 Hello everybody, myself, Mr. Birajdar Bada Saheb, Assistant Professor, Department of Humanities and Sciences, Walshchand Institute of Technology, Sholapur. In this video lecture, we will discuss Laplace Transforms of Integrals. Learning outcome. At the end of this session, students will be able to find Laplace Transform of Integrals of Function F of t. Let us start with statement of Laplace Transform of Integrals. It states that if Laplace Transform of F of t equal to F of s, then Laplace Transform of integration with limit 0 to t F of t dt is equal to 1 upon s into F of s. Now, proof of this statement, let us suppose phi of t equal to u 1 integration with limit 0 to t F of t dt, when we put t equal to 0 in this result, we get phi of 0 equal to 0. And when we differentiate this result with respect to t on both the side, we get phi dash of t equal to now derivative of right hand side integration with respect to t gets cancelled, we get F of t, so that phi dash of t equal to F of t. Now, by property Laplace Transform of derivatives, we can write Laplace of phi dash of t equal to s into Laplace of phi of t minus phi of 0, which is equal to now phi of 0 is equal to 0, we know that so that s into Laplace of phi of t. From this result, we can write Laplace of phi of t equal to 1 upon s into Laplace of phi dash of t, which is equal to 1 by s as it is Laplace of phi dash of t, we can write Laplace of F of t because phi dash of t equal to F of t. Therefore, when we replace phi of t equal to integration with limit 0 to t F of t dt, which is assumed in the left hand side and Laplace of F of t equal to F of s in the right hand side. So, we get Laplace Transform of integration with limit 0 to t F of t dt equal to 1 upon s into F of s, this is the required property, hence the problem. Now, working rule to solve the problems to find Laplace Transform of integration with limit 0 to t F of t dt, we have to follow these steps, step one. First of all note down the value of integrand function F of t, step two. Find the Laplace Transform of this function F of t and denote it by F of s that is Laplace of F of t equal to F of s. Now, step three, write Laplace Transform of integration with limit 0 to t F of t dt equal to 1 upon s into F of s means replace F of t by given function in the left hand side and its Laplace Transform F of s in the right hand side, which gives required answer. Now, let us consider the example, example one. Find the Laplace Transform of integration with limit 0 to t of e raise to minus 2 t into t cube into dt. Here, integrand function e raise to minus 2 t into t cube is denoted by F of t and now, we have to find its Laplace Transform. First of all we find Laplace Transform of t cube, which is equal to gamma of 3 plus 1 upon s raise to 3 plus 1 due to the formula Laplace of t raise to n equal to gamma of n plus 1 upon s raise to n plus 1. Now, we have to simplify right hand side, which is equal to gamma 4 upon s raise to 4, which is equal to gamma 4 is 3 factorial using gamma property upon s raise to 4, which is equal to 3 factorial means 6 upon s raise to 4. Now, by force shifting theorem, we discussed earlier that is Laplace Transform of e raise to minus a t into F of t equal to F of s plus a when Laplace of F of t equal to F of s. By this property, we can write Laplace Transform of e raise to minus 2 t into t cube is equal to in the Laplace Transform of t cube that is 6 upon s raise to 4, we have to replace every s equal to s plus 2 that is equal to 6 upon s plus 2 bracket raise to 4 and this result is denoted by F of s. Now, by property Laplace Transform of integrals that is Laplace of integration with limit 0 to t F of t d t equal to 1 upon s into F of s. So, by this property, we can write Laplace Transform of integration with limit 0 to t of e raise to minus 2 t into t cube into d t equal to 1 upon s as it is into F of s means 6 upon s plus 2 bracket raise to 4 which is equal to 6 divided by s in bracket s plus 2 bracket raise to 4 is the required answer. Now pause the video for a while, write the answer to the question. Question is find Laplace Transform of integration with limit 0 to t hyperbolic cos of 2 t d t. Come back, I hope you have written answer to this question. Here I will going to explain the solution. Question is find the Laplace Transform of integration with limit 0 to t of hyperbolic cos 2 t into d t. Here F of t means hyperbolic cos of 2 t. Now, we have to find Laplace Transform of hyperbolic cos of 2 t which is equal to s upon s square minus 2 square which is equal to s upon s square minus 4 denoted by F of s. So, this is by using formula Laplace of hyperbolic cos a t equal to s upon s square minus a square. Now, by property Laplace Transform of integrals we can write Laplace of integration of F of t d t with limit 0 to t equal to 1 upon s into F of s. By this property we can write Laplace Transform of integration with limit 0 to t of hyperbolic cos 2 t d t equal to 1 upon s as it is and F of s is here s upon s square minus 4. Now, s s gets cancelled with we get answer to this question is 1 upon s square minus 4. Now, example 2 find the Laplace Transform of integration with limit 0 to t t sin 3 t d t. Here F of t equal to t sin 3 t. Now, we have to find Laplace Transform of this using effect of multiplication by t property. For that first we find Laplace Transform of sin of 3 t which is equal to 3 upon s square plus 3 square which is equal to 3 upon s square plus 9 because of the formula Laplace of sin a t equal to a upon s square plus a square. Now, by effect of multiplication by t property we know that Laplace of t into F of t is equal to minus 1 into d by d s of F of s. So, by this property we can write Laplace Transform of t into sin 3 t equal to minus 1 into d by d s of F of s means 3 upon s square plus 9. Now, we have to differentiate this 3 upon s square plus 9 with respect to t s using co-centering. We get minus 1 as it is now keeping s square plus 9 numerator as it is and derivative of numerator 3 0 minus numerator 3 as it is and derivative of denominator is 2 s divided by s square plus 9 bracket square which is equal to minus 1 as it is into now in numerator minus 6 into s divided by s square plus 9 bracket square. Then therefore, Laplace Transform of t into sin 3 t is equal to 6 s upon s square plus 9 bracket square denote this result by F of s. Now, by property Laplace Transform of integrals that is Laplace of integration with limit 0 to t F of t d t equal to 1 upon s into F of s. By this property we can write Laplace Transform of integration with limit 0 to t of t into sin 3 t d t equal to 1 by s as it is. Now, F of s means 6 s upon s square plus 9 bracket square. Now, s s gets cancelled which is equal to 6 upon s square plus 9 bracket square this is the required answer. Now, another example, example 3 find the Laplace Transform of integration with limit 0 to t of e raise to minus t sin 2 t upon t d t. Here F of t is e raise to minus t into sin 2 t upon t. Now, we have to find it is Laplace Transform using suitable properties. First of all we find Laplace Transform of sin 2 t which is equal to 2 upon s square plus 4 because Laplace Transform of sin a t equal to a upon s square plus a square. Now, by effect of division by t property Laplace Transform of F of t upon t is equal to integration with limit s to infinity of F of s d s. So, using this property we can write Laplace Transform of sin 2 t upon t is equal to integration with limit s to infinity of 2 upon s square plus 4 into d s. Now, take 2 is outside and in bracket integration of 1 upon s square plus 4 with respect to 2 s is 1 upon 2 into tan inverse of s by 2 bracket close with limit s to infinity. Now, 2 2 gets cancelled and when we put s equal to upper limit infinity we get tan inverse of infinity minus when s equal to lower limit s we get tan inverse of s by 2. Therefore, Laplace Transform of sin 2 t upon t is equal to now tan inverse of infinity means pi by 2 minus tan inverse of s by 2 which is equal to now pi by 2 minus tan inverse of s by 2 we can write as cot inverse of s by 2. Now, by force shifting theorem that is Laplace Transform of e raise to minus e t into F of t we can write as F of s plus a. Therefore, by this property we can write Laplace Transform of e raise to minus t into sin 2 t upon t is equal to in the Laplace Transform of sin 2 t upon t that is cot inverse of s by 2 we have to replace every s equal to s plus 1 which is equal to cot inverse of s plus 1 by 2. Now, by property Laplace Transform of integrals we know that Laplace of integration with limit 0 to t F of t d t equal to 1 upon s into F of s. By using this property we can write Laplace Transform of u 1 integration with limit 0 to t e to power minus t sin 2 t upon t d t is equal to 1 upon s as it is and here F of s means cot inverse of s plus 1 upon 2. This is the required answer for the u 1 example. To prepare this video lecture I use these two books as references. Thank you.