Integration by Reduction Formulae
Evaluation of sine Function
Reduction Formula
A reduction formula is often used in integration for working out integrals of higher order. It is lengthy and tedious to work across higher-degree expressions, and here the reduction formulas are given as simple expressions with a degree \(n\), to solve these higher-degree expressions. These reduction formulas have been derived from the base formulas of integration and work with the same rules of integration.
Integration by Reduction Formula
Integration by reduction formula is a method relying on recurrence relations. It is used when an expression containing an integer parameter, usually in the form of powers of elementary functions, or products of transcendental functions and polynomials of arbitrary degree, can’t be integrated directly. But using other methods of integration a reduction formula can be set up to obtain the integral of the same or similar expression with a lower integer parameter, progressively simplifying the integral until it can be evaluated. [1] This method of integration is one of the earliest used.
A reduction formula expresses an integral \(I_n\) that depends on some integer \(n\) in terms of another integral \(I_m\) that involves a smaller integer \(m\). If one repeatedly applies this formula, one may then express In in terms of a much simpler integral.
We use integration by parts to establish the reduction formula.
Evaluation of \(sine\) Function
Indefinite
\[ I_n = \int \sin^mx\;dx= \int \sin^{m-1}x\ \text{sin}x \;dx \]
\[ =\sin^{m-1}x\int \text{sin}x\ dx\ + (m-1) \int \text{cos}x\; \sin^{m-1}x \; \text{cos}x\;dx \]
\[ = - \sin^{m-1}x\ \text{cos}x\ + (m-1) \int \sin^{m-2}\ (1-\sin^2x)dx \]
\[ = - \sin^{m-1}x\ \text{cos}x\ + (m-1) \int \sin^{m-2}x\ dx\ - (m-1) \int \sin^mx\ dx \]
\[ = - \sin^{m-1}x\ \text{cosx}\ + (m-1) \int \sin^{m-2}x\ dx\ - (m-1) I_n \]
\[ \therefore I_n = -\dfrac{\sin^{m-1}x\; \text{cos}x}{m}+\dfrac{m-1}{m} \int \sin^{m-2}x\ dx \]
\[ \boxed{\int \sin^m x\ dx=-\dfrac{\sin^{m-1}x\; \text{cos}x}{m}+\dfrac{m-1}{m} \int \sin^{m-2}x\ dx} \]
Definite
\(\text{Now,}\)\[\int_0^{\frac{\pi}{2}} \text{sin}x \ dx = -\bigg[\dfrac{\sin^{m-1}x\; \text{cos}x}{m}\bigg]_0^{\frac{\pi}{2}}+\dfrac{m-1}{m} \int \sin^{m-2}x\ dx \]
\(\text{If}\; m \ge 2, \sin^{m-1}x\; \text{cos}x=0\)
\[ \int_0^{\frac{\pi}{2}} \text{sin}^mx \ dx =\dfrac{m-1}{m} \int_0^{\frac{\pi}{2}} \sin^{m-2}x\ dx \]
\[ \int_0^{\frac{\pi}{2}} \text{sin}^mx \ dx=\dfrac{m-1}{m}.\dfrac{m-3}{m-2} \int_0^{\frac{\pi}{2}} \sin^{m-4}x\ dx \]
\(\text{Case 1:}\)
\(\text{When}\; m\; \text{is even positive integer;}\)
\[ \int_0^{\frac{\pi}{2}} \text{sin}^mx \ dx = \dfrac{m-1}{m}.\dfrac{m-3}{m-2}.\dfrac{m-5}{m-4} \ldots \dfrac{3}{4}.\dfrac{1}{2}.\int_0^{\frac{\pi}{2}}(\text{sin}x)^0dx \]
\[ \hspace{0.9 cm} =\dfrac{m-1}{m}.\dfrac{m-3}{m-2}.\dfrac{m-5}{m-4} \ldots \dfrac{3}{4}.\dfrac{1}{2} \int_0^{\frac{\pi}{2}}dx \]
\[ \;\; =\dfrac{m-1}{m}.\dfrac{m-3}{m-2}.\dfrac{m-5}{m-4} \ldots \dfrac{3}{4}.\dfrac{1}{2}.\dfrac{\pi}{2} \]
\[ \boxed{\int_0^{\frac{\pi}{2}} \text{sin}^mx \ dx = \dfrac{m-1}{m}.\dfrac{m-3}{m-2}.\dfrac{m-5}{m-4} \ldots \dfrac{3}{4}.\dfrac{1}{2}.\dfrac{\pi}{2}} \]
\(\text{Case 2:}\)
\(\text{When}\;m\; \text{is odd positive integer;}\)
\[ \int_0^{\frac{\pi}{2}} \text{sin}^mx \ dx = \dfrac{m-1}{m} \int_0^{\frac{\pi}{2}} \sin^{m-4}x\;dx \]
\[ \int_0^{\frac{\pi}{2}} \text{sin}^mx \ dx=\dfrac{m-1}{m}.\dfrac{m-3}{m-2}.\dfrac{m-5}{m-4} \ldots \dfrac{4}{5}.\dfrac{2}{3}.\int_0^{\frac{\pi}{2}}\text{sin}x\ dx \]
\[ \hspace{0.9 cm}=\dfrac{m-1}{m}.\dfrac{m-3}{m-2}.\dfrac{m-5}{m-4} \ldots \dfrac{4}{5}.\dfrac{2}{3} \]
\[ \boxed{\int_0^{\frac{\pi}{2}} \text{sin}^mx \ dx= \dfrac{m-1}{m}.\dfrac{m-3}{m-2}.\dfrac{m-5}{m-4} \ldots \dfrac{4}{5}.\dfrac{2}{3}} \]
Problem Solving
\(\text{Prove that}\; \int_0^{\frac{\pi}{2}}\sin^{2m} \theta\ d\theta= \frac{(2m)!}{(2^m.m!)^2}.\frac{\pi}{2}\)
General Approach
\(\text{Before going to start this problem we want to show you how $(2m)!$}\) \(\text{can be}\) \(\text{mathematically}\) \(\text{justified;}\)
\((2m)! = 2m.(2m-1).(2m-2).(2m-3).(2m-4).(2m-5).\ldots 6.5.4.3.2.1\)
\(\text{Segregating even and odd terms;}\)
\(=\{2m.(2m-2).(2m-4).\ldots6.4.2\}.\{(2m-1).(2m-3).(2m-5)\ldots 5.3.1\}\)
\(=\{2.m.2(m-1).2.(m-2).\ldots(2.3).(2.2).(2.1)\}.\{1.3.5\ldots(2m-1)\}\)
\(=2^m.\{m(m-1).(m-2).\ldots3.2.1\}.\{1.3.5.\ldots(2m-1)\}\)
\(=2^m.m!.\{1.3.5.\ldots(2m-1)\}\)
\[ \therefore \boxed{(2m)! = 2^m.m!.\{1.3.5.\ldots(2m-1)\}} \]
\(\text{Since 2m is even, so}\)
\[\int_0^{\frac{\pi}{2}} \sin^{2m} \theta\ d\theta= \dfrac{(2m-1).(2m-3).(2m-5)\ldots.3.1}{2m.(2m-2).(2m-4).\ldots4.2}. \dfrac{\pi}{2}\]
\[= \dfrac{(2m-1).(2m-3).(2m-5)\ldots.3.1}{2m.(2m-2).(2m-4).\ldots4.2}\times\dfrac{2m.(2m-2).(2m-4).\ldots4.2}{2m.(2m-2).(2m-4).\ldots4.2} \times \ \dfrac{\pi}{2}\]
\[=\dfrac{2m.(2m-1).(2m-2).(2m-3).(2m-4).(2m-5).\ldots 4.3.2.1}{\{2m.(2m-2).(2m-4).\ldots4.2\}^2}\times\dfrac{\pi}{2}\]
\[= \dfrac{(2m)!}{\{2^m.m(m-1).(m-2).(m-3)\ldots3.2.1\}^2}\times\dfrac{\pi}{2}\]
\[ = \dfrac{(2m)!}{\{2^m.m!\}^2}.\dfrac{\pi}{2} \]
\[\text{Hence,}\; \boxed{\int_0^{\frac{\pi}{2}}\sin^{2m} \theta\ d\theta= \frac{(2m)!}{(2^m.m!)^2}.\frac{\pi}{2}}\]
\(\text{Remarks:}\)
1.
\(\dfrac{(2m)!}{\{2^m.m!\}^2}.\dfrac{\pi}{2}= \dfrac{\frac{\pi}{2 }}{4^m} \dfrac{(2m)!}{m!.m!}= \dfrac{(\pi/2)}{4^m} \binom{2m}{m}\)
2.
\(\dfrac{(2m)!}{\{2^m.m!\}^2}.\dfrac{\pi}{2}= \dfrac{2^m.m!.\{1.3.5.\ldots(2m-1)\}}{2^{2m}.(m!)^2}.\dfrac{\pi}{2}=\dfrac{\{1.3.5\ldots(2n-1)\}}{2^m.m!}.\dfrac{\pi}{2}\)
\[ \boxed{\int_0^{\frac{\pi}{2}}\sin^{2m} \theta\ d\theta= \frac{(2m)!}{(2^m.m!)^2}.\frac{\pi}{2}= \dfrac{(\pi/2)}{4^m} \binom{2m}{m} = \dfrac{\{1.3.5\ldots(2n-1)\}}{2^m.m!}.\dfrac{\pi}{2}} \]
Using Complex Exponential
Here, we will encounter same problem within the interval \(0\) to \(\pi\) then we will draw a conclusion, what if we reduce the interval the interval of integration to \(0\) to \(\pi/2.\)
Here our concern problem is,
\[\int_0^{\pi} \sin^{2m} \theta\ d\theta\; , m=0,1,2,3,\ldots\] \(\text{Since,}\;\ \sin \theta = \dfrac{e^{i\theta}-e^{-i\theta}}{2i}\)
\(\text{Therefore,}\;\; \sin^{2m}\theta = \dfrac{(e^{i\theta}-e^{-i\theta})^{2m}}{(2i)^{2m}}\) \(= \dfrac{(e^{i\theta}-e^{-i\theta})^{2m}}{2^{2m}.i^{2m}}\) \(= \dfrac{(e^{i\theta}-e^{-i\theta})^{2m}}{(-1)^m \ 4^m}\)
\(\text{From the Binomial theorem we have, }\)
\[ (e^{i\theta}-e^{-i\theta})^{2m}=\big\{(- \ e^{i\theta})+(e^{i\theta})\big\}^{2m} \]
\[ \hspace{1.5 cm}\; =\sum_{k=0}^{2m} \binom{2m}{k}\ e^{ki\theta}\ (-\ e^{-i\theta})^{2m-k} \]
\[ \; \; =\sum_{k=0}^{2m} \binom{2m}{k}\ e^{ki\theta}\; \dfrac{(-\ e^{\ -i\theta})^{2m}}{(-\ e^{\ -i\theta})^{k}} \]
\[ \; \; =\sum_{k=0}^{2m} \binom{2m}{k}\ e^{ki\theta}\; \dfrac{(-1)^{2m}\ e^{\ -i2m\theta}}{(-1)\ e^{\ -ik\theta}} \]
\(\text{Since}\;(-1)^{2m} =1, \text{we have}\)
\[ (e^{i\theta}-e^{-i\theta})^{2m} = \sum_{k=0}^{2m} \binom{2m}{k}\ e^{i2k\theta}\; \dfrac{e^{-i2m\theta}}{(-1)^k} = \sum_{k=0}^{2m} \binom{2m}{k}\ \dfrac{e^{i2(k-m)\theta}}{(-1)^k} \]
and so
\[ \int_0^{\pi} \sin^{2m} \theta\ d\theta = \dfrac{1}{(-1)^m4^m} \sum_{k=0}^{2m} \dfrac{\binom{2m}{k}}{(-1)^k} \int_0^{\pi} e^{i2(k-m)\theta} \ d \theta \]
Now concentrate for the moment on the integral at the right. If \(k \neq m,\) then,
\[\int_0^{\pi} e^{i2(k-m)\theta} \ d\theta = \bigg[\dfrac{e^{i2(k-m)\theta}}{i2(k-m)}\bigg]_0^{\pi}=0 \] because, at the lower limit of \(\theta=0, e^{i2(k-m)\theta}=e^0=1\) and the upper limit of \(\theta = \pi, e^{i2(k-m)\theta}\) \(=e^{i(k-m)2\pi}\) \(= e^{\text{integer multiple of}\ 2\pi i}=1,\) too. But if \(k=m\) we can’t use this integration because that gives a division by \(0\). So, set \(k=m\) before doing the integral. Thus, for \(k=m\) we have the integrand reducing to one and so our integral is just
\[\int_0^{\pi} d \theta=\pi\] That is,
\[ \boxed{\int_0^{\pi} e^{i2(k-m)\theta} \ d\theta = \begin{equation*} \left\{ \begin{array}{ll} 0 & \quad k\neq m \ \\ \pi & \quad k=m \end{array} \right.\end{equation*}} \]
Thus, the result of the integration either \(0\) or,
\[ \int_0^{\pi} \sin^{2m} \theta\ d\theta = \dfrac{1}{(-1)^m \ 4^m} \dfrac{\binom{2m}{m}}{(-1)^m}\pi = \dfrac{\pi}{4^m}\binom{2m}{m} \]
Because of the symmetry of \(\sin^{2m}\theta\) over the interval \(0\) to \(\pi\), it is clear that reducing the interval of integration to \(0\) to \(\pi/2\) cuts the value of the integral in half, and so,
\[ \int_0^{\frac{\pi}{2}}\sin^{2m} \theta\ d\theta = \dfrac{(\pi/2)}{4^m} \binom{2m}{m} \]
which is often called Walli’s integral, after the name English mathematician John Wallis. This is a curious naming, since Wallis did not evaluate this integral! His name is nonetheless attached to the integral because integrals of the form \(\int_0^{\pi/2} \sin^m \theta\ d \theta\) can be used to derive Wallis’s famous product formula for \(\pi.\)
References
Mathematical methods for physics and engineering, K.F. Riley, M.P. Hobson, S.J. Bence, Cambridge University Press, 2010, ISBN 978-0-521-86153-3
Further Elementary Analysis, R.I. Porter, G. Bell & Sons Ltd, 1978, ISBN 0-7135-1594-5
http://www.sosmath.com/tables/tables.html -> Indefinite integrals list
http://www.sosmath.com/tables/tables.html -> Indefinite integrals list
