Integration using Trigonometric Properties


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

Worked Example of Integration using Trigonometric Properties

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

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

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

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

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