The remaining examples in this video increase substantially in terms of difficulty, and you should pause the video a lot to check the details and your understanding. You might benefit greatly also from consulting the notes that accompany this video, and make sure that you feel really comfortable with the easier substitution examples. What we illustrate next is a tricky anti-differentiation using a mixture of the circular functions in the integrand. It's not obvious at all what substitution to make that could possibly simplify the calculation. With experience, you realize that if the two powers that appear the even power cos squared x turns out to be the most problematic. You'll see in a moment that the odd power of sine cubed x can be used to our advantage. So, we try making a substitution using this Cos x. So, du-dx equals minus sine x, and du equals minus sine x dx. We now rewrite the integrand to try to exploit this formula for the differential du. We peel off a copy of sine x from the sine cubed x, leaving a factor of sine squared x, which fortuitously can be rewritten as one minus cos squared x, exploiting the circular identity. Then we rearrange the integrand to get minus sine x next to the differential dx, and balance everything out with an extra minus sign out the front. Now comes a magical effect of the substitution. We replace cos x wherever it appears in the integrand by u, and minus sine x dx gets replaced by du. Our original integral is thus transformed into minus the integral of one minus u squared times u squared du. We can expand the integrand. Bring the minus sign back inside to get a simple polynomial u, which is then straightforward to anti-differentiate piece by piece. Remains finally to express everything in terms of x. The original problem is solved. The next example is a difficult integration of a rational function where the integrand is a fraction with a linear polynomial in the numerator, and a quadratic in the denominator. Denominators usually create problems. So, let's try to rewrite the quadratic in a simpler form. By the trick of completing the square, we can rewrite it as x plus one squared, plus one. It will simplify considerably if we make a substitution of u equal to x plus one. So that du-dx equals one, and du equals dx. The numerator can also be expressed in terms of u by a trick of adding and subtracting one to x inside brackets, and rearranging this expression to get 4u plus three. We're working steadily towards expressing everything in terms of u. The integrand becomes 4u plus three divided by u squared plus one, and the differential dx is simply replaced by du. This is substantial progress, and we can simplify things further by splitting up the integrand into pieces and integrating them separately. The first piece we'll come back to in a moment, but the second piece with a constant three at the front of the integral of du of u square plus one, becomes three times inverse tan of u plus C. The reason from an earlier video is that the derivative inverse tan of x becomes the rule one over x squared plus one, describing the curve known as the witch of Maria Agnesi. Expressed as an indefinite integral involving x, which we're applying here with u instead of x. Notice also that the constant of integration is multiplied by three, which becomes another constant, which we can just rename as C again. We can go a step further and rewrite inverse tan of u as inverse tan of x plus one. We still have to figure out the first pie ce. Again the denominator is the problem. So make a new substitution say, v equal to u squared plus one, so the dv-du equals 2u. So that dv equals 2u du. To set things up, we bring the constant two at the front, and then replace the denominator by v, and the new numerator, including the differential, by dv. This produces the much simpler expression two times integral of dv on v. But remember the integral of dx on x is, ln of x plus C, where x is positive, which we can apply here. Notice that the constant of integration is multiplied by two, which we've just renamed by C again. So, we splice this into our earlier answer. There are actually two constants of integration, both of which were called C, which get added together to get another constant, which again we call C. This kind of abuse of notation calling everything C all the time is universally used, and relatively harmless but you should be aware of what you're actually doing when you manipulate the constants. We in fact want an antiderivative in terms of the original variable x, so we convert v into u square plus one, and the remaining u that appears into x plus one, and we've solved the original problem. We'll finish with a very difficult problem which is to anti-differentiate the fraction root x over one plus root x. There are a few different substitutions we could try, but we'll go the whole hog and put u equal to the entire denominator, which can be rewritten as one plus x to the half. Then du-dx becomes a half x to the minus a half, which is one over two root x, so that du equals dx over two root x, which we can rearrange so that dx equals two root x_du. Hence the original integral can be transformed by replacing one plus root x by u, and dx by two root x_du. Then bring the constant two out the front, and simplify the numerator to x, which then becomes u minus one squared. We can now expand the numerator, and split the integrand into pieces. An integer 2u minus two is just u squared over two minus 2u, and an intiderivative of one on u is log of the magnitude of u. We have amalgamated all the constants of integration at the right, which I've called C dash. Not C, for a reason which will become apparent in a moment. Then we carefully write everything in terms of x only. Now, we could just stop here, but notice that we can expand and simplify further in a few steps to obtain x minus two root x, plus two log of one plus root x minus three plus c dash. Note that one plus root x is positive so we can dispense with the magnitude signs and forming the natural logarithm. The expression minus three plus C dash is just a constant which we now call C. Then we end up with this elegant solution of a very difficult problem. In today's video, we introduced and illustrated the method of integration by substitution, which is closely related to the Chain Rule, one of the differentiation rules we learnt in the last module. This method enables one to antidifferentiate complicated functions by exploiting combinations of integrands and differentials, which simplify the integral after making appropriate substitutions and changes of variables. Some of the examples could be solved by a guess and check method. Other examples are very difficult and require considerable experience, even to choose the right substitution that's likely to lead to progress, and then require considerable perseverance, and technical skill to see the calculation through to a successful conclusion. Please read the notes and when you're ready, please attempt exercises. Thank you very much for watching, and I look forward to seeing you again soon.
