Before I start, lemme tell you that this lesson... is going to be the absolue hardest lesson you take in this course to apply, even harder than the chain rule. so you have to practice alot
u-sub is the most key technique u need to solve integrals, it involves substittuting in u instead of x and changinf the refrence from \(dx\) to \(du\), lemme give an example: $$\int \frac{ln(x)}{x} dx$$ now this is completely solvable using an by parts (stay tunned for that), but let's make our lives much much easier, lets change the reference point from x to u, now u already know that the derivative of \(\ln(x) = \frac 1x\), so u can assume that $$u = \ln(x)$$ and by basic implicit differentiation: $$du = \frac{1}{x} dx$$ and then substatute those in: $$\int u du$$ and this is just the power rule: $$\int u du = \frac{u^2}{2} +c$$ and then substatute back in \(\ln(x)\) $$F(x) = \frac{(\ln(x))^2}{2} + c$$ u-sub is the greatest technique of all time and it has many applications in math, not just in calc.
Trust me, this techunique will make ur life so much, and I mean SO MUCH easier..., and it doesn't involve calculus, all the work is done using pre-calc knowledge: for example: $$\int \sqrt{1+\sin(2x)} dx$$ now, if u don't know ur trig, u are basically cooked, but if u do know ur trig, u would notice that: $$1 = \sin(x)^2 +\cos(x)^2$$ AND $$\sin(2x) = 2\sin(x) \cos(x)$$ so we can substatute that in: $$\int \sqrt{\sin(x)^2 + \cos(x)^2 + 2\sin(x) \cos(x)} dx$$ notice anything? look again, now do u see it? do u see the idea? $$\sin(x)^2 + \cos(x)^2 + 2\sin(x) \cos(x) = (\sin(x) + \cos(x))^2$$ the square root is cancelled wit the power, and we end up with: $$\int \sin(x) + \cos(x)$$ and I this you can take it on from here, u can do this with any pre-calc mathimatics concept like long division, log identities, or algibraic maipulation in general
Now we are getting into hard stuff teritory, partail fraction integration is like taking an ugly thing and spliting it into 3 equally-ugly things. so get ready, partual fractions is a technique wher u split a rational fuction into 2 and solve for the ploynomyals of it. so: $$\frac{px + q}{(ax+b)(cx+d)} = \frac{A}{ax+b} + \frac{B}{cx+d}$$ and then solving for A and B, think of it like the reverse of fraction addition. Lemme give You an example to clear things up: $$\int \frac{1}{(x+1)(x-1)} dx$$ $$\frac{1}{(x+1)(x-1)} = \frac{A}{x+1} + \frac{B}{x-1}$$ multiply both sides by \((x+1)(x-1)\) $$A(x-1) + B(x+1) = 1$$ then find the solution of both polynomyals, whcih are 1 and -1 respectively. Pluging in 1: $$A(1-1) + B(1+1) = 1$$ $$2B = 1$$ $$B = \frac12$$ and -1: $$A(-1-1) + B(-1+1) = 1$$ $$-2A = 1$$ $$A = -\frac12$$ then u have the final integral to be: $$\int \frac{-1}{2x+2} + \frac{1}{2x-2} dx$$ which can be solved with normal differentiation rules partial fractions is a very powerful techunique and it is ver complex ith bigger polynomyals, but it is nothing compared to the next 2. from here on, I won't be putting any more "try it"s
disclamer: I won't be able to explain this concept clearly unless I am explaining it in a video, so please watch the video.
So I am going to try to be as clear as possible, in trig sub, u try to think of 3 terms which fit the descriptain of a triangle's sides.
for example:
$$\int \frac{dx}{\sqrt{4 + x^2}}$$
notice anything? the denominator can be the hypotunuse (sorry for the terrible spelling) of a triangle with sides 2 and x;
so we can substatute this value a bit, let's recall what tan is... \(\tan \theta = \frac{opp.}{adj.}\), if we assume that \(\theta\) is the angle between 2 and the hypo., we can deduce that: $$\tan \theta = \frac x2$$ $$\therefore 2\tan \theta = x$$ and don't forget to change the reference $$dx = 2sec^2 (\theta) d\theta$$ subing into the original integral: $$\int \frac{2sec^2 (\theta) d\theta}{\sqrt{4 + 4\tan^2 \theta}}$$ $$\int \frac{2sec^2 (\theta) d\theta}{2\sqrt{1 + \tan^2 \theta}}$$ $$\int \frac{2sec^2 (\theta) d\theta}{2\sqrt{\sec^2 \theta}}$$ $$\int \frac{2sec^2 (\theta) d\theta}{2\sec \theta}$$ simplifying: $$\int sec \theta d\theta$$ to integrate that, we can use algebraic manipulation by adding \(\sec \theta + \tan \theta\) to both sides $$ \int \sec\theta\, d\theta = \int \frac{\sec^2\theta+\sec\theta\tan\theta} {\sec\theta+\tan\theta} \,d\theta. $$ Let $$ u=\sec\theta+\tan\theta, $$ then $$ du=(\sec\theta\tan\theta+\sec^2\theta)\,d\theta. $$ Hence, $$ \int \sec\theta\,d\theta = \int \frac{du}{u} = \ln|u|+C = \ln|\sec\theta+\tan\theta|+C. $$ Back-substitute using $$ \tan\theta=\frac{x}{2}, \qquad \sec\theta=\frac{\sqrt{4+x^2}}{2}, $$ to obtain $$ \int \frac{dx}{\sqrt{4+x^2}} = \ln\left| \frac{\sqrt{4+x^2}}{2} + \frac{x}{2} \right| +C $$ this technique is so useful, yet so dreadful. However, It doesn't compare to what is coming next...
disclamer: I won't be able to explain this concept clearly unless I am explaining it in a video, so please watch the video.
BY PARTS INTEGRATION IS THE MOST DRULING, MOST AGRRESSIVE FORM OF INTEGRATION, IT IS SO TIME CONSUMING THAT IT EAT YOUR SOUL
so I will try to explain it, the general formula is:
$$\int u dv = uv -\int v du$$
appears easy enough, but if you choose the wrong variable u are screwed, lemme give an example:
$$\int \ln x dx$$
there is no standard formula for this so lets divide that into 2 functions \(\ln x\) and \(1 dx\), so we can now assume that
$$u = \ln(x)$$
$$dv = 1 dx $$
but how did we choose which variable is which? well we use smth called differential favorability, basically which function would be
easier to differntiate, and which would be easier to integrate (remember, we differntiate u and integrate dv), and this is decided by
the LIATE rule, which stands for: "logarithmic, inverse trigonometric, algebraic, trigonometric, exponential", they are listed
so that the first (logarithmic) is easiest to differentaite, and the last (exponential) is easiest to integrate. Back to our example, \(ln x\)
is a logarithmic function, and 1 dx is an algebraic function, so we differentaite ln and integrate 1:
$$u = \ln(x)$$
$$dv = 1 dx $$
$$du = \frac{1}{x} dx$$
$$v = x$$
using the standard formula:
$$\int u dv = uv -\int v du$$
$$\ln(x) \cdot x -\int x \frac 1x dx$$
$$\ln(x) \cdot x -\int 1 dx$$
$$\ln(x) \cdot x - x$$
$$x(\ln(x) - 1)$$
this lesson took me like 4h to write + the video T-T