 Hello, Myself, MS Bhasaragam, Assistant Professor, Department of Humanity Design Sciences, Valterist of Technology, Solapur. In this video, we are going to discuss about beta function part 1. Now, learning outcome. At the end of this session, students will be able to evaluate definite integral by beta function. Now, we will see what is beta function. Beta function beta of mn is defined by it is a function of m and n and it is denoted by b of mn. Therefore, beta of mn that is b of mn is equal to integration from 0 to 1 x raise to m minus 1 into bracket 1 minus x raise to n minus 1 dx, where m is greater than 0 and n is greater than 0. The beta function is also known as Euler's integral of first kind, it is a special kind of definite integral. For example, integration from 0 to 1 x square into bracket 1 minus x raise to 1 by 2 dx. Now, here we can compare this integral with the definition of beta function. Here, m minus 1 is 2 and n minus 1 is 1 by 2. Therefore, without integrating we can write the value of this integral as a function of that is beta of 3 and 3 by 2 that is here m is 3 and n is 3 by 2. Now, we will see important properties of beta function, first property is symmetric property beta of mn is equal to beta of nm that is if you interchange m and n the value of the given integral remains same. Now, next very important property is relation between beta and gamma function that is beta of mn is equal to gamma of m into gamma of n divided by gamma of m plus n that is here first of all we have to write the value of the integral in terms of beta function and we have to use the relation between beta and gamma function and we have to express beta in terms of gamma function and making use of the property of gamma function we can write the value of the given integral which is function of beta of mn. Now, we will see the example evaluate the integral integration from 0 to 2 x square into bracket 2 minus x raise to minus 1 by 2 dx. Now, to express this in terms of beta function here we have to bring the term 1 minus x for that the proper substitution is x is equal to 2 t that is dx is equal to 2 d t. Now, here accordingly we have to change the limits when x is equal to 0 we put x equal to 0 we get the value of t as a 0 and when x is equal to 2 we get the value of t is equal to 1. Therefore, the given integral integration from 0 to 2 x square into bracket 2 minus x raise to minus 1 by 2 dx is equal to integration from 0 to 1. Now, here putting x equal to 2 t that is you get 2 t whole square and putting x equal to 2 t then you get 2 minus 2 t raise to minus 1 by 2 and the value of dx as a 2 d t which is equal to here 2 raise to 2 and 2 that is 2 raise to 3 and taking 2 comma from this bracket that is 2 raise to minus of which simplifies which is equal to 2 raise to 5 by 2 integration from 0 to 1 we get t square into bracket 1 minus t raise to minus 1 by 2 d t. Now, which is in terms of beta function if you compare here m minus 1 is 2 and n minus 1 is minus 1 by 2 that we can write in terms of beta function which is equal to 2 raise to 5 by 2 beta of 3 and 1 by 2. Now, here using the relation between beta and gamma function we can write 2 raise to 5 by 2 as it is gamma of 3 into gamma of 1 by 2 divided by gamma of 3 plus 1 by 2 since we know just now we have seen beta of m n equal to gamma of m into gamma of n divided by gamma of m plus n which is equal to 2 raise to 5 by 2 into gamma of 3 into gamma of 1 by 2 divided by sum of this is what 7 by 2 that is what gamma of 7 by 2 which is equal to 2 raise to 5 by 2 and gamma of 3 can be written as 2 factorial that we know gamma of n plus 1 is n factorial keeping gamma of 1 by 2 as it is and here by using the reduction formula of gamma function gamma of 7 by 2 can be written as 5 by 2 into 3 by 2 into 1 by 2 into gamma of 1 by 2. Now, here gamma of 1 by 2 gamma of 1 by 2 will get cancel which is equal to 2 raise to 5 by 2 and this 2 into 2 into 2 that is we get 2 raise to 4 divided by 15 which is equal to we can write simplifying 2 raise to 6 root 2 divided by 15 that which is equal to we can write 64 root 2 by 15 this is what the value of the given integral pause the video for a while and find the value of beta of 5 by 2 3 by 2 I hope you have completed. Now, here we can write beta of 5 by 2 3 by 2 is equal to gamma of 5 by 2 gamma of 3 by 2 divided by gamma of 5 by 2 plus 3 by 2 since we know beta of m n is equal to gamma of m into gamma of n divided by gamma of m plus n. Now, simplifying this which is equal to we can write gamma of 5 by 2 can be written as 3 by 2 into 1 by 2 into gamma of 1 by 2 and gamma of 3 by 2 can be written as 1 by 2 into gamma of 1 by 2 and gamma of 5 by 2 plus 3 by 2 is gamma of 4 by simplifying we get which is equal to 3 by 8 gamma of 1 by 2 whole square and gamma of 4 can be written as 3 factorial which is equal to 3 by 8 and gamma of 1 half is root pi that is root pi whole square divided by 3 factorial means 3 into 2 into 1 that is 3 3 we get cancel which is equal to get pi by 16 which is what the value of beta of 5 by 2 and 3 by 2. Now, we will see the next example evaluate the integral integration from 0 to 2 x raise to 7 into bracket 16 minus x raise to 4 bracket raise to 10 dx. Now, here we have to express this integral in the form of beta function that is integration from 0 to 1 x raise to m minus 1 into 1 minus x raise to n minus 1 dx form for that here the suitable substitution is we have to make use here is x raise to 4 we have to substitute here x raise to 4 equal to 16 t that is solving for x we get x equal to 2 into t raise to 1 by 4. Therefore, dx is equal to 2 into 1 by 4 t raise to minus 3 by 4 dt that is dx is equal to 1 by 2 t raise to minus 3 by 4 dt. Now, according to the substitution you have to change the limits when x is equal to 0 what you get here t equal to 0 and when if you substitute x equal to 2 2 2 will get cancel you get the value of t will be 1. Therefore, the given integral integration from 0 to 2 x raise to 7 16 minus x raise to 4 raise to 10 dx is equal to now here limit will become integration from 0 to 1. Now, here we are substituting x equal to 2 t raise to 1 by 4 that is 2 raise to 7 t raise to 1 by 4 raise to 7 into bracket 16 minus 16 t raise to the bracket 10 and the value of dx as 1 by 2 t raise to minus 3 by 4 dt. Now, which is equal to here 2 raise to 7 divided by 2 means what we get 2 raise to 6 and from this bracket taking 16 common we get 16 raise to 10 integration from 0 to 1 and t raise to 7 by 4 and t raise to minus 3 by 4 means t into bracket 1 minus t bracket complete raise to the power 10 dt which is equal to 2 raise to 6 into 16 raise to 10 can be written as 4 into 16 raise to 11 and now comparing with definition of beta function here m minus 1 is 1 and n minus 1 is 10 therefore, you get m equal to 2 and n equal to 11 which is equal to 4 into 16 raise to 11 into beta of 2 11. Now, again here using the relation between beta and gamma function you get 4 into 16 raise to 11 into gamma of 2 into gamma of 11 divided by gamma of 13. Now, further by using the property of gamma function you can write which is equal to 4 into 16 raise to 11 into gamma of 2 can be written as 1 factorial, gamma of 11 can be written as 10 factorial, gamma of 13 can be written as 12 factorial which is equal to 4 into 16 raise to 11 1 factorial is 1 keeping 10 factorial as it is and writing 12 factorial as 12 into 11 into 10 factorial. Now, here 10 factorial 10 factorial will get cancel which is equal to 16 raise to 11 by 33 which is the value of the given definite integral. Now, you will see one more simple example express in terms of beta function integration from 0 to 1 square root of 1 minus x raise to 4 dx. Now, to get this integral in the form of beta function here power power substitution is you have to make use of x raise to 4 equal to t. Now, solving for x you get x equal to t raise to 1 by 4 that is dx is equal to 1 by 4 t raise to minus 3 by 4 dt. Now, according to the substitution you have to change the limits. Now, when x is equal to 0 you get t equal to 0 and when x is equal to 1 you get t equal to 1. Therefore, the given integral integration from 0 to 1 square root of 1 minus x raise to 4 dx is equal to integration from 0 to 1 1 minus t raise to 1 by 2 and the value of dx as 1 by 4 t raise to minus 3 by 4 dt taking 1 by 4 outside which is equal to 1 by 4 integration from 0 to 1 t raise to minus 3 by 4 into bracket 1 minus t raise to 1 by 2 dt. Now, we express this in terms of beta function and we can write the value of this integral in terms of beta function as which is equal to 1 by 4 as it is beta of 1 by 4 comma 3 by 2. This is what the value of the given integral in terms of beta function.