 Hello, Myself, M.S. Basargaon, Assistant Professor, Department of Humanity and Sciences, Walsh Chandist of Technology, Sholapur. In this video, we are going to discuss beta function part 2. Now, learning outcome, at the end of this session, students will be able to apply trigonometric form of beta function to evaluate definite integrals. Now, we will see trigonometric form of beta function. Already, we have discussed about the beta function in part 1. That is by definition, beta 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. Now, to express this in trigonometric form of beta function, we will substitute here x is equal to sin square theta, that is dx is equal to 2 sin theta cos theta d theta. And accordingly, we will change the limit when x is equal to 0, sin square theta equal to 0 means theta equal to 0. And when x is equal to 1, sin square theta equal to 1 implies theta equal to pi by 2. Therefore, beta of Mn is equal to we can write integration from 0 to pi by 2 putting x equal to sin square theta, that is sin square theta raise to m minus 1 into bracket 1 minus sin square theta raise to n minus 1 and the value of dx is 2 sin theta cos theta d theta. Now, we are taking to outside the integral, which is equal to twice integration from 0 to pi by 2. Now, sin raise to 2 m minus 1 theta and sin theta, that we can write sin raise to 2 m minus 1 theta and 1 minus sin square theta is a cos square theta, that is cos raise to 2 n minus 2 theta into cos theta, that we get cos raise to 2 n minus 1 theta d theta. Now, this can be considered as second form of beta function or trigonometric form of beta function. Now, here we will simplify the formula putting 2 m minus 1 is equal to p and 2 n minus 1 is equal to q, solving for m we get m equal to p plus 1 by 2 and from second one n is equal to q plus 1 by 2. Therefore, beta of p plus 1 by 2 comma q plus 1 by 2 equal to twice 0 to pi by 2 sin raise to p theta cos raise to q theta d theta. Therefore, we can apply the standard formula when evaluating the definite integral of trigonometric function, that is integration from 0 to pi by 2 limit should be 0 to pi by 2 sin raise to p theta and cos raise to q theta d theta is equal to we can write 1 by 2 beta of p plus 1 by 2 comma q plus 1 by 2. Pause the video for a while and evaluate integration from 0 to pi by 2 sin square theta into cos raise to 4 theta d theta. I hope you have completed. Now, by definition of second form of beta function or trigonometric form of beta function, we can write integration from 0 to pi by 2 sin square theta cos raise to 4 theta d theta is equal to one half beta of here p is 2 and q is 4 that is 2 plus 1 by 2 that is p plus 1 by 2 4 plus 1 by 2 that is q plus 1 by 2 which is equal to 1 by 2 into beta of 2 plus 1 by 2 means 3 by 2 and 4 plus 1 by 2 means 5 by 2. Now, which is equal to you can write 1 by 2 gamma of 3 by 2 into gamma of 5 by 2 divided by gamma of 3 by 2 plus 5 by 2 by using the reduction formula of gamma function which is equal to 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 can be written as 3 by 2 into 1 by 2 into gamma of 1 by 2 and gamma of 3 by 2 plus 5 by 2 is gamma of 4. Now, by simplifying we get which is equal to 3 by 16 into gamma of 1 half whole square and gamma of 4 can be written as 3 factorial which is equal to 3 into root pi whole square because you know gamma of 1 half is root pi 16 into 3 factorial means 3 into 2 by canceling 3 we get pi by 32. Now, you will see the example evaluate the integral integration from 0 to pi by 4 cos cube 2 theta into sin dash to 4 4 theta d theta. Now, here we can write integration from 0 to pi by 4 cos cube 2 theta sin dash to 4 4 theta d theta which is equal to now here first of all we have to express this in term of standard form of trigonometric form of beta function that is keeping cos cube 2 theta as it is and sin dash to 4 4 theta can be written as 2 sin 2 theta cos 2 theta dash to 4 d theta by using the formula of sin 2 theta as 2 sin theta cos theta. Now, which is equal to 2 dash to 4 that is 16 we can take outside integration from 0 to pi by 4 that is sin dash to 4 2 theta cos cube 2 theta and cos dash to 4 2 theta means cos dash to 7 2 theta d theta. Now, again here to get in the form of second form of beta function limit should be 0 to pi by 2 for that we are here we can make use of those situation let 2 theta equal to t therefore, d theta equal to 1 by 2 d t. Now, accordingly we will change the limit when theta equal to 0 we get t equal to 0 and putting theta equal to pi by 4 you get t equal to pi by 2. Therefore, the given integral i we can write which is equal to 16 integration from 0 to pi by 2 sin dash to 4 t cos dash to 7 t and value of d theta is 1 by 2 d t. Now, which is second form of beta function with p as a 4 and q as a 7 which is equal to 16 by 2 means 8 that is 1 half of beta of 4 plus 1 by 2 comma 7 plus 1 by 2 which is equal to 4 times beta of 5 by 2 comma 4. Now, using the relation between beta and gamma function we can write which is equal to 4 into gamma 5 by 2 into gamma 4 divided by gamma 5 by 2 plus 4 that is gamma of 13 by 2 and now here we will apply the reduction formula of gamma function which is equal to we can write 4 and gamma of 4 can be written as 3 factorial and gamma keeping gamma of 5 by 2 as it is and gamma of 13 by 2 can be written as 11 by 2 into 9 by 2 into 7 by 2 into 5 by 2 into gamma of 5 by 2. Here gamma of 5 by 2 5 by 2 will get cancelled which is equal to 4 into 96 divided by the multiplication of 11 into 9 into 7 into 5 is 3465 which is equal to 384 divided by 3465 by simplifying we get which is equal to 128 divided by 115 which is the value of the given definite integral. Now, we will see one more example evaluate the integral integration from 0 to pi x into sin raise to 5 x into cos raise to 6x dx. Now, here we have to express now the given integral in terms of a trigonometric form of beta function that is integration from 0 to pi x into sin raise to 5 x into cos raise to 6x dx is equal to we can write integration from 0 to pi pi minus x into sin raise to 5 of pi minus x cos raise to 6 pi minus x dx here we are using the fundamental theorem that is integration from 0 to a f of x dx is equal to integration from 0 to a f of a minus x dx here a is pi which is equal to integration from 0 to pi pi minus x as it is and as we know sin of pi minus x is sin x that we can write sin raise to 5 x and cos of pi minus x is minus cos x and that can be written as cos raise to 6x dx. Therefore, integration from 0 to pi x into sin raise to 5 x into cos raise to 6x dx is equal to now here you see printing into two integrals pi is constant taking outside which is equal to pi into integration from 0 to pi sin raise to 5 x into cos raise to 6x dx minus integration from 0 to pi x into sin raise to 5 x into cos raise to 6x dx. Now, taking this second integral to the left hand side what you get twice integration from 0 to pi x into sin raise to 5 x into cos raise to 6x dx which is equal to pi into integration from 0 to pi sin raise to 5 x into cos raise to 6x dx. Now, which is equal to again here we are getting here the limit from integration from 0 to pi and to apply the trigonometry form of beta function the limit should be 0 to pi by 2 for that again here we will make use of fundamental theorem which we can write this as which is equal to pi into integration from 0 to pi by 2 sin raise to 5 x into cos raise to 6x dx plus pi into integration from 0 to pi by 2 sin raise to 5 pi minus x cos raise to 6 pi minus x dx. Now, here we are using the theorem integration from 0 to 2 a f of x dx is equal to integration from 0 to a f of x dx plus integration from 0 to a f of 2 a minus x dx here 2 a is pi therefore, a we get as a pi by 2. Now, again simplifying this we get therefore, twice integration from 0 to pi x into sin raise to 5 x into cos raise to 6x dx is equal to here again by making use of sin of pi minus x is a sin x and cos of pi minus x is a minus cos x and raise to the power here 6 we get which is equal to 2 pi integration from 0 to pi by 2 sin raise to 5 x into cos raise to 6x dx. Now, here 2 2 cancels from both side therefore, integration from 0 to pi x into sin raise to 5 x into cos raise to 6x dx is equal to pi into 1 by 2 and here p is 5 and q is 6 that is p plus 1 by 2 means 5 plus 1 by 2 and q plus 1 by 2 means 6 plus 1 by 2 which is equal to pi by 2 into beta of 5 plus 1 by 2 means we get 3 and 6 plus 1 by 2 means that is 7 by 2. Now, which is equal to you can write pi by 2 into gamma of 3 into gamma of 7 by 2 divided by gamma of 13 by 2 that is 3 plus 7 by 2 using the relation between beta and gamma function which is equal to you can write pi by 2 into gamma of 3 can be written as 2 factorial keeping gamma of 7 by 2 as it is and gamma of 13 by 2 can be written as 11 by 2 into 9 by 2 into 7 by 2 into gamma of 7 by 2. Now, canceling gamma of 7 by 2 from numerator and denominator we get which is equal to pi into 8 by 693 which is equal to 8 pi by 693 references higher engineering mathematics by Dr. B. S. Gravel.