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The Reduction Formula and Rotations

April 5th, 2009 |

Author: Merrin

The integrals of powers of individual trigonometric functions and hyperbolic functions can be computed with reduction formulas. There are recursion relations which relate integrals of higher powers to integrals of lower powers. The recursion relation brings the power down by two each time. Eventually an integral to the first power or zeroth power remains. The first power integrals are just the fundamental integrals and the zeroth power integral is just x. Recursion relations like these also occur in other situations besides trigonometric and hyperbolic power integrals.

There are 12 formulas in all to derive for this section. Deriving each one individually could be pretty painful, but I will introduce some new integration techniques which generate some of the formulas from others. We show how the formulas involving the nth powers of the trigonometric functions. Any given trigonometric identity involving integrals can be used to derive a corresponding hyperbolic formula. It is possible to jump around in the duality amongst trigonometric functions as well as the duality between trigonometric functions and hyperbolic functions. The method presented is an extension of Osborne’s rule for identities involving trigonometric integral equations. We can’t apply these methods if we don’t have equations to start with so we now derive the first three reduction formula the standard way.

Example 1. Calculate the reduction formula for

$\displaystyle \int \text{sin}\,\, ^{n} xdx$

Solution 1. The basic method is to integrate by parts. The original integral reappears and is solved for algebraically.

$I_{n} = \displaystyle \int \text{sin}\,\, ^{n} xdx$

$u=\text{sin}\,\,^{n-1}x\quad dv = -d(\text{cos}\,\, x)\quad v = -\text{cos}\,\, x$

$du=(n-1)\text{sin}\,\,^{n-2}x\text{cos}\,\, xdx$

$I_n=-\text{sin}\,\,^{n-1}x\text{cos}\,\, x+ (n-1)\displaystyle \int (\text{sin}\,\, x)^{n-2} \text{cos}\,\, ^{2}xdx$

$I_n=-\text{sin}\,\,^{n-1}x\text{cos}\,\, x+ (n-1)\displaystyle \int (\text{sin}\,\, x)^{n-2} (1-\text{sin}\,\,^2x)dx$

$I_n=-\text{sin}\,\,^{n-1}x\text{cos}\,\, x+ (n-1)I_{n-2}-(n-1)I_n$

$I_n=-\cfrac{\text{sin}\,\,^{n-1}x\text{cos}\,\, x}{n}+ \cfrac{(n-1)}{n}I_{n-2}$

Example 2. Calculate the reduction formula for

$\displaystyle \int \text{tan}\,\, ^{n} xdx$

Solution 2. A trigonometric identity solves the integral, instead of integration by parts.

$I_n = {\displaystyle \int \text{tan}\,\, ^{n} xdx }$

$I_n = {\displaystyle \int (\text{tan}\,\, x)^{n-2} ((\text{sec}\,\, x)^{2} -1)dx}$

$I_n ={\displaystyle \int (\text{tan}\,\, x)^{n-2} (\text{sec}\,\, x)^{2} dx -\displaystyle \int (\text{tan}\,\, x)^{n-2} dx }$

$I_n=\displaystyle \int u^{n-2}du -I_{n-2}$

$I_n=\cfrac{(\text{tan}\,\, x)^{n-1}}{{n-1}}-I_{n-2}$

Example 3. Calculate the reduction formula for

$\displaystyle \int \text{sec}\,\, ^{n} xdx$

Solution 3. To chop down high powers of secant we can factor out secant squared which is the derivative of tangent and integrate by parts.

${I_{n} } {=} {\displaystyle \int \text{sec}\,\, ^{n}xdx =\displaystyle \int(\text{sec}\,\, x)^{n-2}(\text{sec}\,\, x)^{2} dx}$

$u=(\text{sec}\,\, x)^{n-2} \quad dv=(\text{sec}\,\, x)^2dx \quad du=(n-2)(\text{sec}\,\, x)^{n-2}(\text{tan}\,\, x)dx$
$v = \text{tan}\,\, x$

$I_n=(\text{sec}\,\, x)^{n-2}\text{tan}\,\, x - (n-2)\displaystyle \int(\text{tan}\,\, x)^2(\text{sec}\,\, x)^{n-2}dx$

$I_n=(\text{sec}\,\, x)^{n-2}\text{tan}\,\, x - (n-2)\displaystyle \int\left((\text{sec}\,\, x)^2-1\right)(\text{sec}\,\, x)^{n-2}dx$

$I_n=(\text{sec}\,\, x)^{n-2}\text{tan}\,\, x - (n-2)I_n+(n-2)I_{n-2}$

$I_n=\cfrac{(\text{sec}\,\, x)^{n-2}\text{tan}\,\, x}{n-1} +\cfrac{n-2}{n-1}I_{n-2}$

Example 4. Given the following three reduction formulas derive the other
three.

$\displaystyle \int\text{sin}\,\,^nxdx=-\cfrac{\text{sin}\,\,^{n-1}x\text{cos}\,\, x}{n}+ \cfrac{(n-1)}{n}\displaystyle \int\text{sin}\,\,^{n-2}xdx$
$\displaystyle \int \text{tan}\,\,^nxdx= \cfrac{(\text{tan}\,\, x)^{n-1}}{{n-1}}-\displaystyle \int \text{tan}\,\,^{n-2}xdx$
$\displaystyle \int \text{sec}\,\,^nxdx=\cfrac{(\text{sec}\,\, x)^{n-2}\text{tan}\,\, x}{n-1} +\cfrac{n-2}{n-1}\displaystyle \int \text{sec}\,\,^{n-2}xdx$

Solution 4. A method to solve for the other three reduction formulas
without much more work is to rotate the equations. The details of the rotation
are as follows.

$x\to x+\pi/2 \qquad dx\to d(x+\pi/2)=dx$

$\text{sin}\,\, x \to \text{sin}\,\,(x+\pi /2)=\text{cos}\,\, x$

$\text{cos}\,\, x \to \text{cos}\,\,(x+\pi /2)=-\text{sin}\,\, x$

To perform this transformation everywhere there is an x including inside dx, replace x with x + pi/2.
The way sine and cosine update is just from the standard trigonometric identities.

$\text{sin}\,(A+B)=\text{sin}\,(A)\text{cos}\,(B)+\text{cos}\,(A)\text{sin}\,\,(B)$

$\text{cos}\,\,(A+B)=\text{cos}\,(A)\text{cos}\,(B)-\text{sin}\,(A)\text{sin}\,(B)$

Now we apply this transformation to the first equation for

$\displaystyle \int \text{sin}\,\,^nxdx$ .
$\displaystyle \int\text{cos}\,\,^nxdx=-\cfrac{(\text{cos}\,\,^{n-1}x)(-\text{sin}\,\, x)}{n}+ \cfrac{(n-1)}{n}\displaystyle \int\text{cos}\,\,^{n-2}xdx$

So we have simply that

$\displaystyle \int\text{cos}\,\,^nxdx=+\cfrac{(\text{cos}\,\,^{n-1}x)(\text{sin}\,\, x)}{n}+ \cfrac{(n-1)}{n}\displaystyle \int\text{cos}\,\,^{n-2}xdx$

Continuing to the next equation for

$\displaystyle \int \text{tan}\,\,^nxdx$
$(-1)^n\displaystyle \int \text{cot}\,\,^nx dx= (-1)^{n-1}\cfrac{(\text{cot}\,\, x)^{n-1}}{{n-1}}-(-1)^{n-2}\displaystyle \int \text{cot}\,\,^{n-2}xdx$
$\displaystyle \int \text{cot}\,\,^nx dx= -\cfrac{(\text{cot}\,\, x)^{n-1}}{{n-1}}-\displaystyle \int \text{cot}\,\,^{n-2}xdx$

Moving onto the last equation,

$\displaystyle \int \text{sec}\,\,^nxdx=\cfrac{(\text{sec}\,\, x)^{n-2}\text{tan}\,\, x}{n-1} +\cfrac{n-2}{n-1}\displaystyle \int \text{sec}\,\,^{n-2}xdx$

$(-1)^n\displaystyle \int \text{csc}\,\,^nxdx=(-1)^{n-1}\cfrac{(\text{csc}\,\, x)^{n-2}\text{cot}\,\, x}{n-1} +(-1)^{n-2}\cfrac{n-2}{n-1}\displaystyle \int \text{csc}\,\,^{n-2}xdx$
$\displaystyle \int \text{csc}\,\,^nxdx=-\cfrac{(\text{csc}\,\, x)^{n-2}\text{cot}\,\, x}{n-1} +\cfrac{n-2}{n-1}\displaystyle \int \text{csc}\,\,^{n-2}xdx$

Example 5. Apply the transformation method to common trigonometric integrals and verify the correct signs and functions result.

Solution 5. We can test the transformation method on the fundamental trigonometric integrals. The integrals we can test are
$\displaystyle \int \text{sin}\,\, x dx = -\text{cos}\,\, x +C \\\displaystyle \int \text{tan}\,\, x dx = +\ln \left| \sec x \right|+C\\\displaystyle \int \text{sec}\,\, x dx = +\ln\left|\text{sec}\,\, x +\text{tan}\,\, x\right| + C$
The transformation x to x +pi/2 then

$\text{cos}\,\, x \to -\text{sin}\,\, x$ ,

$\text{sin}\,\, x \to \text{cos}\,\, x$

and dx to dx results in the following equations

$\displaystyle \int \text{cos}\,\, x dx=+\text{sin}\,\, x +C$
$\displaystyle \int \text{cot}\,\, x dx = -\ln\left|\csc x\right| + C$
$\displaystyle \int \text{csc}\,\, x dx = -\ln \left|\text{csc}\,\, x +\text{cot}\,\, x \right| +C$

All the integrals work out as they should.

To derive the hyperbolic reduction formula, the usual methods of integrating by parts and picking apart the integrands with hyperbolic identities still work. Note that all the hyperbolic identities are known from using Osborne’s rule. Instead we will use a new technique to rotate a trigonometric equation into a hyperbolic equation. To do this requires a complex rotation. So our transformation will be

$x \to ix$

In physics, this transformation is known as a Wick rotation. It is used there to convert a statistical theory into a quantum theory and vice versa.

Here are how the functions transform under the complex rotation.

$dx\to idx \quad \cos x \to \cos (ix)=\text{cosh}\,\,(x) \quad \sin x \to \sin (ix)=i\sinh x$

So let’s try the transformation and derive the hyperbolic reduction formula.

Example 6. Derive a corresponding hyperbolic reduction formula from

$\displaystyle \int\text{sin}\,\,^nxdx=-\cfrac{\text{sin}\,\,^{n-1}x\text{cos}\,\, x}{n}+ \cfrac{(n-1)}{n}\displaystyle \int\text{sin}\,\,^{n-2}xdx$

Solution 6. Using the given transformation we have
$i^n\displaystyle \int \text{sinh}\,\,^nx(idx)=-i^{n-1}\cfrac{\text{sinh}\,\,^{n-1}x\text{cosh}\,\, x}{n}+ i^{n-2}\cfrac{(n-1)}{n}\displaystyle \int\text{sinh}\,\,^{n-2}x(idx)$
Canceling the factors of i, remembering the differentials contribute also,
gives
$\displaystyle \int \text{sinh}\,\,^nxdx=\cfrac{\text{sinh}\,\,^{n-1}x\text{cosh}\,\, x}{n}- \cfrac{(n-1)}{n}\displaystyle \int\text{sinh}\,\,^{n-2}xdx$

Example 7. Find the other hyperbolic reduction formulas

Solution 7.

$\displaystyle \int \text{tan}\,\,^nxdx= \cfrac{(\text{tan}\,\, x)^{n-1}}{{n-1}}-\displaystyle \int \text{tan}\,\,^{n-2}xdx$
$i^{n+1}A=i^{n-1}B-i^{n-1}C$
$A = - B+C$
$\displaystyle \int \text{tanh}\,\,^nxdx= -\cfrac{(\text{tanh}\,\, x)^{n-1}}{{n-1}}+\displaystyle \int \text{tanh}\,\,^{n-2}xdx$
Two sign changes for

$\displaystyle \int\text{tanh}\,\,^n xdx$

$\displaystyle \int \text{sec}\,\,^nxdx=\cfrac{(\text{sec}\,\, x)^{n-2}\text{tan}\,\, x}{n-1} +\cfrac{n-2}{n-1}\displaystyle \int \text{sec}\,\,^{n-2}xdx$
$iA=iB+iC$
$\displaystyle \int \text{sech}\,\,^nxdx=\cfrac{(\text{sech}\,\, x)^{n-2}\text{tanh}\,\, x}{n-1} +\cfrac{n-2}{n-1}\displaystyle \int \text{sech}\,\,^{n-2}xdx$
There are no sign changes for

$\displaystyle \int\text{sech}\,\,^n xdx.$

From playing with the signs up to now it should be obvious how to extend Osborne’s rule to equations involving trigonometric integrals that are converted into equations involving hyperbolic integrals. The rule is simple, every time there is a sine convert it into a hyperbolic sine multiplied by i and if there is an integral involved also multiply by i. Most of these factors cancel leaving real coefficients. The obvious cancellation is i to the power of n-1 then there are only individual factors of negative one to grapple with.
$\displaystyle \int\text{cos}\,\,^nxdx=+\cfrac{(\text{cos}\,\,^{n-1}x)(\text{sin}\,\, x)}{n}+ \cfrac{(n-1)}{n}\displaystyle \int\text{cos}\,\,^{n-2}xdx$
$\displaystyle \int \text{cot}\,\,^nx dx= -\cfrac{(\text{cot}\,\, x)^{n-1}}{{n-1}}-\displaystyle \int \text{cot}\,\,^{n-2}xdx$
$\displaystyle \int \text{csc}\,\,^nxdx=-\cfrac{(\text{csc}\,\, x)^{n-2}\text{cot}\,\, x}{n-1} +\cfrac{n-2}{n-1}\displaystyle \int \text{csc}\,\,^{n-2}xdx$
We can just convert these last three equations in our head.
$\displaystyle \int\text{cosh}\,\,^nxdx=+\cfrac{(\text{cosh}\,\,^{n-1}x)(\text{sinh}\,\, x)}{n}+ \cfrac{(n-1)}{n}\displaystyle \int\text{cosh}\,\,^{n-2}xdx$
$\displaystyle \int \text{coth}\,\,^nx dx= -\cfrac{(\text{coth}\,\, x)^{n-1}}{{n-1}}+\displaystyle \int \text{coth}\,\,^{n-2}xdx$
$\displaystyle \int \text{csch}\,\,^nxdx=+\cfrac{(\text{csch}\,\, x)^{n-2}\text{coth}\,\, x}{n-1} -\cfrac{n-2}{n-1}\displaystyle \int \text{csch}\,\,^{n-2}xdx$

We should also mention how to actually use the reduction formula now that we have derived them all.

Example 8. Using the reduction formula find
$\displaystyle \int (\text{tan}\,\, x)^{7} dx$
Solution 8. Integrals of the nth powers of tangent and cotangent are easy to manipulate.

$\displaystyle \int (\text{tan}\,\, x)^{n} dx =+\frac{1}{n-1} (\text{tan}\,\, x)^{n-1} -\displaystyle \int (\text{tan}\,\, x)^{n-2} dx$

We see that each successive term in the series alternates sign.

$\displaystyle \int (\text{tan}\,\, x)^{n} dx=\frac{1}{n-1} (\text{tan}\,\, x)^{n-1} -\frac{1}{n-3} (\text{tan}\,\, x)^{n-3} +\frac{1}{n-5} (\text{tan}\,\, x)^{n-5}-...$

$\displaystyle \int \text{tan}\,\, ^{7} xdx =\frac{1}{6} \text{tan}\,\, ^{6} x-\frac{1}{4} \text{tan}\,\, ^{4} x+\frac{1}{2} \text{tan}\,\, ^{2} x-x+C$

Example 9. Use the reduction formula to calculate
$I_{7} =\displaystyle \int (\text{sin}\,\, x)^{7} dx$

Solution 9. We first write down the reduction formula identity.

$I_{n} =\displaystyle \int (\text{sin}\,\, x)^{n} dx =-\frac{\text{cos}\,\, x(\text{sin}\,\, x)^{n-1} }{n} +\frac{(n-1)}{n} I_{n-2}$
Now we figure out how many times we have to use the formula. We know that we start with a power of seven and reduce the power by two each time we apply the formula. When we get to seven minus six then we will have to evaluate a fundamental integral. The easiest way to then calculate the integral of the seventh power of $\text{sin}\,\, x$ is to work backwards from the formula for the first power.

$I_{1} =\displaystyle \int \text{sin}\,\, xdx=-\text{cos}\,\, x+C$
Now plug in the formulas recursively and we have.
${I_{3} =-\cfrac{\text{cos}\,\, x(\text{sin}\,\, x)^{2} }{3} -\cfrac{2}{3} \text{cos}\,\, x+C}$

${I_{5} =-\cfrac{\text{cos}\,\, x(\text{sin}\,\, x)^{4} }{5} +\cfrac{4}{5} \left(-\cfrac{\text{cos}\,\, x(\text{sin}\,\, x)^{2} }{3} -\cfrac{2}{3} \text{cos}\,\, x+C\right)}$

$\begin{array}{r}I_{7} =-\cfrac{1}{7} (\text{cos}\,\, x)(\text{sin}\,\, x)^{6} -\cfrac{6}{35} (\text{cos}\,\, x)(\text{sin}\,\, x)^{4} -\cfrac{8}{35} (\text{cos}\,\, x)(\text{sin}\,\, x)^{2}\\ -\cfrac{16}{35} (\text{cos}\,\, x)+C\end{array}$

As demonstrated, getting the numerical coefficients right can be tricky, but working backwards gives a good shot at getting it right.

Example 10. Compute the integral of

$\displaystyle \int \text{sec}\,\,^3xdx$

Solution 10. This is one of my favorite integrals. I remember that the
integral of secant cubed is the average of the derivative of decant and the integral of secant. Let us see how this comes about.

$\displaystyle \int \text{sec}\,\,^nxdx=\cfrac{(\text{sec}\,\, x)^{n-2}\text{tan}\,\, x}{n-1} +\cfrac{n-2}{n-1}\displaystyle \int \text{sec}\,\,^{n-2}xdx$
$\displaystyle \int \text{sec}\,\,^3xdx=\frac{1}{2}\text{sec}\,\, x\text{tan}\,\, x +\frac{1}{2}\displaystyle \int \text{sec}\,\, x dx$
$\displaystyle \int \text{sec}\,\,^3xdx=\frac{1}{2}\text{sec}\,\, x\text{tan}\,\, x +\frac{1}{2}\ln\left|\text{sec}\,\, x +\text{tan}\,\,x \right|+C$

Table of the Trigonometric and Hyperbolic Reduction Formulas

$\displaystyle \int (\text{sin}\,\, x)^{n} dx =-\cfrac{(\text{cos}\,\, x)(\text{sin}\,\, x)^{n-1} }{n} +\cfrac{n-1}{n} \displaystyle \int (\text{sin}\,\, x)^{n-2} dx$

$\displaystyle \int (\text{cos}\,\, x)^{n} dx =+\cfrac{(\text{sin}\,\, x)(\text{cos}\,\, x)^{n-1} }{n} +\cfrac{n-1}{n} \displaystyle \int (\text{cos}\,\, x)^{n-2} dx$

$\displaystyle \int (\text{tan}\,\, x)^{n} dx=+\cfrac{\displaystyle(\text{tan}\,\, x)^{n-1}}{n-1} -\displaystyle \int (\text{tan}\,\, x)^{n-2} dx$

$\displaystyle \int (\text{sec}\,\, x)^{n} dx =+\cfrac{(\text{tan}\,\, x)(\text{sec}\,\, x)^{n-2} }{n-1} +\cfrac{n-2}{n-1} \displaystyle \int (\text{sec}\,\, x)^{n-2} dx$

$\displaystyle \int (\text{cot}\,\, x)^{n} dx =-\cfrac{1}{n-1} (\text{cot}\,\, x)^{n-1} -\displaystyle \int (\text{cot}\,\, x)^{n-2} dx$

$\displaystyle \int (\text{csc}\,\, x)^{n} dx =-\cfrac{(\text{cot}\,\, x)(\text{csc}\,\, x)^{n-2} }{n-1} +\cfrac{n-2}{n-1} \displaystyle \int (\text{csc}\,\, x)^{n-2} dx$

$\displaystyle \int (\text{sinh}\,\, x)^{n} dx=+\cfrac{(\text{sinh}\,\, x)^{n-1} (\text{cosh}\,\, x)}{n} -\cfrac{n-1}{n} \displaystyle \int (\text{sinh}\,\, x)^{n-2} dx$

$\displaystyle \int (\text{cosh}\,\, x)^{n} dx=+\cfrac{(\text{cosh}\,\, x)^{n-1} (\text{sinh}\,\, x)}{n} +\cfrac{n-1}{n} \displaystyle \int (\text{cosh}\,\, x)^{n-2} dx$

$\displaystyle \int (\text{tanh}\,\, x)^{n} dx =-\cfrac{(\text{tanh}\,\, x)^{n-1} }{n-1} +\displaystyle \int (\text{tanh}\,\, x)^{n-2} dx$

$\displaystyle \int (\text{sech}\,\, x)^{n} dx =+\cfrac{(\text{sech}\,\, x)^{n-2} (\text{tanh}\,\, x)}{n-1} +\cfrac{n-2}{n-1} \displaystyle \int (\text{sech}\,\, x)^{n-2} dx$

$\displaystyle \int (\text{coth}\,\, x)^{n} dx =-\cfrac{(\text{coth}\,\, x)^{n-1} }{n-1} +\displaystyle \int (\text{coth}\,\, x)^{n-2} dx$

$\displaystyle \int (\text{csch}\,\, x)^{n} dx =+\cfrac{(\text{csch}\,\, x)^{n-2} (\text{coth}\,\, x)}{n-1} -\cfrac{n-2}{n-1} \displaystyle \int (\text{csch}\,\, x)^{n-2} dx$