NCERT Class 12 Mathematics — Integrals, Exercise 7.3. All 24 questions solved.

Exercise 7.3 is integration using trigonometric identities. Not one integral here can be done as it stands: every single one has to be rewritten first, and the whole exercise is really a test of which identity to reach for.

These five cover almost everything:

$$\sin^2\theta = \frac{1 – \cos 2\theta}{2}, \qquad \cos^2\theta = \frac{1 + \cos 2\theta}{2}$$

$$2\sin \mathrm{A}\cos \mathrm{B} = \sin(\mathrm{A}+\mathrm{B}) + \sin(\mathrm{A}-\mathrm{B})$$

$$2\cos \mathrm{A}\cos \mathrm{B} = \cos(\mathrm{A}+\mathrm{B}) + \cos(\mathrm{A}-\mathrm{B}), \qquad 2\sin \mathrm{A}\sin \mathrm{B} = \cos(\mathrm{A}-\mathrm{B}) – \cos(\mathrm{A}+\mathrm{B})$$

$$\tan^2\theta = \sec^2\theta – 1$$

Key insight. There is one rule of thumb behind all of them: turn products into sums, and even powers into multiple angles. A product of sines and cosines has no antiderivative you can write down; a sum of terms like $\cos 4x$ does. Every question below is that conversion followed by a one-line integration.

Odd powers are the exception — for those, peel off one factor and substitute, as in questions 4 and 5.

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

Questions 1 to 22 — find the integrals.

Question 1

$\sin^2(2x + 5)$

Solution. An even power, so use the double-angle form with $\theta = 2x+5$:

$$\sin^2(2x+5) = \frac{1 – \cos(4x + 10)}{2}$$

$$\int \frac{1 – \cos(4x+10)}{2}\,dx = \frac{x}{2} – \frac{\sin(4x+10)}{8}$$

Question 2

$\sin 3x\cos 4x$

Solution. A product of a sine and a cosine, so use $2\sin \mathrm{A}\cos \mathrm{B} = \sin(\mathrm{A}+\mathrm{B}) + \sin(\mathrm{A}-\mathrm{B})$:

$$\sin 3x\cos 4x = \frac{1}{2}\big[\sin 7x + \sin(-x)\big] = \frac{1}{2}\big[\sin 7x – \sin x\big]$$

$$\int = \frac{1}{2}\left[-\frac{\cos 7x}{7} + \cos x\right]$$

Question 3

$\cos 2x\cos 4x\cos 6x$

Solution. Three factors, so convert two at a time. First $\cos 2x\cos 4x = \tfrac12(\cos 6x + \cos 2x)$:

$$\cos 2x\cos 4x\cos 6x = \frac{1}{2}\big(\cos 6x\cos 6x + \cos 2x\cos 6x\big)$$

Now $\cos^2 6x = \tfrac{1 + \cos 12x}{2}$ and $\cos 2x\cos 6x = \tfrac12(\cos 8x + \cos 4x)$:

$$= \frac{1}{4}\big[1 + \cos 12x + \cos 8x + \cos 4x\big]$$

Integrating term by term:

Question 4

$\sin^3(2x + 1)$

Solution. An odd power, so peel off one factor and use $\sin^2\theta = 1 – \cos^2\theta$ on the rest:

$$\sin^3(2x+1) = \sin(2x+1)\big[1 – \cos^2(2x+1)\big]$$

Put $t = \cos(2x+1)$, so $dt = -2\sin(2x+1)\,dx$:

$$-\frac{1}{2}\int \left(1 – t^2\right)dt = -\frac{1}{2}\left(t – \frac{t^3}{3}\right)$$

Question 5

$\sin^3 x\cos^3 x$

Solution. Both powers are odd, so take one $\sin x$ aside as the differential and convert the rest:

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

With $t = \cos x$, $dt = -\sin x\,dx$:

$$-\int \left(1 – t^2\right)t^3\,dt = -\int \left(t^3 – t^5\right)dt = -\frac{t^4}{4} + \frac{t^6}{6}$$

Question 6

$\sin x\sin 2x\sin 3x$

Solution. Combine the outer two first, using $2\sin \mathrm{A}\sin \mathrm{B} = \cos(\mathrm{A}-\mathrm{B}) – \cos(\mathrm{A}+\mathrm{B})$:

$$\sin x\sin 3x = \frac{1}{2}\big(\cos 2x – \cos 4x\big)$$

So the integrand is $\tfrac12\sin 2x\left(\cos 2x – \cos 4x\right)$. Expanding each product into sums:

$$\sin 2x\cos 2x = \frac{\sin 4x}{2}, \qquad \sin 2x\cos 4x = \frac{1}{2}\big(\sin 6x – \sin 2x\big)$$

$$= \frac{1}{4}\big[\sin 4x – \sin 6x + \sin 2x\big]$$

Integrating each term:

Question 7

$\sin 4x\sin 8x$

Solution. Product of two sines:

$$\sin 4x\sin 8x = \frac{1}{2}\big[\cos 4x – \cos 12x\big]$$

$$\int = \frac{1}{2}\left[\frac{\sin 4x}{4} – \frac{\sin 12x}{12}\right]$$

Question 8

$\dfrac{1 – \cos x}{1 + \cos x}$

Solution. Use the half-angle forms $1 – \cos x = 2\sin^2\tfrac{x}{2}$ and $1 + \cos x = 2\cos^2\tfrac{x}{2}$:

$$\frac{2\sin^2\frac{x}{2}}{2\cos^2\frac{x}{2}} = \tan^2\frac{x}{2} = \sec^2\frac{x}{2} – 1$$

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

Question 9

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

Solution. Add and subtract 1 in the numerator to split off a constant — a standard move whenever the numerator and denominator are close:

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

Now $1 + \cos x = 2\cos^2\tfrac{x}{2}$, so the second term is $\tfrac12\sec^2\tfrac{x}{2}$:

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

Question 10

$\sin^4 x$

Solution. Square the double-angle form and then reduce again:

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

The last term needs the identity a second time: $\cos^2 2x = \tfrac{1 + \cos 4x}{2}$.

$$= \frac{1}{4}\left(\frac{3}{2} – 2\cos 2x + \frac{\cos 4x}{2}\right)$$

Integrating:

Question 11

$\cos^4 2x$

Solution. Identical to question 10 with $\cos$ in place of $\sin$ and the angle doubled:

$$\cos^4 2x = \left(\frac{1 + \cos 4x}{2}\right)^2 = \frac{1}{4}\left(\frac{3}{2} + 2\cos 4x + \frac{\cos 8x}{2}\right)$$

Question 12

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

Solution. Do not reach for a half-angle formula here — the numerator factorises against the denominator. Since $\sin^2 x = 1 – \cos^2 x = (1-\cos x)(1+\cos x)$:

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

$$\int (1 – \cos x)\,dx = x – \sin x$$

Question 13

$\dfrac{\cos 2x – \cos 2\alpha}{\cos x – \cos\alpha}$

Solution. Write both cosines of the double angle in the form $\cos 2\theta = 2\cos^2\theta – 1$, so the numerator becomes a difference of squares:

$$\cos 2x – \cos 2\alpha = 2\cos^2 x – 2\cos^2\alpha = 2(\cos x – \cos\alpha)(\cos x + \cos\alpha)$$

The factor $\cos x – \cos\alpha$ cancels:

$$= 2(\cos x + \cos\alpha)$$

Remember $\alpha$ is a constant, so $\cos\alpha$ integrates to $x\cos\alpha$:

Question 14

$\dfrac{\cos x – \sin x}{1 + \sin 2x}$

Solution. The denominator is a perfect square in disguise, since $\sin 2x = 2\sin x\cos x$:

$$1 + \sin 2x = \sin^2 x + \cos^2 x + 2\sin x\cos x = (\cos x + \sin x)^2$$

Now note that $\cos x – \sin x$ is exactly the derivative of $\cos x + \sin x$. With $t = \cos x + \sin x$:

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

Question 15

$\tan^3 2x\sec 2x$

Solution. Split off one $\tan 2x \sec 2x$ — that pairing is the derivative of $\sec 2x$ up to a constant — and convert the rest:

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

With $t = \sec 2x$, $dt = 2\sec 2x\tan 2x\,dx$:

$$\frac{1}{2}\int \left(t^2 – 1\right)dt = \frac{1}{2}\left(\frac{t^3}{3} – t\right)$$

Question 16

$\tan^4 x$

Solution. Apply $\tan^2 x = \sec^2 x – 1$ twice, peeling off two powers at a time:

$$\tan^4 x = \tan^2 x\left(\sec^2 x – 1\right) = \tan^2 x\sec^2 x – \tan^2 x$$

$$= \tan^2 x\sec^2 x – \sec^2 x + 1$$

The first term integrates by $t = \tan x$:

$$\int = \frac{\tan^3 x}{3} – \tan x + x$$

Question 17

$\dfrac{\sin^3 x + \cos^3 x}{\sin^2 x\cos^2 x}$

Solution. Split the fraction into two and cancel:

$$\frac{\sin^3 x}{\sin^2 x\cos^2 x} + \frac{\cos^3 x}{\sin^2 x\cos^2 x} = \frac{\sin x}{\cos^2 x} + \frac{\cos x}{\sin^2 x}$$

The first is $\sec x\tan x$ and the second is $\operatorname{cosec} x\cot x$, both standard:

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

Question 18

$\dfrac{\cos 2x + 2\sin^2 x}{\cos^2 x}$

Solution. Simplify the numerator before anything else. Using $\cos 2x = 1 – 2\sin^2 x$, the two $\sin^2 x$ terms cancel outright:

$$\cos 2x + 2\sin^2 x = 1 – 2\sin^2 x + 2\sin^2 x = 1$$

So the integrand is just $\sec^2 x$.

Question 19

$\dfrac{1}{\sin x\cos^3 x}$

Solution. Multiply top and bottom by $\sec^4 x$ to express everything in $\tan x$:

$$\frac{1}{\sin x\cos^3 x} = \frac{\sec^4 x}{\tan x} = \frac{\left(1 + \tan^2 x\right)\sec^2 x}{\tan x}$$

using $\sec^2 x = 1 + \tan^2 x$. With $t = \tan x$, $dt = \sec^2 x\,dx$:

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

Question 20

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

Solution. The numerator factorises as a difference of squares, and the denominator expands as in question 14:

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

One factor of $(\cos x + \sin x)$ cancels:

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

which is $\tfrac{f’}{f}$ with $f = \cos x + \sin x$.

Question 21

$\sin^{-1}(\cos x)$

Solution. This looks like an inverse-trig integral but is really an algebraic one. Using $\cos x = \sin\left(\tfrac{\pi}{2} – x\right)$:

$$\sin^{-1}(\cos x) = \sin^{-1}\left(\sin\left(\frac{\pi}{2} – x\right)\right) = \frac{\pi}{2} – x$$

on the range where the inverse undoes the sine. So the integrand is a simple linear function:

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

Question 22

$\dfrac{1}{\cos(x – a)\cos(x – b)}$

Solution. The trick is to manufacture the numerator you need. Note that $(x – b) – (x – a) = a – b$ is a constant, so multiply and divide by $\sin(a – b)$ and expand it:

$$\sin(a-b) = \sin\big[(x-b) – (x-a)\big] = \sin(x-b)\cos(x-a) – \cos(x-b)\sin(x-a)$$

Dividing this by $\cos(x-a)\cos(x-b)$ splits into two tangents:

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

Integrating, and using $\int \tan\theta\,d\theta = -\log|\cos\theta|$:

$$= \frac{1}{\sin(a-b)}\big[-\log|\cos(x-b)| + \log|\cos(x-a)|\big]$$

Questions 23 and 24 — choose the correct answer.

Question 23

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

Solution. Split the fraction term by term:

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

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

The two minus signs — one from the integrand, one from $\int\operatorname{cosec}^2 x\,dx = -\cot x$ — cancel. Compare with question 39 of Exercise 7.2, where the numerator was $+1$ and the answer came out as $\tan x – \cot x$.

Question 24

$\displaystyle\int \frac{e^x(1 + x)}{\cos^2(e^x x)}\,dx$ equals

Solution. Look at what appears inside the cosine: $x e^x$. Differentiating it by the product rule:

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

which is exactly the numerator. So with $t = x e^x$:

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

Common mistakes

• Trying to integrate a product of sines and cosines directly. There is no such standard integral. Convert to a sum first — questions 2, 3, 6 and 7 are all this one step.
• Using the wrong reduction for an odd power. Even powers use the double-angle formula (questions 1, 10, 11); odd powers peel off one factor and substitute (questions 4, 5). Applying the even method to an odd power leads nowhere.
• Reaching for half-angle formulas when factorising is quicker. Question 12 cancels to $1 – \cos x$ in one line, and question 18’s numerator collapses to $1$. Simplify before you integrate.
• Missing that $1 + \sin 2x$ is a perfect square. It is $(\cos x + \sin x)^2$, which turns questions 14 and 20 into one-line substitutions.
• Forgetting $\alpha$, $a$ and $b$ are constants. In question 13, $\int\cos\alpha\,dx = x\cos\alpha$, not $\sin\alpha$. Question 22 likewise.
• Confusing the two similar MCQs. Exercise 7.2 question 39 gives $\tan x – \cot x$; this exercise’s question 23 gives $\tan x + \cot x$. The numerator differs, and so does the sign.

Practise next

• Exercise 7.4 — integrals of some particular functions, where the standard forms with $x^2 \pm a^2$ take over.
• Exercise 7.2 — worth another pass, since almost every question here ends in a substitution once the identity has been applied.