Trigonometry Made Easy: Integration by Parts Demystified

Integration by Parts is made of the product rule of differentiation. The derivative of \(uv\) is \(u’v + uv’\) and integrates both sides.

\( \begin{aligned} \require{color}
(uv)’ &= u’v + uv’ \\
uv &= \int u’v + \int uv’ \\
\int u’v &= uv-\int uv’ \text{ or } \int uv’ = uv-\int u’v
\end{aligned} \)

Worked Examples of Integration by Parts

\( \displaystyle
\text{Let }A_n = \int_{0}^{\frac{\pi}{2}} \cos^{2n} x dx \text{ and } B_n = \int_{0}^{\frac{\pi}{2}} x^2 \cos^{2n} x dx, \text{where } n \text{ is an integer, } n \ge 0 \text{ and } A_n > 0, \ B_n > 0. \\
\displaystyle \text{The recurrence formula is } n A_n = \frac{2n-1}{2} A_{n-1}.
\)

\( \displaystyle
(a) \ \ \ \text{Show that } A_n = 2n \int_{0}^{\frac{\pi}{2}} x \sin x \cos^{2n-1} x dx \text{ for } n\ge 1. \)

\( \begin{aligned} \require{color}
\color{green} u &= \cos^{2n}x, v’ = 1 \\
\color{green} u’ &= 2n \cos^{2n-1}x (-\sin x), v = x \\
\end{aligned} \\ \)
\( \begin{aligned} \require{color} \displaystyle
A_n &= \Big[x \cos^{2n}x\Big]_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}} 2n \cos^{2n-1} (-\sin x) x dx &\color{red} \int uv’ = uv-\int u’v \\
\therefore A_n &= 2n \int_{0}^{\frac{\pi}{2}} x \sin x \cos^{2n-1} x dx
\end{aligned} \)

\( \displaystyle
(b) \ \ \ \text{Show that } \frac{A_n}{n^2} = \frac{2n-1}{n} B_{n-1}-2 B_n \text{ for } n \ge 1. \\
\)

\( \begin{aligned} \require{color} \displaystyle
u &= \sin x \cos^{2n-1} x \\
u’ &= \cos x \cos^{2n-1} x + \sin x (2n-1) \cos^{2n-2} x (-\sin x) \\
&= \cos^{2n} x-(2n-1) \cos^{2n-2} x \sin^2 x \\
&= \cos^{2n} x-(2n-1) \cos^{2n-2} x (1-\cos^2 x) \\
&= \cos^{2n} x-(2n-1) \cos^{2n-2} x + (2n-1) \cos^{2n} x \\
&= 2n \cos^{2n} x-(2n-1) \cos^{2n-2} x \\
v’ &= x \\
v &= \frac{1}{2} x^2 \\
\color{green} \int uv’ &\color{red}= uv-\int u’v \\
A_n &= 2n \Big[\sin x \cos^{2n-1} x \times \frac{1}{2} x^2 \Big]_{0}^{\frac{\pi}{0}}-2n \int_{0}^{\frac{\pi}{2}} \Big[2n \cos^{2n} x-(2n-1) \cos^{2n-2} x \Big] \times \frac{1}{2} x^2 dx \\
&= -n \int_{0}^{\frac{\pi}{2}} 2n x^2 \cos^{2n} x dx + n \int_{0}^{\frac{\pi}{2}} (2n-1) x^2 \cos^{2n-2} x dx \\
&= -2n^2 \int_{0}^{\frac{\pi}{2}} x^2 \cos^{2n} x dx + (2n-1)n \int_{0}^{\frac{\pi}{2}} x^2 \cos^{2n-2} x dx \\
A_n &= -2n^2 B_n + (2n-1)n B_{n-1} \\
\therefore \frac{A_n}{n^2} &= -2 B_n + \frac{2n-1}{n} B_{n-1} \color{red} \cdots (1)
\end{aligned} \)

\( \displaystyle
(c) \ \ \ \text{Show that } \frac{1}{n^2} = \frac{2 B_{n-1}}{A_{n-1}}-\frac{2 B_n}{A_n} \text{ for } n \ge 1. \\ \)

\( \begin{aligned} \require{color} \displaystyle
n A_n &= \frac{2n-1}{2} A_{n-1} &\color{red} \text{given} \\
\frac{2 A_n}{A_{n-1}} &= \frac{2n-1}{n} \\
\frac{A_n}{n^2} &= -2 B_n + \frac{2 A_n}{A_{n-1}} B_{n-1} &\color{red} \text{substitute to } (1) \\
\therefore \frac{1}{n^2} &= \frac{2 B_{n-1}}{A_{n-1}}-\frac{2 B_n}{A_n} &\color{red} \text{divide both sides by } n^2
\end{aligned} \)

\( \displaystyle
(d) \ \ \ \text{Show that } \sum_{k=1}^{n} \frac{1}{k^2} = \frac{\pi^2}{6}-\frac{2 B_n}{A_n} \\
\)

\( \begin{aligned} \require{AMSsymbols} \require{color} \displaystyle
\sum_{k=1}^{n} \frac{1}{k^2} &= \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots + \frac{1}{n^2} \\
&= \frac{2 B_0}{A_0}-\frac{2 B_1}{A_1} + \frac{2 B_1}{A_1}-\frac{2 B_2}{A_2} + \frac{2 B_2}{A_2} – \frac{2 B_3}{A_3} + \cdots + \frac{2 B_{n-1}}{A_{n-1}}-\frac{2 B_n}{A_n} \\
&= \frac{2 B_0}{A_0}-\frac{2 B_n}{A_n} \color{red} \cdots (2) \\
B_0 &= \int_{0}^{\frac{\pi}{2}} x^2 \cos^{0} x dx \\
&= \int_{0}^{\frac{\pi}{2}} x^2 dx = \Big[\frac{x^3}{3} \Big]_{0}^{\frac{\pi}{2}} = \frac{\pi^3}{24} \\
A_0 &= \int_{0}^{\frac{\pi}{2}} \cos^0 x dx \\
&= \int_{0}^{\frac{\pi}{2}} dx = \big[x \big]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2} \\
\newcommand\ddfrac[2]{\frac{\displaystyle #1}{\displaystyle #2}}
\color{green} (2) &= \frac{2 \ddfrac{\pi^3}{24}}{\ddfrac{\pi}{2}}-\frac{2 B_n}{A_n} \\
\therefore \sum_{k=1}^{n} \frac{1}{k^2} &= \frac{\pi^2}{6}-\frac{2 B_n}{A_n}
\end{aligned} \)

\( \displaystyle
(e) \ \ \ \text{Given } \sin x \ge \frac{2x}{\pi} \text{ for } 0 \le x \le \frac{\pi}{2} \text{ show that } B_n \le \int_{0}^{\frac{\pi}{2}} x^2 \Bigg(1-\frac{4x^2}{\pi^2} \Bigg)^n dx. \\
\)

\( \begin{aligned} \require{color} \displaystyle
\sin^2 x &\ge \frac{4x^2}{\pi^2} \\
-\sin^2 x &\le-\frac{4x^2}{\pi^2} \\
1-\sin^2 x &\le 1-\frac{4x^2}{\pi^2} \\
\cos^2 x &\le 1-\frac{4x^2}{\pi^2} \\
\cos^{2n} x &\le \Bigg(1-\frac{4x^2}{\pi^2}\Bigg)^n \\
x^2 \cos^{2n} x &\le x^2 \Bigg(1-\frac{4x^2}{\pi^2}\Bigg)^n \\
\int_{0}^{\frac{\pi}{2}} x^2 \cos^{2n} x dx &\le \int_{0}^{\frac{\pi}{2}} x^2 \Bigg(1-\frac{4x^2}{\pi^2}\Bigg)^n dx \\
\therefore B_n &\le \int_{0}^{\frac{\pi}{2}} x^2 \Bigg(1-\frac{4x^2}{\pi^2}\Bigg)^n dx
\end{aligned} \)

\( \displaystyle
(f) \ \ \ \text{Show that } \int_{0}^{\frac{\pi}{2}} x^2 \Bigg(1-\frac{4x^2}{\pi^2}\Bigg)^n dx = \frac{\pi^2} {8(n+1)} \int_{0}^{\frac{\pi}{2}} \Bigg(1-\frac{4x^2}{\pi^2}\Bigg)^{n+1} dx.
\)

\( \begin{aligned} \require{AMSsymbols} \require{color} \displaystyle
\int_{0}^{\frac{\pi}{2}} x^2 \Bigg(1-\frac{4x^2}{\pi^2}\Bigg)^n dx &= \int_{0}^{\frac{\pi}{2}} x \times x \Bigg(1-\frac{4x^2}{\pi^2}\Bigg)^n dx \color{green} \cdots (3) \\
\color{red} u &\color{red}= x, u’ = 1 \\
\color{red} v’ &\color{red}= x \Bigg(1-\frac{4x^2}{\pi^2}\Bigg)^n \\
\frac{d}{dx} \Bigg(1-\frac{4x^2}{\pi^2}\Bigg)^{n+1} &= (n+1) \Bigg(1-\frac{4x^2}{\pi^2}\Bigg)^n \times \Big(-\frac{8x}{\pi^2}\Big) \\
\frac{-\pi^2}{8(n+1)}\frac{d}{dx} \Bigg(1-\frac{4x^2}{\pi^2}\Bigg)^{n+1} &= x \Bigg(1-\frac{4x^2}{\pi^2}\Bigg)^n \\
\color{red} v &\color{green}= \frac{-\pi^2}{8(n+1)} \Bigg(1-\frac{4x^2}{\pi^2}\Bigg)^{n+1} \\
\color{red} (3) &= \frac{-\pi^2}{8(n+1)} \Bigg[x\Bigg(1-\frac{4x^2}{\pi^2}\Bigg)^{n+1}\Bigg]_{0}^{\frac{\pi}{2}} + \frac{\pi^2}{8(n+1)} \int_{0}^{\frac{\pi}{2}} \Bigg(1-\frac{4x^2}{\pi^2}\Bigg)^{n+1} dx \\
\therefore \int_{0}^{\frac{\pi}{2}} x^2 \Bigg(1-\frac{4x^2}{\pi^2}\Bigg)^n dx &= \frac{\pi^2} {8(n+1)} \int_{0}^{\frac{\pi}{2}} \Bigg(1-\frac{4x^2}{\pi^2}\Bigg)^{n+1} dx
\end{aligned} \)

\( \displaystyle
(g) \ \ \ \text{Show that } B_n \le \frac{\pi^3}{16(n+1)} \int_{0}^{\frac{\pi}{2}} \cos^{2n+3} t dt \le \frac{\pi^3}{16(n+1)} A_n \text{ using the substitution } x = \frac{\pi}{2} \sin t. \)

\( \text{From } (e) \text{ and } (f), \)
\( \begin{aligned} \require{color} \displaystyle
B_n &\le \int_{0}^{\frac{\pi}{2}} x^2 \Bigg(1-\frac{4x^2}{\pi^2}\Bigg)^n dx = \frac{\pi^2} {8(n+1)} \int_{0}^{\frac{\pi}{2}} \Bigg(1-\frac{4x^2}{\pi^2}\Bigg)^{n+1} dx \\
B_n &\le \frac{\pi^2} {8(n+1)} \int_{0}^{\frac{\pi}{2}} \Bigg(1-\frac{4x^2}{\pi^2}\Bigg)^{n+1} dx \color{green} \cdots (4) \\
dx &= \frac{\pi}{2} \cos t dt \\
x &= 0 \rightarrow t = 0 \\
x &= \frac{\pi}{2} \rightarrow t = \frac{\pi}{2} \\
\color{red} (4) \cdots B_n &\le \frac{\pi^2}{8(n+1)} \int_{0}^{\frac{\pi}{2}} \Bigg(1-\frac{4}{\pi^2} \frac{\pi^2}{2^2} \sin^2 t \Bigg)^{n+1} \frac{\pi}{2} \cos t dt \\
&= \frac{\pi^3}{16(n+1)} \int_{0}^{\frac{\pi}{2}} \big(1-\sin^2 t\big)^{n+1} \cos t dt \\
&= \frac{\pi^3}{16(n+1)} \int_{0}^{\frac{\pi}{2}} \big(\cos^2 t\big)^{n+1} \cos t dt \\
&= \frac{\pi^3}{16(n+1)} \int_{0}^{\frac{\pi}{2}} \cos^{2n+3} t dt \\
\therefore B_n &\le \frac{\pi^3}{16(n+1)} \int_{0}^{\frac{\pi}{2}} \cos^{2n+3} t dt \color{red} \cdots (5) \\
\int_{0}^{\frac{\pi}{2}} \cos^{2n+3} t dt &\le \int_{0}^{\frac{\pi}{2}} \cos^{2n} t dt \color{red} \text{ for } 0 \le t \le \frac{\pi}{2}\\
\int_{0}^{\frac{\pi}{2}} \cos^{2n+3} t dt &\le A_n \\
\therefore \frac{\pi^3}{16(n+1)} \int_{0}^{\frac{\pi}{2}} \cos^{2n+3} t dt &\le \frac{\pi^3}{16(n+1)} A_n \color{red} \cdots (6) \\
\therefore B_n &\le \frac{\pi^3}{16(n+1)} \int_{0}^{\frac{\pi}{2}} \cos^{2n+3} t dt \le \frac{\pi^3}{16(n+1)} A_n \color{red} \text{ by } (5) \text{ and } (6)
\end{aligned} \)

\( \displaystyle
(h) \ \ \ \text{Show that } \frac{\pi^2}{6}-\frac{\pi^3}{8(n+1)} \le \sum_{k=1}^{n} \frac{1}{k^2} \lt \frac{\pi^2}{6}. \)

\( \begin{aligned} \require{AMSsymbols} \require{color} \displaystyle
\frac{2 B_n}{A_n} &= \frac{\pi^2}{6}-\sum_{k=1}^{n} \frac{1}{k^2} \color{red} \cdots (7) &\color{red} \text{by } (d) \\
B_n &\le \frac{\pi^3}{16(n+1)} A_n &\color{red} \text{by } (g) \\
\frac{B_n}{A_n}B_n &\le \frac{\pi^3}{16(n+1)} \\
\frac{2 B_n}{A_n}B_n &\le \frac{\pi^3}{8(n+1)} \color{red} \cdots (8) \\
\frac{\pi^2}{6}-\sum_{k=1}^{n} \frac{1}{k^2} &\le \frac{\pi^3}{8(n+1)} &\color{red} \text{by } (7) \text{ and } (8) \\
\frac{\pi^2}{6}-\frac{\pi^3}{8(n+1)} &\le \sum_{k=1}^{n} \frac{1}{k^2} &\color{red} \cdots (9) \\
\sum_{k=1}^{n} \frac{1}{k^2} &= \frac{\pi^2}{6}-\frac{2 B_n}{A_n} &\color{red} \text{by } (7) \\
\sum_{k=1}^{n} \frac{1}{k^2} &< \frac{\pi^2}{6} \color{red} \cdots (10) &\color{red} \frac{2 B_n}{A_n} > 0 \\
\therefore \frac{\pi^2}{6}-\frac{\pi^3}{8(n+1)} &\le \sum_{k=1}^{n} \frac{1}{k^2} \lt \frac{\pi^2}{6} &\color{red} \text{by } (9) \text{ and } (10)
\end{aligned} \)

\( \displaystyle
(i) \ \ \ \text{Find } \lim_{n\to\infty} \sum_{k=1}^{n} \frac{1}{k^2}. \)

\( \begin{aligned} \require{AMSsymbols} \require{color} \displaystyle
\lim_{n\to\infty} \frac{\pi^3}{8(n+1)} &= 0 \\
\frac{\pi^2}{6}-0 &\le \lim_{n\to\infty} \sum_{k=1}^{n} \frac{1}{k^2} \lt \frac{\pi^2}{6} \\
\therefore \lim_{n\to\infty} \sum_{k=1}^{n} \frac{1}{k^2} &= \frac{\pi^2}{6}
\end{aligned} \)

Related Links

Unlock your full learning potential—download our expertly crafted slide files for free and transform your self-study sessions!

Discover more enlightening videos by visiting our YouTube channel!

 

Algebra Algebraic Fractions Arc Binomial Expansion Capacity Common Difference Common Ratio Differentiation Double-Angle Formula Equation Exponent Exponential Function 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 Pythagoras Theorem Quadratic Quadratic Factorise Ratio Rational Functions Sequence Sketching Graphs Surds Time Transformation Trigonometric Functions Trigonometric Properties Volume




Related Articles

Responses

Your email address will not be published. Required fields are marked *