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 \(\displaystyle A =\int_{1}^{3}\frac{\cos^2{\ddfrac{\pi x}{8}}}{x(4-x)} dx \).
\( \begin{aligned} \displaystyle \require{AMSsymbols} \require{color}
\text{Let } u &= 4-x \text{ , then } du = -dx. \\
x &= 3 \Rightarrow u = 4-3 = 1 \\
x &= 1 \Rightarrow u = 4-1 = 3
\end{aligned} \)
\( \begin{aligned} \displaystyle \require{color}
A &=\int_{3}^{1}\frac{\cos^2{\ddfrac{\pi(4-u)}{8}}}{(4-u)u} (-du) &\color{red} \text{replace by } u = 4 -x \\
&=\int_{3}^{1}\frac{\cos^2{\Big(\ddfrac{\pi}{2} – \ddfrac{\pi u}{8}}\Big)}{(4-u)u} (-du) \\
&= \int_{1}^{3}\frac{\sin^2{\ddfrac{\pi u}{8}}}{(4-u)u} du &\color{red} \cos\Big(\frac{\pi}{2}-\theta \Big) = \sin{\theta} \\
2A &= \int_{1}^{3}\frac{\cos^2{\ddfrac{\pi x}{8}}}{x(4-x)} dx + \int_{1}^{3}\frac{\sin^2{\ddfrac{\pi u}{8}}}{(4-u)u} du \\
&= \int_{1}^{3}\frac{\cos^2{\ddfrac{\pi x}{8}}}{x(4-x)} dx + \int_{1}^{3}\frac{\sin^2{\ddfrac{\pi x}{8}}}{(4-x)x} dx \\
&= \int_{1}^{3}\frac{\cos^2{\ddfrac{\pi x}{8}} + \sin^2{\ddfrac{\pi x}{8}}}{x(4-x)} dx \\
&= \int_{1}^{3}\frac{1}{x(4-x)} dx &\color{red} \sin^2{\theta} + \cos^2{\theta} = 1 \\
&= \frac{1}{4} \int_{1}^{3}\Big( \frac{1}{x} + \frac{1}{4-x} \Big) dx &\color{red} \text{partial fractions by (a)} \\
&= \frac{1}{4} \int_{1}^{3}\Big( \frac{1}{x}-\frac{-1}{4-x} \Big) dx \\
&= \frac{1}{4} \Big[\log_e{x}-\log_e{(4-x)} \Big]_{1}^{3} \\
&= \frac{1}{4} \Big[(\log_e{3}-\log_e{1})-(\log_e{1}-\log_e{3}) \Big] \\
&= \frac{1}{2} \log_e{3} \\
\therefore A &= \frac{1}{4} \log_e{3}
\end{aligned} \)
Algebra Algebraic Fractions Arc Binomial Expansion Capacity Common Difference Common Ratio Differentiation Double-Angle Formula Equation Exponent Exponential Function Factorials Factorise Functions Geometric Sequence Geometric Series Index Laws Inequality Integration Kinematics Length Conversion Logarithm Logarithmic Functions Mass Conversion Mathematical Induction Measurement Perfect Square Perimeter Prime Factorisation Probability Product Rule Proof Quadratic Quadratic Factorise Ratio Rational Functions Sequence Sketching Graphs Surds Time Transformation Trigonometric Functions Trigonometric Properties Volume
