Ok Now I looked through your document properly. The first time, I only looked at the bits where you said you got stuck. Now I see you have some other problems :P
Now With substitution, we use it when we see 2 things in an integral. The first is a function. The second is its derivative. The aim is to simplify the expression to something even a bit easier, hopefully to something we recognize how to do. Even if its something subtle.
EG \int \tan (x^5+5) x^4 dx
Now Look at this. The Function x^5 + 5 is in the integral, even though it isn't alone. And its even what is being "tan-ed". And its derivative isn't even in there! Or is it?...
The derivative of x^5 + 5 is 5x^4. 5x^4 isn't in there, but x^4 is! So how can we make that into what we want? We multiply the Integrand ( The Function being integrated, in this case tan (x^5 +5) x^4 ) by 5. Now we can't just put in a constant factor in where we like and let it be. We have to balance the integral, by putting a factor of 1/5 outside the integral. Because as you should know, constant factors can be taken in and out of integrals.
So with the new integral, 1/5 \int \tan (x^5+5) 5x^4 dx, we know its equal to the original one because when we take the factor of 5 out, 5/5=1, same integral! Now its ready for substitution :)
We see a function, in this case x^5+5, and its derivative in the integrand now. When we see such an occurence, let u = The Function, for this one x^5 +5 . That is because since its derivative is in there it simplies the whole integral is this fashion : \int \tan (u) \frac{du}{dx} dx.
You can see why that's good, the dx's cancel out leaving you without any x's!
Now its a simpler integral! \int \tan (u) du Even if you don't know how to find the integral of tan, its simpler than the integral before.
In fact there's an exercise for you, find the integral of tan. Remember tan x = (sin x)/(cos x). Rewrite tan in that way. Hint: What is the derivative of cos x?
Now I Will take you through each question in your document step by step :)
Question 1.
\int \frac{(\ln x)^2}{x} dx
Well I see ln x in there, and its derivative 1/x.
So let u = ln x, and therefore du/dx = 1/x, or du = (1/x) dx.
So now putting those into the integral, it becomes \int u^2 \frac{du}{dx} dx. The dx's cancel out, and the remaining integral is easy!
\frac{u^3}{3} + C.
Watch out though, we aren't done! The question wasn't in terms of u, we just made it so to make it easier to do! We have to replace the u with the x terms again. In this case u = ln x, so the integral is \frac{(\ln x)^3}{3} + C.
Question 2.
\int \sin^6 \theta \cos \theta d\theta
Now that one could fool some people. Cos is there, and its derivative -sin could be there if we balanced the factors of -1.
However, since sin^6 is there, u=cos theta doesn't help Us. That is because we want its derivative, du/d(thtea), to cancel out with the d(theta)! But That won't happen since its sin^6, which is (du/dtheta)^6.
So you do the other option, we see Sin, and its derivative cos is there all nice and alone.
u = \sin \theta, du/d\theta=\cos \theta.
Putting those into the integral, it becomes \int u^6 \frac{du}{d\theta} d\theta.
dx's Cancel out, remanining integral is easy, u^7/7 +C, but remember to but back in terms of theta.
\frac{(\sin \theta)^7}{7} +C
It didn't really matter if you made the wrong choice of substitution at the beginning, you can always start again when it doesn't work and try the other option. With time and experience you will usually get the best way first :)
Question 3.
\int \frac{z^2}{(1+z^3)^{\frac{1}{3}}} dz (Cube root is the same as power 1/3).
We see z^3 +1, and we almost see its derivative, 3z^2. So we put a factor of 3 inside the integral, balancing with a 1/3 outside.
\frac{1}{3}\int \frac{1}{(1+z^3)^{\frac{1}{3}}}3z^2 dz.
So now, u = z^3 +1, du/dz =3z^2.
\frac{1}{3}\int \frac{1}{u^{1/3}} \frac{du}{dz} dz
dz's cancel out, and When you flip a fraction, its the same as to the power of -1. And when you have a power inside brackets, and then a power outside, you multiply the powers. So we get
\frac{1}{3} \int u^{-1/3} du.
To finish that off we use the "Power Rule for Integrals":
\int x^n dx = \frac{x^{n+1}}{n+1} + C.
Use the Power Rule for Integrals, \frac{1}{3} \int u^{-1/3} du = \frac{1}{3} \frac{u^{2/3}}{2/3} +C.
Some simple algebra and not forgetting to replace u back into terms of z can simplify the integral to (z^3+1)^{2/3}/2 +C.
Question 4.
\int \frac{1}{x} \frac{1}{\ln x} dx
We see ln x and its derivative!
u=ln x, du/dx = 1/x
\int \frac{1}{u} \frac{du}{dx} dx
dx's cancel, and you know the remaining Integral to be ln u +C!
Hold on, replace u in terms of x again!
\ln (\ln x) + C.
Look at that, a log inside a log :)Now that we have finished those questions, here are some simple ones for you to do yourself:
1. \int \sin^3 x \cos x dx
2. \int x (x^2+1)^{10} dx
3. \int 3x \sqrt{1+x^2} dx
4. \int \frac{x}{\sqrt{1-x^4}} dx
That ones a bit harder, so I tell you the substitution you use. u = x^2
And use this rule, you will learn it later : \int \frac{1}{\sqrt{1-v^2}} dv = \arcsin v +C. Remember v is just any letter.
5. \int \frac{x+3}{\sqrt{x^2+6x-5}} dx
Hint: What do we see when we put in a factor of 2 in the integral?
6. \int 4x e^{x^2} dx
7. \int \frac{\cos x}{(1+2\sin x)^2} dx
Hint: u = 1 + 2 sin xI Think that's more then enough, good luck!
