 Namaste. Myself, Mr. Viradhar Bala Sahib, Assistant Professor, Department of Humanities and Sciences, Walchand Institute of Technology, Solapu. In this video lecture, we will discuss Laplace transform using the property effect of division by T. Learning outcome, at the end of this session students will be able to apply the property effect of division by T to find Laplace transform of given function. Let us start with the property effect of division by T. It states that if Laplace transform of f of t equal to f of s, then Laplace transform of f of t upon t is equal to integration with limit s to infinity f of s ds. Now, working rule to solve the problems, when given function is of the type f of t upon t, we have to follow the following steps. Step one, first find Laplace transform of numerator function f of t and denote it by f of s. Step two, using effect of division by T property, we write Laplace transform of f of t upon t equal to integration with limit s to infinity of f of s into ds. In this result, substitute the value of f of t in the left side and its Laplace transform f of s in the right hand side. Step three, integrate right hand side integration with respect to ds between the limit s to infinity. We get required answer. Now, consider examples. Example one, find the Laplace transform of 1 minus e raise to minus t upon t. Here, in the numerator function 1 minus e raise to minus t is denoted by f of t and now we have to find its Laplace transform. Therefore, Laplace transform of f of t which is equal to Laplace transform of 1 minus e raise to minus t which is equal to Laplace transform of 1 minus Laplace transform of e raise to minus t. This is using linear property. Therefore, Laplace transform of f of t equal to Laplace transform of 1 minus e raise to minus t which is equal to. Now, Laplace transform of 1 is 1 by s minus Laplace transform of e raise to minus t is 1 upon s plus 1. This is we can write using the formula Laplace transform of e raise to minus a t equal to 1 upon s plus a. So, this Laplace transform of f of t answer is denoted by f of s. Now, by property of effect of division by t that is Laplace of f of t upon t equal to integration with the limit s to infinity f of s d s. Now, in this result in the left hand side replace f of t by 1 minus e raise to minus t and in the right hand side f of s by 1 upon s minus 1 upon s plus 1. Therefore, Laplace transform of 1 minus e raise to minus t upon t equal to integration with limit s to infinity of 1 upon s minus 1 upon s plus 1 into d s which is equal to in bracket. Now, separate the integration for each term we get integration of 1 upon s into d s minus integration of 1 upon s plus 1 into d s bracket close limit s to infinity which is equal to in bracket integration of 1 by s with respect to t s is log s minus integration of 1 upon s plus 1 with respect to t s is log of s plus 1 bracket close with limit s to infinity. Now, using logarithm property log a minus log b equal to log of a by b we can write this right hand side as log of s upon s plus 1 bracket close with limit s to infinity which is equal to in bracket log of s upon from denominator we can take s common in bracket 1 plus 1 by s and bracket close limit s to infinity which is equal to. Now, s s gets cancelled in bracket log of 1 upon 1 plus 1 by s with limit s to infinity. Now, we have to substitute the limits of s when we put s equal to upper limit infinity and using the result 1 by infinity equal to 0 we get log of 1 minus when s equal to lower limit s we get log of 1 upon 1 plus 1 by s which is equal to log 1 is 0 minus. Now, log of 1 upon 1 plus 1 by s can be simplified as log of s upon s plus 1. Therefore, Laplace transform of 1 minus e raise to minus t upon t is equal to we can write log of s plus 1 upon s this is the required answer to the question. Now, pause the video for a while and write the answer to the question question is find Laplace transform of sin t upon t come back I hope you have written answer to this question. Now, here I will going to give the solution question is find the Laplace transform of sin t upon t here numerator function sin t is denoted by f of t and it is Laplace transform is equal to 1 upon s square plus 1 and denoted by f of s again by property that is effect of division by t stated as Laplace of f of t upon t equal to integration with limit s to infinity f of s d s by this property we can write Laplace of sin t upon t equal to integration with limit s to infinity of 1 upon s square plus 1 into d s. Now, integration of 1 upon s square plus 1 with respect to 2 s is tan inverse of s with limit s to infinity now substitute the limits we get when s is upper limit infinity tan inverse of infinity minus when s equal to lower limit s we get tan inverse of s. Therefore, Laplace transform of sin t upon t is equal to now tan inverse of infinity is pi by 2 minus tan inverse of s which is equal to pi by 2 minus tan inverse of s is equal to cot inverse of s. So, this is the identity from inverse trigonometric function this is the required answer to the given question now let us consider another example find the Laplace transform of sin square t upon t here numerator function f of t equal to sin square t now we have to find Laplace transform of sin square t, but it is not in standard form first we have to simplify it as 1 minus cos 2 t upon t from trigonometric function formula. Now, by taking Laplace transform on both this side we get Laplace transform of f of t which is equal to Laplace transform of sin square t which is equal to now in the right hand side 1 by 2 is a constant coefficient right first outside the bracket and in bracket Laplace transform of 1 minus cos 2 t we can separate into 2 term that is Laplace transform of 1 minus Laplace transform of cos of 2 t using linear property. Therefore, Laplace transform of sin square t equal to 1 by 2 into bracket now Laplace transform of 1 is 1 upon s minus Laplace transform of cos of 2 t is s upon s square plus 2 square, but 2 square we can write as a 4 and bracket close. So, this result is denoted by f of s. Now, by effect of division by t Laplace of f of t upon t is equal to integration with limit s to infinity f of s t s. Therefore, by this property we can write Laplace transform of sin square t upon t is equal to now in the right hand side f of s we have to substitute and coefficient 1 by 2 I can write outside the integration and into integration with limit s to infinity in bracket 1 by s minus s upon s square plus 4 bracket close into d s which is equal to 1 by 2 as it is in bracket separate the integration for each term first term integration of 1 by s into d s minus for second term we can adjust 2 numerator and denominator to get in standard for. Therefore, 1 by 2 into integration of 2 s upon s square plus 4 into d s bracket close with limit s to infinity. Which is equal to 1 by 2 as it is in bracket integration of 1 by s is log s minus 1 by 2 as it is into integration of 2 s upon s square plus 4 with respect to t s is log of s square plus 4 bracket close with limit s to infinity which is equal to now from the bracket we can take 1 by 2 is outside. So, that 1 by 2 into 1 by 2 is 1 by 4 into bracket 2 into log s minus log of s square plus 4 bracket close with limit s to infinity which is equal to 1 by 4 as it is and by property of logarithm 2 log s we can write log s square and minus log of s square plus 4 we can write log of s square upon s square plus 4 and bracket close with limit s to infinity which is equal to 1 by 4 as it is in bracket log s square upon from denominator s square is taken common and in bracket 1 plus 4 upon s square bracket close and curly bracket close with limit s to infinity. Now, s square s square cancel therefore, log plus transform of sine square t upon t equal to 1 by 4 into bracket log of 1 upon 1 plus 4 upon s square with limit s to infinity. Now, we have to substitute the limits of s which is equal to 1 by 4 in bracket when we put s equal to upper limit infinity and using the result 4 upon infinity tends to 0 we get log 1 minus when s equal to lower limit s we get log of 1 upon 1 plus 4 upon s square which is equal to 1 by 4 in bracket log 1 0 minus log of 1 upon 1 plus 4 upon s square can be written as log of s square upon s square plus 4 which is equal to 1 by 4 in bracket minus log of s square upon s square plus 4 bracket close which is equal to 1 by 4 log of s square plus 4 upon s square. So, this is the required answer to the given example. Now, to prepare this video lecture I refer these two books as references. Thank you.