Reduction Formula

A reduction formula is regarded as an important method of integration. Integration by reduction formula always helps to solve complex integration problems. It can be used for powers of elementary functions, trigonometric functions, products of two are more complex functions, etc. These are the functions that cannot be integrated easily. Therefore for easing the process of integration, we will discuss here Reduction Formula for integration with examples. Let us learn the important concept!

Reduction Formula

Concept of Reduction Formula in Integration:

The reduction formula can be applied to different functions with combinations of different types of functions in a single problem. The formula for the reduction can be divided into various categories as given below:

• Exponential functions
• Trigonometric functions
• Inverse trigonometric functions
• Hyperbolic trigonometric functions
• Algebraic functions

Therefore to get the solution of integrals we can use the reduction formulas. These formulas will enable us to reduce the degree of the integrand and calculate the integrals in a finite number of steps. Below are the important reduction formulas for integrals involving the most common functions.

Some Important Formulas

Reduction Formula for Trigonometric Functions

• \(\int sin^{n}(x)dx=\frac{-Sin^{n-1}(x)Cos(x)}{n}+\frac{n-1}{n}Sin^{n-2}(x)dx\)
• \(\int tan^{n}(x)dx=\frac{-tan^{n-1}(x)}{n-1}-\int tan^{n-2}(x)dx\)
• \(\int sin^{n}(x)\: cos^{m}(x)dx=\frac{sin^{n+1}(x)cos^{m-1}(x)}{n+m}+\frac{m-1}{n+m}\: \int sin^{n}(x)\: cos^{m-2}(x)dx\)
• \(\int x^{n}cos(x)dx=x^{n}sin(x)-n\int x^{n-1}sin(x)dx\)
• \(\int x^{n}sin(x)dx=-x^{n}cos(x)+n\int x^{n-1}cos(x)dx\)

Reduction Formula for Logarithmic Functions

• \({\large\int\normalsize} {{\ln^n}x\,dx} ={\large\frac{{{x^{n + 1}}{\ln^m}x}}{{n + 1}}\normalsize}-\;{\large\frac{m}{{n + 1}}\int\normalsize} {{x^n}{\ln^{m – 1}}x\,dx}\)
• \(\int \frac{ln^m x}{x^n}\,=\,\frac{ln^m x}{\left ( n-1 \right )x^{n+1}}\,+\, \frac{m}{n-1}\int \frac{ln^{m-1}x}{x^n}\,dx,\;n\neq1\)

Reduction Formula for Algebraic Functions

• \({\large\int\normalsize} {\large{\frac{{{x^n}}}{{a{x^n} + b}}\normalsize}} \,dx = {\large\frac{x}{a}\normalsize} – {\large\frac{b}{a}\int {\frac{{dx}}{{a{x^n} + b}}}\normalsize}\)
• \(\int \frac{dx}{\left ( ax^2 +bx+c \right )^n}\,=\, \frac{-2ax-b}{\left ( n-1 \right )\left ( b^2 -4ac \right )\left ( ax^2 +bx+c \right )^{n-1}}\,-\, \frac{2\left ( 2n-3 \right )a}{\left ( n-1 \right )\left ( b^2 – 4ac \right )}\int \frac{dx}{\left ( ax^2 +bx + c \right )^{n-1}},\,n\neq1\)
• \(\int \frac{dx}{\left ( x^2+a^2 \right )^n}\,=\, \frac{x}{2\left ( n-1 \right )a^2 \left ( x^2 + a^2 \right )^{n-1}}\,+\, \frac{2n-3}{2\left ( n-1 \right )a^2}\int \frac{dx}{\left ( x^2 + a^2 \right )^{n-1}},\,n\neq1\)
• \(\int \frac{dx}{\left ( x^2-a^2 \right )^n}\,=\, \frac{x}{2\left ( n-1 \right )a^2 \left ( x^2 – a^2 \right )^{n-1}}\,-\, \frac{2n-3}{2\left ( n-1 \right )a^2}\int \frac{dx}{\left ( x^2 – a^2 \right )^{n-1}},\,n\neq1\)

Reduction Formula for Exponential Functions

• \({\large\int\normalsize} {{x^n}{e^{mx}}dx} ={\large\frac{1}{m}\normalsize}{x^n}{e^{mx}}-\; {\large\frac{n}{m}\normalsize} {\large\int\normalsize} {{x^{n – 1}}{e^{mx}}dx}\)
• \({\large\int\normalsize} {{\large\frac{{{e^{mx}}}}{{{x^n}}}\normalsize} dx} =– {\large\frac{{{e^{mx}}}}{{\left( {n – 1} \right){x^{n – 1}}}}\normalsize}+\; {\large\frac{m}{{n – 1}}\normalsize} {\large\int\normalsize} {{\large\frac{{{e^{mx}}}}{{{x^{n – 1}}}}\normalsize} dx}\)
• \({\large\int\normalsize} {{{\sinh }^n}x\,dx} =– {\large\frac{1}{n}\normalsize}{\sinh ^{n – 1}}x\cosh x-\; {\large\frac{{n – 1}}{n}\normalsize} {\large\int\normalsize} {{{\sinh }^{n – 2}}x\,dx}\)
• \({\large\int\normalsize} {\large\frac{{dx}}{{{{\sinh }^n}x}}\normalsize} = – {\large\frac{1}{n}\normalsize}{\sinh ^{n – 1}}x\cosh x +\;{\large\frac{{n – 2}}{{n – 1}}\int\normalsize} {\large\frac{{dx}}{{{{\sinh }^{n – 2}}x}}\normalsize}\)

Solved Examples

Q.1:  Evaluate the integral:

\(\displaystyle \int{{{z^7}{{\left( {8 + 3{z^4}} \right)}^8}\,dz}}\)

Solution: First,

\(u = 8 + 3{z^4}\hspace{0.25in} \to \hspace{0.25in}du = 12{z^3}dz\hspace{0.25in}\,\,\,\,\,\, \to \hspace{0.25in}\,\,\,\,\,{z^3}dz = \frac{1}{{12}}du\)

Let’s do a quick rewrite of the integrand,

\(\int{{{z^7}{{\left( {8 + 3{z^4}} \right)}^8}\,dz}} = \int{{{z^4}{z^3}{{\left( {8 + 3{z^4}} \right)}^8}\,dz}} = \int{{{z^4}{{\left( {8 + 3{z^4}} \right)}^8}\,{z^3}dz}}\)

Now, notice that we can convert all of the z’s in the integrand except apparently for the z^4 that is in the front. We can notice from the substitution that we can solve it for z^4 to get,

\({z^4} = \frac{1}{3}\left( {u – 8} \right)\)

Now, with this we can do the substitution and evaluate the integral.

\(\begin{align*}\int{{{z^7}{{\left( {8 + 3{z^4}} \right)}^8}\,dz}} & = \frac{1}{{12}}\int{{\frac{1}{3}\left( {u – 8} \right){u^8}du}} = \frac{1}{{36}}\int{{{u^9} – 8{u^8}du}} = \frac{1}{{36}}\left( {\frac{1}{{10}}{u^{10}} – \frac{8}{9}{u^9}} \right) + c\\ &  = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{{36}}\left( {\frac{1}{{10}}{{\left( {8 + 3{z^4}} \right)}^{10}} – \frac{8}{9}{{\left( {8 + 3{z^4}} \right)}^9}} \right) + c}}\end{align*}\)