Introduction to Integration Reduction Formula:
There are many functions whose integrals cannot be reduced to one or more the other of the well known standard forms of integration . However , in some cases these integrals can be connected algebraically with integrals of other expressions which can either be directly integrable or which may be easier to integrate than the original functions . Such connecting algebraic relations are called 'reduction formulae' . These formu;ae connect an integral with another which is of the same type , but is of lower degree or order or at any rate easier to integrate than the original one .
Example of Integration Reduction Formula
Ex:1 Find the reduction formula for `int` xn eax dx , n being a positive integer and hence evaluate `int` x^3 eax dx .
Sol: Let In = `int` xneax dx .
On using the formulae for integration by parts , we get
In = `(x^n e^(ax))/(a)` - `int` n xn-1 `(e^(ax))/(a)` dx
= `(x^n e^(ax))/(a)` - `(n)/(a)` `int` xn-1 eax dx .
= `(x^n e^(ax))/(a)` - `(n)/(a)` In-1 .
This is called reduction formulae for `int` xn eax dx . Now In-1 in turn can be connected to In-2 . By successive reduction of n , the original integral In finally depends on I0 , where I0 = `int` eax dx = `(e^(ax))/(a)` .
To evaluate `int` x^3 eax dx , we take a = 5 and use the reduction formula for n = 3 , 2 , 1 in that order . Then we have I have recently faced lot of problem while learning how to solve linear equations word problems, But thank to online resources of math which helped me to learn myself easily on net.
I3 = `int` x^3 e5x dx = `(x^3 5^(ax))/(5)` - `(3)/(5)` I2 .
I2 = `(x^2 e^(5x))/(5)` - `(2)/(5)` I1
I1 = `(xe^(5x))/(5)` - `(1)/(5)` I0
I0 = `(e^(5x))/(5)` + c
Hence I3 = `(x^3 e^(5x))/(5)` - `(3)/(5^2)` x2e5x + `(6)/(5^3)` xe5x - `(6)/(5^4)` e5x + c
Integration Reduction Formula- Example
Q:1 Find the reduction formula for `int` tannx dx for an integer n`>=` 2 and hence find `int` tan6x dx .
Sol : Let In = `int` tannx dx
= `int` tann-2x tan2x dx
= `int` tann-2x sec2x dx - `int` tann-2x dx
= `(tan^(n-1)x)/(n-1)` - In-2 ,
which is the required reduction formula .
When n is even , In will finally depend on
I0 = `int` dx = x + c1
When n is odd , In will finally depend on
I1 = `int` tanx dx = log (secx) + c2
Now , I6 = `int` tan6x dx
= `(tan^5x)/(5)` - `int` tan4x dx
= `(tan^5x)/(5)` - `(tan^3x)/(3)` + `int` tan2x dx
= `(tan^5x)/(x)` - `(tan^3x)/(3)` + tanx - x + c .
There are many functions whose integrals cannot be reduced to one or more the other of the well known standard forms of integration . However , in some cases these integrals can be connected algebraically with integrals of other expressions which can either be directly integrable or which may be easier to integrate than the original functions . Such connecting algebraic relations are called 'reduction formulae' . These formu;ae connect an integral with another which is of the same type , but is of lower degree or order or at any rate easier to integrate than the original one .
Example of Integration Reduction Formula
Ex:1 Find the reduction formula for `int` xn eax dx , n being a positive integer and hence evaluate `int` x^3 eax dx .
Sol: Let In = `int` xneax dx .
On using the formulae for integration by parts , we get
In = `(x^n e^(ax))/(a)` - `int` n xn-1 `(e^(ax))/(a)` dx
= `(x^n e^(ax))/(a)` - `(n)/(a)` `int` xn-1 eax dx .
= `(x^n e^(ax))/(a)` - `(n)/(a)` In-1 .
This is called reduction formulae for `int` xn eax dx . Now In-1 in turn can be connected to In-2 . By successive reduction of n , the original integral In finally depends on I0 , where I0 = `int` eax dx = `(e^(ax))/(a)` .
To evaluate `int` x^3 eax dx , we take a = 5 and use the reduction formula for n = 3 , 2 , 1 in that order . Then we have I have recently faced lot of problem while learning how to solve linear equations word problems, But thank to online resources of math which helped me to learn myself easily on net.
I3 = `int` x^3 e5x dx = `(x^3 5^(ax))/(5)` - `(3)/(5)` I2 .
I2 = `(x^2 e^(5x))/(5)` - `(2)/(5)` I1
I1 = `(xe^(5x))/(5)` - `(1)/(5)` I0
I0 = `(e^(5x))/(5)` + c
Hence I3 = `(x^3 e^(5x))/(5)` - `(3)/(5^2)` x2e5x + `(6)/(5^3)` xe5x - `(6)/(5^4)` e5x + c
Integration Reduction Formula- Example
Q:1 Find the reduction formula for `int` tannx dx for an integer n`>=` 2 and hence find `int` tan6x dx .
Sol : Let In = `int` tannx dx
= `int` tann-2x tan2x dx
= `int` tann-2x sec2x dx - `int` tann-2x dx
= `(tan^(n-1)x)/(n-1)` - In-2 ,
which is the required reduction formula .
When n is even , In will finally depend on
I0 = `int` dx = x + c1
When n is odd , In will finally depend on
I1 = `int` tanx dx = log (secx) + c2
Now , I6 = `int` tan6x dx
= `(tan^5x)/(5)` - `int` tan4x dx
= `(tan^5x)/(5)` - `(tan^3x)/(3)` + `int` tan2x dx
= `(tan^5x)/(x)` - `(tan^3x)/(3)` + tanx - x + c .
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