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Integration by Reduction

Definition and Notation

Integration by reduction is an integration technique which involves expressing the integral in the form of a recurrence relation. It is useful for integration problems involving an integer parameter $n$, such as functions raised to a power, which cannot be integrated directly.

The $n$th integral is denoted $I_n$. Integration by parts is used to find a reduction formula for $I_n$, expressed in terms of $I_k$ where $k<n$.

Worked Example

Example 1

Find a reduction formula for $\displaystyle{I_n=\int\cos^n{x} \mathrm{d} x}$ $(n \geq 1)$ and hence find \[\int_0^{\frac{\pi}{2} }\cos^6{x}\mathrm{d} x.\]

Solution

Note that $\cos^n{x}=\cos^{n-1}{x}\cos{x}$. The integral can then be written in a form suitable for integration by parts:

\[I_n=\int\cos^{n-1}{x}\cos{x} \, \mathrm{d}x.\]

Recall the formula for integration by parts: $\displaystyle\int uv' \, \mathrm{d}x = uv - \int u'v\mathrm{d} x$.

Since the aim is to find a reduction formula, we require the integral on the right-hand side to involve $\cos{x}$ raised to a power less than $n$. We therefore set $u=\cos^{n-1}{x}$, since the derivative of $u$ is $u'=(n-1)\cos^{n-2}{x}(-\sin{x})$ (by the chain rule). Then set $v'=\cos{x}$ and integrate to obtain $v=\sin{x}$.

We have:

\begin{align} u &= \cos^{n-1} x, & v &= \sin x. \\ u' &= (n-1) \cos^{n-2}{x}(-\sin{x}) , & v' &= \cos x. \end{align}

The integral therefore becomes:

\begin{align} I_n &= \cos^{n-1}{x}\sin{x}-\int(n-1)\cos^{n-2}{x}(-\sin{x})\sin{x}\mathrm{d} x \\ &=\cos^{n-1}{x}\sin{x}+(n-1)\int\sin^2{x}\cos^{n-2}{x}\mathrm{d} x. \end{align}

Recall the trigonometric identity $\cos^2{x}+\sin^2{x}=1$. Rearranging this gives $\sin^2{x}=1-\cos^2{x}$. Substitute this into the integral:

\begin{align} I_n &= \cos^{n-1}{x}\sin{x}+(n-1)\int(1-\cos^2{x})\cos^{n-2}{x}\mathrm{d} x \\ &= \cos^{n-1}{x}\sin{x}+(n-1)\int\cos^{n-2}{x}-\cos^n{x}\mathrm{d} x \\ &= \cos^{n-1}{x}\sin{x}+(n-1)\int\cos^{n-2}{x}\mathrm{d} x -(n-1)\int\cos^n{x}\mathrm{d} x \end{align}

Note that both of the integrals on the right-hand side consist of the cosine function raised to some power. Since \[I_n=\int\cos^n{x} \mathrm{d} x,\] we have \[I_{n-2} = \int \cos^{n-2} \mathrm{d} x.\]

Substituting these expressions gives:

\[I_n=\cos^{n-1}{x}\sin{x}+(n-1)I_{n-2}-(n-1)I_n\]

This can be rearranged to give a recurrence relation for $I_n$:

\begin{align} I_n+(n-1)I_n &= \cos^{n-1}{x}\sin{x}+(n-1)I_{n-2} \\ nI_n &= \cos^{n-1}{x}\sin{x}+(n-1)I_{n-2} \\ I_n &= \dfrac{1}{n}\Bigl(\cos^{n-1}{x}\sin{x}+(n-1)I_{n-2}\Bigl). \end{align}

The recurrence relation for $\begin{align}I_6=\int\cos^6{x}\mathrm{d} x\end{align}$ is given by:

\begin{align} I_6 &= \dfrac{1}{6}\Bigl(\cos^{6-1}{x}\sin{x}+(6-1)I_{6-2}\Bigl) \\ &=\dfrac{1}{6}\Bigl(\cos^{5}{x}\sin{x}+5I_{4}\Bigl). \end{align}

Applying the limits gives:

\begin{align} I_6 &= \dfrac{1}{6}\Bigl[\cos^5{x}\sin{x}\Bigl]_0^{\frac{\pi}{2} } +\dfrac{5}{6}I_4 \\ &=\dfrac{1}{6}(0^5\cdot1-1^5\cdot0)+\dfrac{5}{6}I_4 \\ &=\dfrac{5}{6}I_4 \end{align}

Now apply the recurrence relation to $I_4$:

\begin{align} I_4 &= \dfrac{1}{4}\Bigl(\cos^{4-1}{x}\sin{x}+(4-1)I_{4-2}\Bigl) \\ &=\dfrac{1}{4}\Bigl(\cos^{3}{x}\sin{x}+3I_{2}\Bigl). \end{align}

This can be evaluated without further use of the recurrence relation by applying the limits and noting that $\displaystyle I_2=\int_0^{\frac{\pi}{2} }\cos^2{x}\mathrm{d} x.$ can be evaluated directly by using the identity $\cos^2{x}=\dfrac{1}{2}(1+\cos{2x})$:

\begin{align} I_4 &= \dfrac{1}{4}\Bigl[\cos^3{x}\sin{x}\Bigl]_0^{\frac{\pi}{2} }+\dfrac{3}{4}\int_0^{\frac{\pi}{2} }\cos^2{x}\mathrm{d} x \\ &= \dfrac{1}{4}(0^3\cdot1-1^3\cdot0)+\dfrac{3}{4}\int_0^{\frac{\pi}{2} }\dfrac{1}{2}(1+\cos{2x})\mathrm{d} x \\ &= \dfrac{3}{4}\cdot\dfrac{1}{2}\Bigl[x+\dfrac{\sin{2}x}{2}\Bigl]_0^{\frac{\pi}{2} } \\ &=\dfrac{3}{8}\Bigl(\dfrac{\pi}{2}+\dfrac{\sin{\pi} }{2}\Bigl)-\dfrac{3}{8}\Bigl(0+\dfrac{\sin{0} }{2}\Bigl) \\ &= \dfrac{3\pi}{16}. \end{align}

Substituting this into the expression for $I_6$ gives:

\begin{align} I_6 &= \dfrac{5}{6}I_4 \\ &= \dfrac{5}{6}\cdot\dfrac{3\pi}{16} \\ &= \dfrac{15\pi}{96} \\ &= \dfrac{5\pi}{32}. \end{align}

Hence,

\[\int_0^{\frac{\pi}{2} }\cos^6{x}\mathrm{d} x= \dfrac{5\pi}{32}.\]

Video Example

Example 1

Prof. Robin Johnson finds a reduction formula for $\begin{align}I_n=\int\tan^n{x}\;\mathrm{d}x,\,(n\geq2)\end{align}$ and hence computes $\begin{align}\int_0^{\large{\pi/4}}\tan^6{x}\;\mathrm{d}x\end{align}$.

Whiteboard maths

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