Integrating sin, cos and sec² | CAIE 9709 Pure Mathematics 2 Notes

Chapters

Pure Mathematics 2

Mathematics9709

Reverse three radian derivatives

Three linear-angle trig families

cos u→

+ sin u

divide by a

sin u→

− cos u

keep the minus

sec² u→

+ tan u

divide by a

u = ax + b · every answer also needs + C

Reverse the radian derivatives; sine is the family member that introduces a minus.

Secant is the reciprocal of cosine. In symbols, $\sec x=1/\cos x$ , so $\sec^2x$ means $1/\cos^2x$ .

\int\cos(ax+b)\,dx

=\frac1a\sin(ax+b)+C

\int\sin(ax+b)\,dx

=-\frac1a\cos(ax+b)+C

\int\sec^2(ax+b)\,dx

=\frac1a\tan(ax+b)+C

Common mistake

Sine integrates to negative cosine. The inside coefficient must still be divided out after fixing that sign.

Handle each term as its own family

Worked example

Two different inside coefficients

Integrate $4\cos2x+3\sin(3x-1)$ term by term.

$4\cos2x$ becomes $2\sin2x$ .
$3\sin(3x-1)$ becomes $-\cos(3x-1)$ .

2\sin2x-\cos(3x-1)+C

Differentiating both terms restores the original coefficients 4 and 3.

Use radians and exact values at the limits

Worked example

A short exact definite integral

\begin{aligned}I&=\int_0^{\pi/4}2\cos2x\,dx\\&\quad+\int_0^{\pi/4}\sec^2x\,dx\end{aligned}

=\left[\sin2x+\tan x\right]_0^{\pi/4}

=1+1=2

Your turnIndependent check [4 marks]

Evaluate $\displaystyle\int_0^{\pi/6}6\sin3x\,dx$ exactly.

Show worked answer

\int6\sin3x\,dx=-2\cos3x

\left[-2\cos3x\right]_0^{\pi/6}=0-(-2)

Answer

2

Reverse three radian derivatives

Three linear-angle trig families

cos u→

+ sin u

divide by a

sin u→

− cos u

keep the minus

sec² u→

+ tan u

divide by a

u = ax + b · every answer also needs + C

Reverse the radian derivatives; sine is the family member that introduces a minus.

Secant is the reciprocal of cosine. In symbols, $\sec x=1/\cos x$ , so $\sec^2x$ means $1/\cos^2x$ .

\int\cos(ax+b)\,dx

=\frac1a\sin(ax+b)+C

\int\sin(ax+b)\,dx

=-\frac1a\cos(ax+b)+C

\int\sec^2(ax+b)\,dx

=\frac1a\tan(ax+b)+C

Common mistake

Sine integrates to negative cosine. The inside coefficient must still be divided out after fixing that sign.

Handle each term as its own family

Worked example

Two different inside coefficients

Integrate $4\cos2x+3\sin(3x-1)$ term by term.

$4\cos2x$ becomes $2\sin2x$ .
$3\sin(3x-1)$ becomes $-\cos(3x-1)$ .

2\sin2x-\cos(3x-1)+C

Differentiating both terms restores the original coefficients 4 and 3.

Use radians and exact values at the limits

Worked example

A short exact definite integral

\begin{aligned}I&=\int_0^{\pi/4}2\cos2x\,dx\\&\quad+\int_0^{\pi/4}\sec^2x\,dx\end{aligned}

=\left[\sin2x+\tan x\right]_0^{\pi/4}

=1+1=2

Your turnIndependent check [4 marks]

Evaluate $\displaystyle\int_0^{\pi/6}6\sin3x\,dx$ exactly.

Show worked answer

\int6\sin3x\,dx=-2\cos3x

\left[-2\cos3x\right]_0^{\pi/6}=0-(-2)

Answer

2