Mathematics 9709 · Pure Mathematics 2

Integration of sin, cos and sec² of (ax + b)

The same move a third time: reverse the three trig derivatives, and divide by a to put back the chain-rule factor.
The one new thing to watch is a sign: integrating $\sin$ gives a minus cosine.
Keep all angles in radians.

Reverse the derivative, then divide by $a$ (the number in front of $x$ inside): $\int\cos(ax+b)\,dx=\tfrac1a\sin(ax+b)+c$ , $\int\sin(ax+b)\,dx=-\tfrac1a\cos(ax+b)+c$ , $\int\sec^2(ax+b)\,dx=\tfrac1a\tan(ax+b)+c$ .

The three rules

$\int\cos(ax+b)\,dx=\tfrac1a\sin(ax+b)+c$

$\int\sin(ax+b)\,dx=-\tfrac1a\cos(ax+b)+c$

$\int\sec^2(ax+b)\,dx=\tfrac1a\tan(ax+b)+c$

A quick way to never get the sign wrong

Differentiation goes $\sin\to\cos\to-\sin\to-\cos\to\sin$ round a wheel. Integration goes the same wheel backwards. So $\int\cos=\sin$ (one step back from $\sin$ ). And $\int\sin$ steps back to $-\cos$ . The minus comes from the wheel, so you don't have to remember it.

Worked examples

Worked example

Find $\int\sin 2x\,dx$ and $\int\cos 3x\,dx$

$\int\sin2x\,dx$ : reverse to $-\cos2x$ , divide by $a=2$ → $-\tfrac12\cos2x+c$ .
$\int\cos3x\,dx$ : reverse to $\sin3x$ , divide by $a=3$ → $\tfrac13\sin3x+c$ .

Answer

-\tfrac12\cos2x+c\ ,\quad \tfrac13\sin3x+c

Worked example

Exact value of $\int_0^{\pi/4}(3-2\sin 2x)\,dx$ (modelled on a P3 trig integral)

Integrate term by term. The constant $3$ gives $3x$ ; for $-2\sin2x$ , use $\int\sin2x=-\tfrac12\cos2x$ , so $-2\cdot(-\tfrac12\cos2x)=\cos2x$ . Antiderivative $\big[\,3x+\cos2x\,\big]_0^{\pi/4}$ .
Top limit $x=\tfrac{\pi}{4}$ : $\tfrac{3\pi}{4}+\cos\tfrac{\pi}{2}=\tfrac{3\pi}{4}+0$ (calculator in radians).
Bottom limit $x=0$ : $0+\cos0=1$ . Top minus bottom: $\tfrac{3\pi}{4}-1$ .

Answer

\tfrac{3\pi}{4}-1

Where the marks go

Common mistake

\sin

integrates to

-\cos

(that minus is the whole point), and you still need the

\tfrac1a

\int\sin2x=-\tfrac12\cos2x

, not

-\cos2x

Common mistake

These rules only work in radians. Put your calculator in radians before you work out

\cos\tfrac{\pi}{2}

: in degree mode you get a wrong, often non-zero, value.

Now you try

Find $\int 4\sec^2\!\big(2x-\tfrac{\pi}{3}\big)\,dx$ . (9709-style)

Your turn— tap to reveal the worked answer (9709-style)

$\sec^2$ reverses to $\tan$ , and here $a=2$ , so $4\cdot\tfrac12\tan\!\big(2x-\tfrac{\pi}{3}\big)+c$ :

Answer

2\tan\!\big(2x-\tfrac{\pi}{3}\big)+c

PreviousIntegration of 1/(ax + b)NextFurther integration of trigonometric functions

Integration of sin, cos and sec² of (ax + b)

The same move a third time: reverse the three trig derivatives, and divide by a to put back the chain-rule factor.
The one new thing to watch is a sign: integrating $\sin$ gives a minus cosine.
Keep all angles in radians.

The three rules

$\int\cos(ax+b)\,dx=\tfrac1a\sin(ax+b)+c$

$\int\sin(ax+b)\,dx=-\tfrac1a\cos(ax+b)+c$

$\int\sec^2(ax+b)\,dx=\tfrac1a\tan(ax+b)+c$

A quick way to never get the sign wrong

Worked examples

Worked example

Find $\int\sin 2x\,dx$ and $\int\cos 3x\,dx$

$\int\sin2x\,dx$ : reverse to $-\cos2x$ , divide by $a=2$ → $-\tfrac12\cos2x+c$ .
$\int\cos3x\,dx$ : reverse to $\sin3x$ , divide by $a=3$ → $\tfrac13\sin3x+c$ .

Answer

-\tfrac12\cos2x+c\ ,\quad \tfrac13\sin3x+c

Worked example

Exact value of $\int_0^{\pi/4}(3-2\sin 2x)\,dx$ (modelled on a P3 trig integral)

Integrate term by term. The constant $3$ gives $3x$ ; for $-2\sin2x$ , use $\int\sin2x=-\tfrac12\cos2x$ , so $-2\cdot(-\tfrac12\cos2x)=\cos2x$ . Antiderivative $\big[\,3x+\cos2x\,\big]_0^{\pi/4}$ .
Top limit $x=\tfrac{\pi}{4}$ : $\tfrac{3\pi}{4}+\cos\tfrac{\pi}{2}=\tfrac{3\pi}{4}+0$ (calculator in radians).
Bottom limit $x=0$ : $0+\cos0=1$ . Top minus bottom: $\tfrac{3\pi}{4}-1$ .

Answer

\tfrac{3\pi}{4}-1

Where the marks go

Common mistake

\sin

integrates to

-\cos

(that minus is the whole point), and you still need the

\tfrac1a

\int\sin2x=-\tfrac12\cos2x

, not

-\cos2x

Common mistake

These rules only work in radians. Put your calculator in radians before you work out

\cos\tfrac{\pi}{2}

: in degree mode you get a wrong, often non-zero, value.

Now you try

Find $\int 4\sec^2\!\big(2x-\tfrac{\pi}{3}\big)\,dx$ . (9709-style)

Your turn— tap to reveal the worked answer (9709-style)

$\sec^2$ reverses to $\tan$ , and here $a=2$ , so $4\cdot\tfrac12\tan\!\big(2x-\tfrac{\pi}{3}\big)+c$ :

Answer

2\tan\!\big(2x-\tfrac{\pi}{3}\big)+c