NCERT Class 12 Mathematics — Integrals, Exercise 7.2. All 39 questions solved.

Exercise 7.2 is integration by substitution — the reverse of the chain rule. Where Exercise 7.1 tidied the integrand algebraically, here you change the variable itself. Put $t = g(x)$, so that $dt = g'(x)\,dx$, and the integral

$$\int f\big(g(x)\big)\,g'(x)\,dx \quad\text{becomes}\quad \int f(t)\,dt$$

Two standard forms account for well over half the exercise and are worth committing to memory:

$$\int \frac{f'(x)}{f(x)}\,dx = \log|f(x)| + \mathrm{C}, \qquad \int \big[f(x)\big]^n f'(x)\,dx = \frac{[f(x)]^{n+1}}{n+1} + \mathrm{C}$$

Key insight. The substitution is never a guess. Look through the integrand for a function and a constant multiple of its own derivative. That pairing is what the substitution is for, and once you spot it the choice of $t$ makes itself. In question 1 the numerator $2x$ is exactly the derivative of the denominator $1 + x^2$; in question 29 the $\cot x$ is the derivative of $\log\sin x$. Find the pair first, then write the substitution.

Every answer below has been checked by differentiating it back to the original integrand.

Questions 1 to 37 — integrate the functions.

Question 1

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

Solution. The numerator is precisely the derivative of the denominator, so this is the standard $\tfrac{f’}{f}$ form. Put $t = 1 + x^2$, giving $dt = 2x\,dx$:

$$\int \frac{dt}{t} = \log|t| + \mathrm{C}$$

Question 2

$\dfrac{(\log x)^2}{x}$

Solution. $\tfrac{1}{x}$ is the derivative of $\log x$, so put $t = \log x$ and $dt = \tfrac{1}{x}dx$:

$$\int t^2\,dt = \frac{t^3}{3} + \mathrm{C}$$

Question 3

$\dfrac{1}{x + x\log x}$

Solution. Factor the denominator first — the $x$ is common:

$$\frac{1}{x(1 + \log x)}$$

Now $t = 1 + \log x$ gives $dt = \tfrac{1}{x}dx$, and the integral is $\int \tfrac{dt}{t}$.

Question 4

$\sin x \sin(\cos x)$

Solution. Put $t = \cos x$, so $dt = -\sin x\,dx$ — note the minus sign, which is where this one is usually lost.

$$\int \sin t \,(-dt) = \cos t + \mathrm{C}$$

Question 5

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

Solution. Rather than substituting, use the identity $2\sin\theta\cos\theta = \sin 2\theta$ to halve the work:

$$\sin(ax+b)\cos(ax+b) = \frac{1}{2}\sin\big(2(ax+b)\big)$$

Integrating, and dividing by the $2a$ that the chain rule produces:

$$\frac{1}{2}\cdot\frac{-\cos\big(2(ax+b)\big)}{2a}$$

Question 6

$\sqrt{ax + b}$

Solution. Put $t = ax + b$, so $dt = a\,dx$:

$$\frac{1}{a}\int t^{1/2}\,dt = \frac{1}{a}\cdot\frac{2}{3}t^{3/2}$$

Question 7

$x\sqrt{x + 2}$

Solution. Here the $x$ outside is not the derivative of $x + 2$, so express it in terms of the substitution instead. Put $t = x + 2$, so $x = t – 2$ and $dx = dt$:

$$\int (t-2)\sqrt{t}\,dt = \int \left(t^{3/2} – 2t^{1/2}\right)dt = \frac{2}{5}t^{5/2} – \frac{4}{3}t^{3/2}$$

Question 8

$x\sqrt{1 + 2x^2}$

Solution. Unlike question 7, the $x$ here is a multiple of the derivative of $1 + 2x^2$. Put $t = 1 + 2x^2$, so $dt = 4x\,dx$:

$$\frac{1}{4}\int t^{1/2}\,dt = \frac{1}{4}\cdot\frac{2}{3}t^{3/2} = \frac{t^{3/2}}{6}$$

Question 9

$(4x + 2)\sqrt{x^2 + x + 1}$

Solution. $\tfrac{d}{dx}(x^2 + x + 1) = 2x + 1$, and $4x + 2$ is exactly twice that. Put $t = x^2 + x + 1$, $dt = (2x+1)dx$:

$$2\int t^{1/2}\,dt = 2\cdot\frac{2}{3}t^{3/2}$$

Question 10

$\dfrac{1}{x – \sqrt{x}}$

Solution. Factor out $\sqrt{x}$ from the denominator:

$$\frac{1}{\sqrt{x}\left(\sqrt{x} – 1\right)}$$

Now put $t = \sqrt{x} – 1$, so $dt = \tfrac{1}{2\sqrt{x}}dx$ and $\tfrac{dx}{\sqrt{x}} = 2\,dt$:

$$2\int \frac{dt}{t} = 2\log|t|$$

Question 11

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

Solution. As in question 7, rewrite $x$ in terms of $t = x + 4$, so $x = t – 4$:

$$\int \frac{t – 4}{\sqrt{t}}\,dt = \int \left(t^{1/2} – 4t^{-1/2}\right)dt = \frac{2}{3}t^{3/2} – 8t^{1/2}$$

Factoring out $\tfrac{2}{3}\sqrt{t}$:

$$= \frac{2}{3}\sqrt{t}\,(t – 12) = \frac{2}{3}\sqrt{x+4}\,\big((x+4) – 12\big)$$

Question 12

$\left(x^3 – 1\right)^{1/3} x^5$

Solution. $\tfrac{d}{dx}(x^3 – 1) = 3x^2$, so split $x^5 = x^3 \cdot x^2$ and put $t = x^3 – 1$, giving $x^3 = t + 1$ and $dt = 3x^2 dx$:

$$\frac{1}{3}\int t^{1/3}(t+1)\,dt = \frac{1}{3}\int \left(t^{4/3} + t^{1/3}\right)dt$$

$$= \frac{1}{3}\left(\frac{3}{7}t^{7/3} + \frac{3}{4}t^{4/3}\right)$$

Question 13

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

Solution. $t = 2 + 3x^3$ gives $dt = 9x^2 dx$:

$$\frac{1}{9}\int t^{-3}\,dt = \frac{1}{9}\cdot\frac{t^{-2}}{-2} = -\frac{1}{18t^2}$$

Question 14

$\dfrac{1}{x(\log x)^m}$, $x > 0$, $m \neq 1$

Solution. $t = \log x$, $dt = \tfrac{1}{x}dx$:

$$\int t^{-m}\,dt = \frac{t^{1-m}}{1-m}$$

The condition $m \neq 1$ is exactly what stops the denominator vanishing — at $m = 1$ the integral would be $\log|\log x|$ instead.

Question 15

$\dfrac{x}{9 – 4x^2}$

Solution. $\tfrac{d}{dx}(9 – 4x^2) = -8x$, so put $t = 9 – 4x^2$ and $dt = -8x\,dx$:

$$-\frac{1}{8}\int \frac{dt}{t} = -\frac{1}{8}\log|t|$$

Question 16

$e^{2x + 3}$

Solution. $t = 2x + 3$, $dt = 2\,dx$:

$$\frac{1}{2}\int e^t\,dt = \frac{e^t}{2}$$

Question 17

$\dfrac{x}{e^{x^2}}$

Solution. Write the integrand as $x e^{-x^2}$. Then $t = -x^2$ gives $dt = -2x\,dx$:

$$-\frac{1}{2}\int e^t\,dt = -\frac{e^{-x^2}}{2}$$

Question 18

$\dfrac{e^{\tan^{-1}x}}{1 + x^2}$

Solution. $\tfrac{1}{1+x^2}$ is the derivative of $\tan^{-1}x$, so $t = \tan^{-1}x$ and the integral is simply $\int e^t\,dt$.

Question 19

$\dfrac{e^{2x} – 1}{e^{2x} + 1}$

Solution. Divide numerator and denominator by $e^x$ — this is the step that makes the structure visible:

$$\frac{e^x – e^{-x}}{e^x + e^{-x}}$$

Now the numerator is the derivative of the denominator, so with $t = e^x + e^{-x}$ the integral is $\int \tfrac{dt}{t}$.

Question 20

$\dfrac{e^{2x} – e^{-2x}}{e^{2x} + e^{-2x}}$

Solution. Here the numerator is already half the derivative of the denominator, since $\tfrac{d}{dx}\left(e^{2x} + e^{-2x}\right) = 2\left(e^{2x} – e^{-2x}\right)$:

$$\frac{1}{2}\int \frac{dt}{t}, \qquad t = e^{2x} + e^{-2x}$$

Question 21

$\tan^2(2x – 3)$

Solution. There is no direct integral for $\tan^2$, so use $\tan^2\theta = \sec^2\theta – 1$ first — the same move as question 19 of Exercise 7.1:

$$\int \left[\sec^2(2x-3) – 1\right]dx = \frac{\tan(2x-3)}{2} – x$$

Question 22

$\sec^2(7 – 4x)$

Solution. $t = 7 – 4x$ gives $dt = -4\,dx$, so the chain rule factor is negative:

$$-\frac{1}{4}\int \sec^2 t\,dt = -\frac{\tan t}{4}$$

Question 23

$\dfrac{\sin^{-1}x}{\sqrt{1 – x^2}}$

Solution. $\tfrac{1}{\sqrt{1-x^2}}$ is the derivative of $\sin^{-1}x$, so $t = \sin^{-1}x$ and the integral is $\int t\,dt$.

Question 24

$\dfrac{2\cos x – 3\sin x}{6\cos x + 4\sin x}$

Solution. Take a factor of 2 out of the denominator: $6\cos x + 4\sin x = 2(3\cos x + 2\sin x)$. Now $\tfrac{d}{dx}(2\sin x + 3\cos x) = 2\cos x – 3\sin x$, which is exactly the numerator. So with $t = 2\sin x + 3\cos x$:

$$\frac{1}{2}\int \frac{dt}{t}$$

Question 25

$\dfrac{1}{\cos^2 x\,(1 – \tan x)^2}$

Solution. Rewrite $\tfrac{1}{\cos^2 x}$ as $\sec^2 x$, which is the derivative of $\tan x$. Putting $t = 1 – \tan x$ gives $dt = -\sec^2 x\,dx$:

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

Question 26

$\dfrac{\cos\sqrt{x}}{\sqrt{x}}$

Solution. $t = \sqrt{x}$ gives $dt = \tfrac{1}{2\sqrt{x}}dx$, so $\tfrac{dx}{\sqrt{x}} = 2\,dt$:

$$2\int \cos t\,dt = 2\sin t$$

Question 27

$\sqrt{\sin 2x}\,\cos 2x$

Solution. $t = \sin 2x$ gives $dt = 2\cos 2x\,dx$:

$$\frac{1}{2}\int t^{1/2}\,dt = \frac{1}{2}\cdot\frac{2}{3}t^{3/2} = \frac{t^{3/2}}{3}$$

Question 28

$\dfrac{\cos x}{\sqrt{1 + \sin x}}$

Solution. $t = 1 + \sin x$, $dt = \cos x\,dx$:

$$\int t^{-1/2}\,dt = 2t^{1/2}$$

Question 29

$\cot x \log\sin x$

Solution. The derivative of $\log\sin x$ is $\tfrac{\cos x}{\sin x} = \cot x$ — the two pieces of the integrand are a function and its own derivative. Put $t = \log\sin x$:

$$\int t\,dt = \frac{t^2}{2}$$

Question 30

$\dfrac{\sin x}{1 + \cos x}$

Solution. $\tfrac{d}{dx}(1 + \cos x) = -\sin x$, so $t = 1 + \cos x$ gives $dt = -\sin x\,dx$:

$$-\int \frac{dt}{t} = -\log|t|$$

Question 31

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

Solution. Same substitution as question 30, but now the power is $-2$:

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

Question 32

$\dfrac{1}{1 + \cot x}$

Solution. Write $\cot x = \tfrac{\cos x}{\sin x}$ and clear the compound fraction:

$$\frac{1}{1 + \frac{\cos x}{\sin x}} = \frac{\sin x}{\sin x + \cos x}$$

Now use the standard trick for this family: write the numerator as a combination of the denominator and its derivative. Since $\tfrac{d}{dx}(\sin x + \cos x) = \cos x – \sin x$,

$$\sin x = \frac{1}{2}\big[(\sin x + \cos x) – (\cos x – \sin x)\big]$$

so the integral splits into a constant and a $\tfrac{f’}{f}$ term:

$$\frac{1}{2}\int dx – \frac{1}{2}\int \frac{\cos x – \sin x}{\sin x + \cos x}dx = \frac{x}{2} – \frac{1}{2}\log\left|\sin x + \cos x\right|$$

Question 33

$\dfrac{1}{1 – \tan x}$

Solution. The mirror image of question 32:

$$\frac{1}{1 – \frac{\sin x}{\cos x}} = \frac{\cos x}{\cos x – \sin x}$$

With $\tfrac{d}{dx}(\cos x – \sin x) = -\sin x – \cos x$, write

$$\cos x = \frac{1}{2}\big[(\cos x – \sin x) – (-\sin x – \cos x)\big]$$

giving

$$\frac{x}{2} – \frac{1}{2}\log\left|\cos x – \sin x\right|$$

Question 34

$\dfrac{\sqrt{\tan x}}{\sin x\cos x}$

Solution. Multiply numerator and denominator by $\tfrac{1}{\cos^2 x}$ to manufacture a $\sec^2 x$:

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

Now $t = \tan x$ gives $dt = \sec^2 x\,dx$:

$$\int t^{-1/2}\,dt = 2\sqrt{t}$$

Question 35

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

Solution. $t = 1 + \log x$, $dt = \tfrac{1}{x}dx$:

$$\int t^2\,dt = \frac{t^3}{3}$$

Question 36

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

Solution. Split the awkward factor: $\tfrac{x+1}{x} = 1 + \tfrac{1}{x}$, which is exactly $\tfrac{d}{dx}(x + \log x)$. So $t = x + \log x$:

$$\int t^2\,dt = \frac{t^3}{3}$$

Question 37

$\dfrac{x^3\sin\left(\tan^{-1}x^4\right)}{1 + x^8}$

Solution. Two layers here. Note $1 + x^8 = 1 + (x^4)^2$, and $\tfrac{d}{dx}\tan^{-1}(x^4) = \tfrac{4x^3}{1 + x^8}$ — which is four times the $\tfrac{x^3}{1+x^8}$ sitting in the integrand. So $t = \tan^{-1}(x^4)$:

$$\frac{1}{4}\int \sin t\,dt = -\frac{\cos t}{4}$$

Questions 38 and 39 — choose the correct answer.

Question 38

$\displaystyle\int \frac{10x^9 + 10^x \log_e 10}{x^{10} + 10^x}\,dx$ equals

Solution. Differentiate the denominator and compare with the numerator:

$$\frac{d}{dx}\left(x^{10} + 10^x\right) = 10x^9 + 10^x\log_e 10$$

That is the numerator exactly, so the integrand is $\tfrac{f’}{f}$ and the integral is $\log|f|$.

Question 39

$\displaystyle\int \frac{dx}{\sin^2 x\cos^2 x}$ equals

Solution. Replace the numerator $1$ by $\sin^2 x + \cos^2 x$ — a standard move that splits the fraction into two easy pieces:

$$\frac{\sin^2 x + \cos^2 x}{\sin^2 x\cos^2 x} = \frac{1}{\cos^2 x} + \frac{1}{\sin^2 x} = \sec^2 x + \operatorname{cosec}^2 x$$

$$\int \left(\sec^2 x + \operatorname{cosec}^2 x\right)dx = \tan x – \cot x + \mathrm{C}$$

The minus sign comes from $\int \operatorname{cosec}^2 x\,dx = -\cot x$, which is the whole difference between (A) and (B).

Common mistakes

• Losing the sign when the derivative is negative. $t = \cos x$ gives $dt = -\sin x\,dx$; $t = 1 + \cos x$ likewise. Questions 4, 22, 30 and 31 all turn on it.
• Forgetting to divide by the chain-rule constant. If $dt = 4x\,dx$ then $x\,dx = \tfrac{dt}{4}$, and that $\tfrac14$ must appear in the answer. Questions 8, 20 and 37 each carry one.
• Substituting when the outside factor is not the derivative. In questions 7 and 11 the $x$ is not a multiple of the derivative, so you rewrite $x$ in terms of $t$ instead. Compare with question 8, where it is.
• Attempting $\tan^2$ or $\tfrac{1}{\sin^2 x\cos^2 x}$ directly. Neither has a standard integral. Convert first: $\tan^2\theta = \sec^2\theta – 1$ (question 21), and $1 = \sin^2 x + \cos^2 x$ (question 39).
• Leaving the answer in terms of $t$. The question was asked in $x$; substitute back at the end, every time.
• Dropping the modulus in a logarithm. $\int \tfrac{f’}{f} = \log|f| + C$. It matters wherever $f$ can be negative — questions 3, 10, 15, 24, 30, 32 and 33.

Practise next

• Exercise 7.3 — integration using trigonometric identities, where the rewriting comes before the substitution.
• Exercise 7.1 — worth revisiting if the standard integrals used here are not yet automatic.