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Integration: Exercise 2

by Batool Akmal
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    Have a look at our next example where you have a function 2x minus 1 that’s being multiplied with a sin of something. So, it’s a sin function. We have decided that we are going to use integration by parts, but usually when you have 2 functions multiplying together, you have only 2 options. You either use the substitution method or integration by parts. So we have sine of 3x dx. Since I have decided that we are going to be doing integration by parts here, I’m going to write my formula. So I have uv' dx equals to uv minus the integral of vdu dx. I’m also going to remind you a little bit about trig and differentiating and integrating. So, sine goes to cos and cos goes to minus sine. Differentiate in that direction, integrate in that direction. So I’ve got all the information or the things my mind could perhaps slip up on, on the side here before I start this question. So let’s split this into 2. U equals to 2X minus 1. V' equals the sine of 3X. This, we’re going to differentiate. This, we’re going to integrate to get your u' and v. U' gives me 2 and v gives me -- remember we are integrating that, so sine integrates to minus cos. So that’s going to go to minus cos 3X and when you integrate, you divide by the differential. We put this all into the formula. So you have u times v from here firstly. You multiply u with v to give you minus and we just put this minus on the outside and then I have 2x minus 1 multiplied by cos of 3x over 3, minus the integral of vdu dx So, the integral of those two terms here. So...

    About the Lecture

    The lecture Integration: Exercise 2 by Batool Akmal is from the course Advanced Integration.


    Author of lecture Integration: Exercise 2

     Batool Akmal

    Batool Akmal


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