Integrating Using Trigonometric Identities

Only three trigonometric functions have integrals you can write down directly: $\sin$, $\cos$ and $\sec^2$. Squares such as $\sin^2 x$, $\cos^2 2x$ and $\tan^2 x$, and products such as $\sin 3x\cos x$, are not in the table. The syllabus asks you to "use trigonometrical relationships in carrying out integration": rewrite the integrand with an identity until every term is standard, then integrate. These questions are common in P3, often as the second half of a question whose first half proves the identity you need.

The idea: lower the power, raise the angle

The double angle formula for cosine has three forms:

$$\cos 2x = \cos^2 x - \sin^2 x = 2\cos^2 x - 1 = 1 - 2\sin^2 x$$

Rearranging the last two gives $\cos^2 x$ and $\sin^2 x$ in terms of $\cos 2x$, which has no square. A squared function of $x$ becomes a first power of a function of $2x$, and a first power of $\cos 2x$ is easy to integrate.

To remember which sign goes with which, test $x = 0$: $\cos^2 0 = 1$ and $\tfrac{1}{2}(1 + \cos 0) = 1$, so cosine takes the plus sign.

The identities for $\tan^2$ and $\cot^2$ come from dividing $\sin^2 x + \cos^2 x = 1$ by $\cos^2 x$ and by $\sin^2 x$. They work because $\sec^2 x$ and $\operatorname{cosec}^2 x$ are derivatives you know:

$$\frac{d}{dx}\tan x = \sec^2 x, \qquad \frac{d}{dx}\cot x = -\operatorname{cosec}^2 x$$

so $\displaystyle\int \sec^2 x\, dx = \tan x + c$ and $\displaystyle\int \operatorname{cosec}^2 x\, dx = -\cot x + c$. The second is not in the formula list, but it follows from differentiating $\cot x = \dfrac{\cos x}{\sin x}$ with the quotient rule (see Differentiating trigonometric functions).

Squares of sine and cosine

$$\int \cos^2 x\, dx = \int \tfrac{1}{2}(1 + \cos 2x)\, dx = \tfrac{1}{2}x + \tfrac{1}{4}\sin 2x + c$$ $$\int \sin^2 x\, dx = \int \tfrac{1}{2}(1 - \cos 2x)\, dx = \tfrac{1}{2}x - \tfrac{1}{4}\sin 2x + c$$

Do not memorise these results; memorise the identities and derive the integrals each time. With a different angle the numbers change: $\displaystyle\int \sin^2 3x\, dx = \int \tfrac{1}{2}(1 - \cos 6x)\, dx = \tfrac{1}{2}x - \tfrac{1}{12}\sin 6x + c$.

The graph shows why the answer contains $\tfrac{1}{2}x$. The curve $y = \sin^2 x$ oscillates between $0$ and $1$ about the line $y = \tfrac{1}{2}$, so on average it contributes $\tfrac{1}{2}$ per unit of $x$. The $\cos 2x$ term is the wobble around that average.

$y = \sin^2 x$ oscillates about $y = \tfrac{1}{2}$, twice as fast as $\sin x$. The shaded area from $0$ to $\pi$ is exactly $\tfrac{\pi}{2}$, half of the rectangle of height $1$.

Higher even powers

For $\cos^4 x$ or $\sin^4 x$, apply the identity twice. Square the identity for $\cos^2 x$, and you meet $\cos^2 2x$, which needs the identity again with angle $2x$:

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

Questions usually ask you to prove an identity like this first, then use it.

Products of the two

$\sin^2 x\cos^2 x$ looks hard but collapses: $\sin x\cos x = \tfrac{1}{2}\sin 2x$, so $\sin^2 x\cos^2 x = \tfrac{1}{4}\sin^2 2x = \tfrac{1}{8}(1 - \cos 4x)$.

Odd powers such as $\sin^3 x$ or $\cos^5 x$, and products such as $\sin^2 x\cos x$, are handled differently: by recognition or substitution, since $\cos x$ is the derivative of $\sin x$. See Integrating f'(x)/f(x) and related forms and Integration by substitution.

Squares of tangent and cotangent

$\tan^2 x$ has no simple integral as it stands, but $\sec^2 x$ does:

$$\int \tan^2 x\, dx = \int\left(\sec^2 x - 1\right) dx = \tan x - x + c$$

Similarly $\displaystyle\int \cot^2 x\, dx = \int\left(\operatorname{cosec}^2 x - 1\right) dx = -\cot x - x + c$.

Some integrands become $\sec^2$ after a double angle step. Since $1 + \cos 2x = 2\cos^2 x$,

$$\frac{1}{1 + \cos 2x} = \frac{1}{2\cos^2 x} = \tfrac{1}{2}\sec^2 x, \qquad\text{so}\qquad \int\frac{1}{1 + \cos 2x}\, dx = \tfrac{1}{2}\tan x + c$$

Products of sines and cosines with different angles

$\sin 3x\cos x$ is a product of two different angles. Adding the compound angle formulae

$$\sin(A + B) = \sin A\cos B + \cos A\sin B, \qquad \sin(A - B) = \sin A\cos B - \cos A\sin B$$

gives $\sin(A + B) + \sin(A - B) = 2\sin A\cos B$. Similar sums and differences give the other two forms.

These are not on the formula list and you are not expected to quote them. A question that needs one either asks you to derive it from the compound angle formulae or gives it to you. Know how to derive them in two lines.

Using the R formula

An expression $a\cos x + b\sin x$ can be written as $R\cos(x - \alpha)$ (see The R method). That turns an integrand like $\dfrac{1}{(\sqrt{3}\cos x + \sin x)^2}$ into a multiple of $\sec^2(x - \alpha)$, which integrates to a tangent. Look for this whenever the question has just asked you to express something in the form $R\cos(x \pm \alpha)$ or $R\sin(x \pm \alpha)$.

Worked examples

Practice

Answers

1. $\cos^2 4x = \tfrac{1}{2}(1 + \cos 8x)$, so the integral is $\tfrac{1}{2}x + \tfrac{1}{16}\sin 8x + c$.

2. $2\sin^2\tfrac{1}{2}x = 1 - \cos x$, so the integral is $x - \sin x + c$.

3. $\displaystyle\int_0^{\pi/4}\left(\sec^2 x - 1\right) dx = \Big[\tan x - x\Big]_0^{\pi/4} = 1 - \dfrac{\pi}{4}$.

4. $(\sin x + \cos x)^2 = \sin^2 x + \cos^2 x + 2\sin x\cos x = 1 + \sin 2x$. Then $\Big[x - \tfrac{1}{2}\cos 2x\Big]_0^{\pi/2} = \left(\tfrac{\pi}{2} + \tfrac{1}{2}\right) - \left(0 - \tfrac{1}{2}\right) = \dfrac{\pi}{2} + 1$.

5. $1 + \cos 2x = 1 + (2\cos^2 x - 1) = 2\cos^2 x$, so $\dfrac{1}{1 + \cos 2x} = \dfrac{1}{2\cos^2 x} = \tfrac{1}{2}\sec^2 x$. Then $\Big[\tfrac{1}{2}\tan x\Big]_0^{\pi/4} = \dfrac{1}{2}$.

6. $\sin 2x\cos 2x = \tfrac{1}{2}\sin 4x$. Then $\Big[-\tfrac{1}{8}\cos 4x\Big]_0^{\pi/6} = -\tfrac{1}{8}\cos\tfrac{2\pi}{3} + \tfrac{1}{8} = \tfrac{1}{16} + \tfrac{1}{8} = \dfrac{3}{16}$.

7. $\sin^2 x\cos^2 x = (\sin x\cos x)^2 = \left(\tfrac{1}{2}\sin 2x\right)^2 = \tfrac{1}{4}\sin^2 2x = \tfrac{1}{4}\cdot\tfrac{1}{2}(1 - \cos 4x) = \tfrac{1}{8}(1 - \cos 4x)$. Then $\tfrac{1}{8}\Big[x - \tfrac{1}{4}\sin 4x\Big]_0^{\pi/4} = \tfrac{1}{8}\left(\tfrac{\pi}{4} - \tfrac{1}{4}\sin\pi\right) = \dfrac{\pi}{32}$.

8. $\cos(5x + 3x) = \cos 5x\cos 3x - \sin 5x\sin 3x$ and $\cos(5x - 3x) = \cos 5x\cos 3x + \sin 5x\sin 3x$. Adding: $\cos 8x + \cos 2x = 2\cos 5x\cos 3x$. Then $\tfrac{1}{2}\Big[\tfrac{1}{8}\sin 8x + \tfrac{1}{2}\sin 2x\Big]_0^{\pi/4} = \tfrac{1}{2}\left(\tfrac{1}{8}\sin 2\pi + \tfrac{1}{2}\sin\tfrac{\pi}{2}\right) = \tfrac{1}{2}\cdot\tfrac{1}{2} = \dfrac{1}{4}$.

9. $\sin^4 x = \tfrac{1}{4}(1 - \cos 2x)^2 = \tfrac{1}{4}\left(1 - 2\cos 2x + \cos^2 2x\right) = \tfrac{1}{4}\left(1 - 2\cos 2x + \tfrac{1}{2} + \tfrac{1}{2}\cos 4x\right) = \tfrac{3}{8} - \tfrac{1}{2}\cos 2x + \tfrac{1}{8}\cos 4x$. Then $\Big[\tfrac{3}{8}x - \tfrac{1}{4}\sin 2x + \tfrac{1}{32}\sin 4x\Big]_0^{\pi/4} = \tfrac{3\pi}{32} - \tfrac{1}{4}\sin\tfrac{\pi}{2} + \tfrac{1}{32}\sin\pi = \tfrac{3\pi}{32} - \tfrac{1}{4} = \dfrac{3\pi - 8}{32}$.

10. (a) $R\cos(x + \alpha) = R\cos x\cos\alpha - R\sin x\sin\alpha$, so $R\cos\alpha = 1$ and $R\sin\alpha = 1$. Hence $R = \sqrt{2}$, $\alpha = \tfrac{\pi}{4}$: $\cos x - \sin x = \sqrt{2}\cos\left(x + \tfrac{\pi}{4}\right)$. (b) The integrand is $\dfrac{1}{2\cos^2\left(x + \frac{\pi}{4}\right)} = \tfrac{1}{2}\sec^2\left(x + \tfrac{\pi}{4}\right)$, and $x + \tfrac{\pi}{4}$ runs from $\tfrac{\pi}{4}$ to $\tfrac{5\pi}{12}$, where the cosine is not zero. The integral is $\tfrac{1}{2}\Big[\tan\left(x + \tfrac{\pi}{4}\right)\Big]_0^{\pi/6} = \tfrac{1}{2}\left(\tan\tfrac{5\pi}{12} - \tan\tfrac{\pi}{4}\right)$. By the compound angle formula, $\tan\tfrac{5\pi}{12} = \tan\left(\tfrac{\pi}{4} + \tfrac{\pi}{6}\right) = \dfrac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}} = \dfrac{\sqrt{3} + 1}{\sqrt{3} - 1} = \dfrac{(\sqrt{3} + 1)^2}{2} = 2 + \sqrt{3}$. So the integral is $\tfrac{1}{2}\left(2 + \sqrt{3} - 1\right) = \dfrac{1 + \sqrt{3}}{2}$.