Reduction formula is regarded as a method of integration. Integration by reduction formula helps to solve the powers of elementary functions, polynomials of arbitrary degree, products of transcendental functions and the functions that cannot be integrated easily, thus, easing the process of integration and its problems.
Formulas for Reduction in Integration
The reduction formula can be applied to different functions including trigonometric functions like sin, cos, tan, etc., exponential functions, logarithmic functions, etc. Here, the formula for reduction is divided into 4 types:
- For exponential functions
- For trigonometric functions
- For inverse trigonometric functions
- For hyperbolic trigonometric functions
- For algebraic functions
Reduction Formula for Exponential Functions
- ∫xn emx dx = [(1/m) xnemx ]− [(n/m) ∫xn−1 emx  ]dx
- ∫emx/xn dx = −[emx/(n−1)xn−1 ]+ [(m/n−1) ∫emx/xn−1]dx, n≠1
Reduction Formula for Hyperbolic Trigonometric Functions
- ∫sinhnx dx = −(1/n) sinhn−1 x cosh x − (n−1/n) ∫sinhn−2x dx
- ∫coshnx dx = (1/n) sinh x coshn−1 x − (n−1/n) ∫coshn−2x dx
- ∫tanhnx dx = −(1/n−1) tanhn−1x + ∫tanhn−2x dx, n≠1
- ∫cothnx dx = −(1/n−1)cothn−1x + ∫cothn−2x dx, n≠1
- ∫sechnx dx = \(\begin{array}{l}\frac{sech^{n-2}x\;tanh\,x}{n-1}\;+\;\frac{n-2}{n-1}\,\int sech^{n-2}x\,dx, n\neq1\end{array} \)
- \(\begin{array}{l}\int\frac{dx}{sinh^n\, x}\,=\, -\frac{cosh\,x}{\left ( n-1 \right )sinh^{n-1}\,x}\,+\,\frac{\left ( n-2 \right )}{\left ( n-1 \right )}\int \frac{dx}{sinh^{n-2}\,x},\; n\neq 1\end{array} \)
- \(\begin{array}{l}\int \frac{dx}{cosh^n\, x}\,=\, -\frac{sinh\,x}{\left ( n-1 \right )cosh^{n-1}\,x}\,+\,\frac{\left ( n-2 \right )}{\left ( n-1 \right )}\int \frac{dx}{cosh^{n-2}\,x},\,n\neq1\end{array} \)
- ∫sinhnx coshmx dx = \(\begin{array}{l}\frac{sinh^{n+1}x\, cosh^{m-1}x}{n+m}\,+\, \frac{\left ( m-1 \right )}{\left ( n+m \right )}\int sinh^n x cosh^{m-2}x\, dx\end{array} \)
Reduction Formula for Trigonometric Functions
- \(\begin{array}{l} \int sin^{n}(x)dx=\frac{-Sin^{n-1}(x)Cos(x)}{n}+\frac{n-1}{n} \int Sin^{n-2}(x)dx\end{array} \)
- \(\begin{array}{l} \int tan^{n}(x)dx=\frac{tan^{n-1}(x)}{n-1}-\int tan^{n-2}(x)dx\end{array} \)
- \(\begin{array}{l} \int sin^{n}(x)\: cos^{m}(x)dx=\frac{sin^{n+1}(x)cos^{m-1}(x)}{n+m}+\frac{m-1}{n+m}\: \int sin^{n}(x)\: cos^{m-2}(x)dx\end{array} \)
- \(\begin{array}{l} \int x^{n}cos(x)dx=x^{n}sin(x)-n\int x^{n-1}sin(x)dx\end{array} \)
- \(\begin{array}{l} \int x^{n}sin(x)dx=-x^{n}cos(x)+n\int x^{n-1}cos(x)dx\end{array} \)
- \(\begin{array}{l}\int \frac{dx}{sin^n x}\,=\,- \frac{cos\,x}{\left ( n-1 \right )sin^{n-1}x}\,+\,\frac{\left ( n-2 \right )}{\left ( n-1 \right )}\int \frac{dx}{sin^{n-2}x},\,n\neq1\end{array} \)
- \(\begin{array}{l}\int \frac{dx}{cos^n x}\,=\, \frac{sin\,x}{\left ( n-1 \right )cos^{n-1}x}\,+\,\frac{\left ( n-2 \right )}{\left ( n-1 \right )}\int \frac{dx}{cos^{n-2}x},\,n\neq1\end{array} \)
Reduction Formula for Logarithmic Functions
- ∫xn lnmx dx = \(\begin{array}{l}\frac{x^{n+1}\,ln^m\,x}{n+1}\,-\,\frac{m}{n+1}\int x^n\, ln^{m-1}x\,dx\end{array} \)
- \(\begin{array}{l}\int \frac{ln^m x}{x^n}\,=\,-\frac{ln^m x}{\left ( n-1 \right )x^{n+1}}\,+\, \frac{m}{n-1}\int \frac{ln^{m-1}x}{x^n}\,dx,\;n\neq1\end{array} \)
Reduction Formula for Inverse Trigonometric Functions
- ∫xn arcsin x dx = \(\begin{array}{l}\frac{x^{n+1}}{n+1}\, arcsin \,x\, -\, \frac{1}{n+1}\int \frac{x^{n+1}}{\sqrt{1-x^2}}\,dx\end{array} \)
- ∫xn arccos x dx = \(\begin{array}{l}\frac{x^{n+1}}{n+1}\, arccos \,x\, +\, \frac{1}{n+1}\int \frac{x^{n+1}}{\sqrt{1-x^2}}\,dx\end{array} \)
- ∫xn arctan x dx = \(\begin{array}{l}\frac{x^{n+1}}{n+1}\, arctan \,x\, -\, \frac{1}{n+1}\int \frac{x^{n+1}}{\sqrt{1+x^2}}\,dx\end{array} \)
Reduction Formula for Algebraic Functions
- \(\begin{array}{l}\int \frac{dx}{\left ( ax^2 +bx+c \right )^n}\,=\, \frac{-2ax-b}{\left ( n-1 \right )\left ( b^2 -4ac \right )\left ( ax^2 +bx+c \right )^{n-1}}\,-\, \frac{2\left ( 2n-3 \right )a}{\left ( n-1 \right )\left ( b^2 – 4ac \right )}\int \frac{dx}{\left ( ax^2 +bx + c \right )^{n-1}},\,n\neq1\end{array} \)
- \(\begin{array}{l}\int \frac{dx}{\left ( x^2+a^2 \right )^n}\,=\, \frac{x}{2\left ( n-1 \right )a^2 \left ( x^2 + a^2 \right )^{n-1}}\,+\, \frac{2n-3}{2\left ( n-1 \right )a^2}\int \frac{dx}{\left ( x^2 + a^2 \right )^{n-1}},\,n\neq1\end{array} \)
- \(\begin{array}{l}\int \frac{dx}{\left ( x^2-a^2 \right )^n}\,=\, \frac{x}{2\left ( n-1 \right )a^2 \left ( x^2 – a^2 \right )^{n-1}}\,-\, \frac{2n-3}{2\left ( n-1 \right )a^2}\int \frac{dx}{\left ( x^2 – a^2 \right )^{n-1}},\,n\neq1\end{array} \)
Learn More:
| Exponential Functions | Trigonometric Functions |
| Logarithmic Functions | Inverse Trigonometric Functions |
| Algebra | Hyperbolic Function Formula |
Solved Example Questions
Question: Evaluate the integral: ∫tan5(2x) dx
Solution:
Use: ∫tann(u) du = (1/ n−1) tann−1(u) − ∫tann−2(u) du
Substitution:
a = 2x
∴ ½ da = dx
Hence,
∫tan5 (2x) dx = ½ [∫tan5(a) da]
= ½ [¼ tan4(a) − ∫tan3(a) da]
= ½ [¼ tan4(a) – [½ tan2(a) – ∫tan(a) da]]
= 1/8 tan4(a) − ¼ tan2(a) + ½ ln sec(a) + C
= 1/8 tan4(2x) − ¼ tan2(2x) + ½ ln sec(2x) + C
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