Integrating Trig Functions (Cambridge (CIE) IGCSE Additional Maths): Revision Note

• Remember to add 'c', the constant of integration, for any indefinite integrals 

a) Find, in the form$\mathrm{f}\left(x\right)+c$, an expression for each integral

i. $\int \cos x\mathrm{d}x$

ii. $\int {\sec }^2(3x-\frac{\mathrm{\pi }}{3})\mathrm{d}x$

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

$\int \cos x\mathrm{d}x=\mathbf{sin}\mathit{x}\mathbf{+}\mathit{c}$  

ii.   Use the standard result: $\int {\sec }^2\left(ax+b\right)dx=\frac{1}{a}\tan \left(ax+b\right)+c.$  

$\int {\sec }^2(3x-\frac{\mathrm{\pi }}{3})\mathrm{d}x=\frac{\mathbf{1}}{\mathbf{3}}\mathbf{tan}\mathbf{(}\mathbf{3}\mathit{x}\mathbf{-}\frac{\mathbf{\pi }}{\mathbf{3}}\mathbf{)}\mathbf{+}\mathit{c}$

b) A curve has the following equation:

$y=\int 2\sin \left(2x+\frac{\mathrm{\pi }}{6}\right)\mathrm{d}x$

The curve passes through the point with coordinates $\left(\frac{\mathrm{\pi }}{3},\sqrt{3}\right).$

Find an expression for $y.$

Factor out the 2 in front of the sin first, and then use the standard result: $\int \sin \left(ax+b\right)dx=-\frac{1}{a}\cos \left(ax+b\right)+c.$   

$$\begin{gathered} y=2\int \sin \left(2x+\frac{\mathrm{\pi }}{6}\right)dx \\ y=2\left(-\frac{1}{2}\cos \left(2x+\frac{\mathrm{\pi }}{6}\right)+c\right) \end{gathered}$$  

When multiplying the arbitrary constant $c$, by 2, it is still just an arbitrary constant and so it is still just written as $c.$   

$y=-\cos \left(2x+\frac{\mathrm{\pi }}{6}\right)+c$  

Substitute in the given coordinate, and evaluate, to find $c.$   

At $x=\frac{\mathrm{\pi }}{3},y=\sqrt{3}$

$$\begin{gathered} \sqrt{3}=-\cos \left(\frac{2\mathrm{\pi }}{3}+\frac{\mathrm{\pi }}{6}\right)+c \\ \sqrt{3}=-\cos \left(\frac{5\mathrm{\pi }}{6}\right)+c \\ \sqrt{3}=\frac{\sqrt{3}}{2}+c \\ c=\frac{\sqrt{3}}{2} \end{gathered}$$

  Rewrite the full expression for $y.$  

$\mathit{y}\mathbf{=}\mathbf{-}\mathbf{cos}\mathbf{(}\mathbf{2}\mathit{x}\mathbf{+}\frac{\mathbf{\pi }}{\mathbf{6}}\mathbf{)}\mathbf{+}\frac{\sqrt{\mathbf{3}}}{\mathbf{2}}$