Integrating Trig Functions

$\int \sin x d x = - \cos x + c$

$\int \cos x d x = \sin x + c$

where $c$ is the constant of integration

$\int \sec^{2} x d x = \tan x + c$

$\int \sin (a x + b) d x = - \frac{1}{a} \cos (a x + b) + c$

$\int \cos (a x + b) d x = \frac{1}{a} \sin (a x + b) + c$

$\int \sec^{2} (a x + b) d x = \frac{1}{a} \tan (a x + b) + c$

For calculus with trigonometric functions angles must be measured in radians

Find, in the form

f (x) + c

, an expression for each integral

\int \cos x d x

ii.

\int \sec^{2} (3 x - \frac{π}{3}) d x

i. This is a result you should be able to recall from memory.

\int \cos x d x = \sin x + c

ii. Use the standard result:

\int \sec^{2} (a x + b) d x = \frac{1}{a} \tan (a x + b) + c .

\int \sec^{2} (3 x - \frac{π}{3}) d x = \frac{1}{3} \tan (3 x - \frac{π}{3}) + c

A curve has the following equation:

y = \int 2 \sin (2 x + \frac{π}{6}) d x

The curve passes through the point with coordinates

(\frac{π}{3}, \sqrt{3}) .

Find an expression for

y .

Factor out the 2 in front of the sin first, and then use the standard result:

\int \sin (a x + b) d x = - \frac{1}{a} \cos (a x + b) + c .

y = 2 \int \sin (2 x + \frac{π}{6}) d x y = 2 (- \frac{1}{2} \cos (2 x + \frac{π}{6}) + c)

When multiplying the arbitrary constant

c

, by 2, it is still just an arbitrary constant and so it is still just written as

c .

y = - \cos (2 x + \frac{π}{6}) + c

Substitute in the given coordinate, and evaluate, to find

c .

x = \frac{π}{3}, y = \sqrt{3}

\sqrt{3} = - \cos (\frac{2 π}{3} + \frac{π}{6}) + c \sqrt{3} = - \cos (\frac{5 π}{6}) + c \sqrt{3} = \frac{\sqrt{3}}{2} + c c = \frac{\sqrt{3}}{2}

Rewrite the full expression for

y .

y = - \cos (2 x + \frac{π}{6}) + \frac{\sqrt{3}}{2}

Integrating Trig Functions (Cambridge O Level Additional Maths)