrac{x^2 + x + 1}{(x+1)^2(x+2)}$

Solution. A repeated linear factor needs two fractions of its own:

$$\frac{x^2+x+1}{(x+1)^2(x+2)} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x+2}$$

Putting $x = -1$ gives $B = 1$; $x = -2$ gives $C = 3$; matching $x^2$ coefficients gives $A + C = 1$, so $A = -2$.

Question 22

$\tan^{-1}\sqrt{\dfrac{1-x}{1+x}}$

Solution. Put $x = \cos 2\theta$. Then

$$\frac{1 – \cos2\theta}{1 + \cos2\theta} = \tan^2\theta \quad\Longrightarrow\quad \tan^{-1}\sqrt{\cdots} = \theta = \frac12\cos^{-1}x$$

So the problem reduces to $\tfrac12\displaystyle\int\cos^{-1}x\,dx$, which is integration by parts with $1$ as the second function:

$$\int\cos^{-1}x\,dx = x\cos^{-1}x – \sqrt{1-x^2}$$

Question 23

$\dfrac{\sqrt{x^2+1}\left[\log\left(x^2+1\right) – 2\log x\right]}{x^4}$

Solution. Combine the two logarithms first — that is what reveals the substitution:

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

Also $\sqrt{x^2+1} = x\sqrt{1 + \tfrac{1}{x^2}}$, so the integrand is

$$\frac{1}{x^3}\sqrt{1+\frac{1}{x^2}}\,\log\left(1+\frac{1}{x^2}\right)$$

Now $t = 1 + \tfrac{1}{x^2}$ gives $\dfrac{dx}{x^3} = -\dfrac{dt}{2}$, leaving $-\tfrac12\int\sqrt{t}\log t\,dt$ — integration by parts with $\log t$ as the first function.

Evaluate the definite integrals in Exercises 24 to 31.

Question 24

$\displaystyle\int_{\pi/2}^{\pi} e^x\left(\frac{1 – \sin x}{1 – \cos x}\right) dx$

Solution. Convert to half angles: $1 – \cos x = 2\sin^2\tfrac{x}{2}$ and $\sin x = 2\sin\tfrac{x}{2}\cos\tfrac{x}{2}$, so the bracket is

$$\frac{1}{2}\operatorname{cosec}^2\frac{x}{2} – \cot\frac{x}{2}$$

Take $f(x) = -\cot\tfrac{x}{2}$; then $f'(x) = \tfrac12\operatorname{cosec}^2\tfrac{x}{2}$, and the bracket is exactly $f(x) + f'(x)$. Hence the antiderivative is $-e^x\cot\tfrac{x}{2}$. At $x = \pi$ the cotangent is $0$; at $x = \tfrac{\pi}{2}$ it is $1$.

Question 25

$\displaystyle\int_{0}^{\pi/4} \frac{\sin x\cos x}{\cos^4 x + \sin^4 x}\,dx$

Solution. Divide numerator and denominator by $\cos^4 x$:

$$\frac{\tan x\sec^2 x}{1 + \tan^4 x}$$

Put $t = \tan^2 x$, so $dt = 2\tan x\sec^2 x\,dx$, with limits $0$ to $1$:

$$\frac12\int_0^1\frac{dt}{1+t^2} = \frac12 \cdot \frac{\pi}{4}$$

Question 26

$\displaystyle\int_{0}^{\pi/2} \frac{\cos^2 x\,dx}{\cos^2 x + 4\sin^2 x}$

Solution. Divide through by $\cos^2 x$ and then multiply above and below by $\sec^2 x$:

$$\frac{\sec^2 x}{\left(1 + \tan^2 x\right)\left(1 + 4\tan^2 x\right)}$$

With $t = \tan x$ the limits become $0$ to $\infty$, and partial fractions in $t^2$ give

$$\int_0^\infty\left(\frac{4/3}{1+4t^2} – \frac{1/3}{1+t^2}\right) dt = \frac43 \cdot \frac{\pi}{4} – \frac13 \cdot \frac{\pi}{2}$$

Question 27

$\displaystyle\int_{\pi/6}^{\pi/3} \frac{\sin x + \cos x}{\sqrt{\sin 2x}}\,dx$

Solution. The numerator is the derivative of $\sin x – \cos x$, and the denominator can be written in terms of the same quantity:

$$(\sin x – \cos x)^2 = 1 – \sin 2x \quad\Longrightarrow\quad \sin 2x = 1 – t^2$$

with $t = \sin x – \cos x$. The limits become $\tfrac{1-\sqrt3}{2}$ and $\tfrac{\sqrt3-1}{2}$ — symmetric about zero:

$$\int\frac{dt}{\sqrt{1-t^2}} = \left[\sin^{-1}t\right]$$

Because $\sin^{-1}$ is odd, the two evaluations add rather than cancel.

Question 28

$\displaystyle\int_{0}^{1} \frac{dx}{\sqrt{1+x} – \sqrt{x}}$

Solution. Rationalise: the conjugate turns the denominator into $(1+x) – x = 1$, so the integrand is simply $\sqrt{1+x} + \sqrt{x}$:

$$\frac23\left[(1+x)^{3/2} + x^{3/2}\right]_0^1 = \frac23\left[2\sqrt2 + 1 – 1\right]$$

Question 29

$\displaystyle\int_{0}^{\pi/4} \frac{\sin x + \cos x}{9 + 16\sin 2x}\,dx$

Solution. Same substitution as question 27: $t = \sin x – \cos x$ gives $\sin 2x = 1 – t^2$, so the denominator becomes $25 – 16t^2$. The limits run from $-1$ to $0$:

$$\int_{-1}^{0}\frac{dt}{25 – 16t^2} = \frac{1}{16}\int_{-1}^{0}\frac{dt}{\left(\frac54\right)^2 – t^2} = \frac{1}{40}\left[\log\left|\frac{5+4t}{5-4t}\right|\right]_{-1}^{0}$$

At $t = 0$ the logarithm is $0$; at $t = -1$ it is $\log\tfrac19 = -\log 9$.

Question 30

$\displaystyle\int_{0}^{\pi/2} \sin 2x\,\tan^{-1}(\sin x)\,dx$

Solution. Write $\sin 2x = 2\sin x\cos x$ so that $t = \sin x$ absorbs the $\cos x\,dx$:

$$2\int_0^1 t\tan^{-1}t\,dt$$

Now integrate by parts with $\tan^{-1}t$ as the first function:

$$\int t\tan^{-1}t\,dt = \frac{t^2}{2}\tan^{-1}t – \frac12\int\frac{t^2}{1+t^2}\,dt = \frac{t^2}{2}\tan^{-1}t – \frac12\left(t – \tan^{-1}t\right)$$

The step that matters is $\dfrac{t^2}{1+t^2} = 1 – \dfrac{1}{1+t^2}$. At $t = 1$ the bracket is $\tfrac{\pi}{4} – \tfrac12$.

Question 31

$\displaystyle\int_{1}^{4}\Big[|x – 1| + |x – 2| + |x – 3|\Big] dx$

Solution. Handle the three moduli separately, splitting each at its own zero within $[1, 4]$:

$$\int_1^4 |x-1|\,dx = \int_1^4 (x-1)\,dx = \frac92$$

$$\int_1^4 |x-2|\,dx = \int_1^2 (2-x)\,dx + \int_2^4 (x-2)\,dx = \frac12 + 2 = \frac52$$

$$\int_1^4 |x-3|\,dx = \int_1^3 (3-x)\,dx + \int_3^4 (x-3)\,dx = 2 + \frac12 = \frac52$$

The first needs no split because $x – 1 \ge 0$ throughout the interval.

Prove the following (Exercises 32 to 37).

Question 32

Prove that $\displaystyle\int_{1}^{3} \frac{dx}{x^2(x+1)} = \frac{2}{3} + \log\frac{2}{3}$

Solution. Partial fractions with a repeated factor $x^2$:

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

Integrating gives $\log\left|\dfrac{x+1}{x}\right| – \dfrac1x$. Evaluating:

$$\left(\log\frac43 – \frac13\right) – \left(\log 2 – 1\right) = \log\frac{4}{3 \cdot 2} + \frac23 = \frac23 + \log\frac23$$

Question 33

Prove that $\displaystyle\int_{0}^{1} x\,e^x\,dx = 1$

Solution. Integration by parts with $x$ as the first function gives $\int x e^x dx = xe^x – e^x$:

$$\Big[xe^x – e^x\Big]_0^1 = (e – e) – (0 – 1) = 1$$

The upper limit contributes nothing — the whole value comes from $-e^0$ at the lower limit.

Question 34

Prove that $\displaystyle\int_{-1}^{1} x^{17}\cos^4 x\,dx = 0$

Solution. Check the parity of each factor. $x^{17}$ is odd, and $\cos^4 x$ is even because $\cos(-x) = \cos x$. Odd $\times$ even is odd, so

$$f(-x) = (-x)^{17}\cos^4(-x) = -x^{17}\cos^4 x = -f(x)$$

By $\mathbf{P_7}$ the integral of an odd function over $[-1, 1]$ is zero. No antiderivative is required — and just as well, since $x^{17}\cos^4 x$ would take several rounds of parts.

Question 35

Prove that $\displaystyle\int_{0}^{\pi/2} \sin^3 x\,dx = \frac{2}{3}$

Solution. An odd power of sine: peel off one factor for the differential.

$$\sin^3 x = \left(1 – \cos^2 x\right)\sin x$$

With $t = \cos x$ and $dt = -\sin x\,dx$, the limits run from $1$ to $0$:

$$\int_0^1\left(1 – t^2\right) dt = 1 – \frac13$$

Question 36

Prove that $\displaystyle\int_{0}^{\pi/4} 2\tan^3 x\,dx = 1 – \log 2$

Solution. Split off two powers of tangent so the Pythagorean identity applies:

$$\tan^3 x = \tan x\left(\sec^2 x – 1\right) = \tan x\sec^2 x – \tan x$$

The first term integrates to $\tfrac12\tan^2 x$ and the second to $-\log|\sec x|$. Between $0$ and $\tfrac{\pi}{4}$:

$$2\left[\frac{\tan^2 x}{2} – \log|\sec x|\right]_0^{\pi/4} = 2\left(\frac12 – \log\sqrt2\right)$$

Question 37

Prove that $\displaystyle\int_{0}^{1} \sin^{-1}x\,dx = \frac{\pi}{2} – 1$

Solution. Integration by parts with $1$ as the second function — the standard move for an inverse trigonometric function on its own:

$$\int\sin^{-1}x\,dx = x\sin^{-1}x – \int\frac{x}{\sqrt{1-x^2}}\,dx = x\sin^{-1}x + \sqrt{1-x^2}$$

At $x = 1$ this is $\tfrac{\pi}{2} + 0$; at $x = 0$ it is $0 + 1$.

Choose the correct answers in Exercises 38 to 40.

Question 38

$\displaystyle\int \frac{dx}{e^x + e^{-x}}$ is equal to

Solution. Multiply above and below by $e^x$ to clear the negative exponent:

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

Now $t = e^x$ gives $\displaystyle\int\frac{dt}{1+t^2} = \tan^{-1}t$.

Option (D) is the trap for anyone who sees a sum in the denominator and reaches for a logarithm — but the numerator would have to be the derivative $e^x – e^{-x}$, not $1$.

Question 39

$\displaystyle\int \frac{\cos 2x}{(\sin x + \cos x)^2}\,dx$ is equal to

Solution. Factor the numerator using $\cos 2x = \cos^2 x – \sin^2 x$:

$$\frac{(\cos x – \sin x)(\cos x + \sin x)}{(\sin x + \cos x)^2} = \frac{\cos x – \sin x}{\sin x + \cos x}$$

The numerator is now exactly the derivative of the denominator, so the integral is a logarithm of it. Option (C) has the sign inside the modulus wrong, which is what you get by cancelling the wrong factor.

Question 40

If $f(a + b – x) = f(x)$, then $\displaystyle\int_{a}^{b} x f(x)\,dx$ is equal to

Solution. Let $I = \displaystyle\int_a^b x f(x)\,dx$ and apply $\mathbf{P_3}$:

$$I = \int_a^b (a + b – x)\,f(a+b-x)\,dx = \int_a^b (a+b-x)f(x)\,dx$$

using the given condition. Splitting the bracket:

$$I = (a+b)\int_a^b f(x)\,dx – I \quad\Longrightarrow\quad 2I = (a+b)\int_a^b f(x)\,dx$$

Option (C) is the answer if you use $b – a$, the length of the interval, where $a + b$, twice its midpoint, is what the algebra produces.

Common mistakes

• Starting a technique before simplifying. Questions 8, 10, 15 and 16 are one-liners after rewriting, and long dead ends if you substitute first. Always read the integrand once for algebra before choosing a method.
• Forgetting the second fraction for a repeated factor. $(x+1)^2$ in question 21 needs both $\dfrac{A}{x+1}$ and $\dfrac{B}{(x+1)^2}$; one alone cannot represent the function.
• Rationalising only the denominator’s sign, not the whole expression. In questions 2 and 28 the conjugate must multiply both numerator and denominator.
• Treating $\sqrt{x^2+1}$ as $x + 1$ or similar. Question 23 works because $\sqrt{x^2+1} = x\sqrt{1 + x^{-2}}$ exactly; no approximation is involved.
• Using an antiderivative where a property is intended. Question 34 is a parity argument. Attempting $\int x^{17}\cos^4 x\,dx$ is possible in principle and hopeless in practice.
• Losing the constant from a linear substitution. In question 17 the $\dfrac{1}{a}$ comes from $t = f(ax+b)$; dropping it is the commonest slip in the whole exercise.
• Splitting a modulus at a point outside the interval. In question 31 the factor $|x-1|$ needs no split on $[1,4]$, because $x – 1$ never changes sign there. Splitting anyway is harmless but wastes time; failing to split $|x-2|$ and $|x-3|$ is not.

Practise next

• Exercise 8.1 — the definite integrals of this chapter applied to areas bounded by curves.
• Exercise 7.10 — the properties of definite integrals used in questions 34 and 40 here.