 Hello and welcome to the session. In this session we will discuss evaluation of definite integrals by substitution. We have a definite integral a to b fx dx. Now let's see the various steps involved in evaluating this definite integral by substitution. In the first step we consider this integral without limits that is in this form and we substitute y equal to fx or x equal to gy to reduce the given integral to a known form. In the next step we integrate the new integrand with respect to the new variable without mentioning the constant of integration. Then we re-substitute for the new variable and write the answer in terms of the original variable. Then lastly we find the values of answers obtained in previous step at the given limits of integral and find the differences of the values at the upper and lower limits. We have one more method in which the step 3 would be changed that is in the second method the integral will be kept in the new variable itself and the limits of the integral will accordingly be changed. Let's evaluate the integral i equal to integral limits 1 to 2 x upon x square plus 1 dx. Here we substitute x square plus 1 equal to t this gives us 2x dx equal to dt or x dx equal to 1 upon 2 dt. So we get integral x over x square plus 1 dx is written as 1 upon 2 integral dt upon t which is equal to 1 upon 2 log modulus t. Now we substitute the value for t that is x square plus 1 so this gives us 1 upon 2 log modulus x square plus 1 so now integral 1 to 2 x over x square plus 1 dx is equal to 1 upon 2 log modulus x square plus 1 limits 1 to 2. This would be equal to 1 upon 2 log modulus 2 square plus 1 minus log modulus 1 square plus 1 which is equal to 1 upon 2 log modulus 5 minus log modulus 2. This is equal to 1 upon 2 log 5 upon 2. This is the value for the given integral i. Now we shall discuss some properties of definite integrals. The properties that we shall discuss now would be very useful in evaluating the definite integrals more easily. The first property is integral a to b fx dx is equal to integral a to b ft dt. Next one is integral a to b fx dx is equal to minus of integral b to a fx dx or this can also be given as integral a to a fx dx is equal to 0. Next property is integral a to b fx dx is equal to integral a to c fx dx plus integral c to b fx dx. Next property is integral a to b fx dx is equal to integral a to b f of a plus b minus x dx. And the next one which is the particular case of this property that is the fourth property is 0 to a fx dx is equal to integral 0 to a f of a minus x dx. Then we have integral 0 to 2 a fx dx is equal to 0 to a fx dx plus integral 0 to a f of 2 a minus x dx. Next integral 0 to 2 a fx dx is equal to 2 multiplied by integral 0 to a fx dx. If we have f of 2 a minus x is equal to fx and this integral is equal to 0 if f of 2 a minus x is equal to minus fx. Then we have integral minus a to a fx dx is equal to 2 multiplied by 0 to a fx dx if f is an even function that is f of minus x is equal to fx and this integral would be equal to 0 if f is an odd function that is f of minus x is equal to minus fx. So these are the important properties of definite integrals that we will use by evaluating the definite integrals. Consider i equal to integral minus 1 to 1 sine cube x cos 4x dx. Now we have to evaluate this integral using the above properties. Here the function fx is equal to sine cube x into cos 4x now f of minus x is equal to sine cube minus x into cos to the power 4 minus x. This is equal to minus sine cube x into cos to the power 4x that is we get f of minus x is equal to minus fx. So this implies that fx is an odd function and so from this 8th property in which we have integral minus a to a fx dx is equal to 0 if f of minus x is equal to minus fx. So the value of i would be equal to 0. So this completes our session. Hope you have understood how we evaluate the definite integrals by substitution and by using the properties of definite integrals.