# Integration using Trigonometric Properties

Trigonometric properties such as the sum of squares of sine and cosine with the same angle are one,
$$\displaystyle \sin^2{\theta} + \cos^2{\theta} = 1 \\ \cos\Big(\frac{\pi}{2}-\theta \Big) = \sin{\theta}$$
can simplify harder integration.

### Worked on an Example of Integration using Trigonometric Properties

(a)   Find $a$ and $b$ for $\displaystyle \frac{1}{x(4-x)} = \frac{a}{x} + \frac{b}{4-x}$.

$\newcommand\ddfrac[2]{\frac{\displaystyle #1}{\displaystyle #2}}$
\begin{aligned} \displaystyle \require{AMSsymbols} \require{color} \frac{a}{x} + \frac{b}{4-x} &= \frac{a(4-x) + bx}{x(4-x)} &\color{red} \text{single fraction} \\ &= \frac{4a-ax+bx}{x(4-x)} &\color{red} \text{expand the numerator} \\ &= \frac{(-a+b)x + 4a}{x(4-x)} &\color{red} \text{rearrange} \\ \text{Solve } -a+b &=0 \text{ and } 4a = 1 \text{, we get } a = \frac{1}{4} \text{ and } b = \frac{1}{4}. \\ \therefore \frac{1}{x(4-x)} &= \frac{1}{4} \Big(\frac{1}{x} + \frac{1}{4-x} \Big) \end{aligned}

(b)   Evaluate $\displaystyle A =\int_{1}^{3}\frac{\cos^2{\ddfrac{\pi x}{8}}}{x(4-x)} dx$.

\begin{aligned} \displaystyle \require{AMSsymbols} \require{color} \text{Let } u &= 4-x \text{ , then } du = -dx. \\ x &= 3 \Rightarrow u = 4-3 = 1 \\ x &= 1 \Rightarrow u = 4-1 = 3 \end{aligned}
\begin{aligned} \displaystyle \require{color} A &=\int_{3}^{1}\frac{\cos^2{\ddfrac{\pi(4-u)}{8}}}{(4-u)u} (-du) &\color{red} \text{replace by } u = 4 -x \\ &=\int_{3}^{1}\frac{\cos^2{\Big(\ddfrac{\pi}{2} – \ddfrac{\pi u}{8}}\Big)}{(4-u)u} (-du) \\ &= \int_{1}^{3}\frac{\sin^2{\ddfrac{\pi u}{8}}}{(4-u)u} du &\color{red} \cos\Big(\frac{\pi}{2}-\theta \Big) = \sin{\theta} \\ 2A &= \int_{1}^{3}\frac{\cos^2{\ddfrac{\pi x}{8}}}{x(4-x)} dx + \int_{1}^{3}\frac{\sin^2{\ddfrac{\pi u}{8}}}{(4-u)u} du \\ &= \int_{1}^{3}\frac{\cos^2{\ddfrac{\pi x}{8}}}{x(4-x)} dx + \int_{1}^{3}\frac{\sin^2{\ddfrac{\pi x}{8}}}{(4-x)x} dx \\ &= \int_{1}^{3}\frac{\cos^2{\ddfrac{\pi x}{8}} + \sin^2{\ddfrac{\pi x}{8}}}{x(4-x)} dx \\ &= \int_{1}^{3}\frac{1}{x(4-x)} dx &\color{red} \sin^2{\theta} + \cos^2{\theta} = 1 \\ &= \frac{1}{4} \int_{1}^{3}\Big( \frac{1}{x} + \frac{1}{4-x} \Big) dx &\color{red} \text{partial fractions by (a)} \\ &= \frac{1}{4} \int_{1}^{3}\Big( \frac{1}{x}-\frac{-1}{4-x} \Big) dx \\ &= \frac{1}{4} \Big[\log_e{x}-\log_e{(4-x)} \Big]_{1}^{3} \\ &= \frac{1}{4} \Big[(\log_e{3}-\log_e{1})-(\log_e{1}-\log_e{3}) \Big] \\ &= \frac{1}{2} \log_e{3} \\ \therefore A &= \frac{1}{4} \log_e{3} \end{aligned}

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