 Hello and welcome to the session. My name is Mansi and I'm going to help you with the following question. The question says evaluate the following definite integral that is integral to secant square x plus x cube plus 2 dx where lower limit of integration is 0 and upper limit of integration is pi by 4. So let us start with the solution to this question. Now the integral given to us is 0 to pi by 4 here the limit goes from 0 to pi by 4 to secant square x plus x cube plus 2 dx. Now on separating these three integrants we can write it as limit goes from 0 to pi by 4 to secant square x dx plus integral x cube dx where limit goes from 0 to pi by 4 plus integral 2 dx where limit goes from 0 to pi by 4. Now we see that integral of secant square x is tan x so we have 2 tan x here limit goes from 0 to pi by 4 plus integral x cube dx is x 4 by 4 x raise to power 4 by 4 here also limit goes from 0 to pi by 4 plus here we have 2 x where limit goes from 0 to pi by 4. Now this is equal to 2 into tan pi by 4 minus tan 0 plus here we will have pi by 4 the whole raise to power 4 minus 0 this whole multiplied by 1 by 4 plus 2 into pi by 4 minus 0 this is equal to now tan pi by 4 is 1 so 2 into 1 minus tan 0 is 0 plus 1 by 4 into pi raise to power 4 divided by 256 plus pi by 2 because 2 gets cancelled with 4 to give a 2 here now we have this this can be written as 2 plus pi by 2 plus pi raise to power 4 by 1024. So our answer to this question is pi raise to power 4 divided by 1024 plus pi by 2 plus 2 I hope that you understood the question and enjoyed the session have a good day.