Question

@SolomonZelman

Answer 1

\[\int\limits \dfrac{tanx}{cosx}~dx\]

Answer 2

And we have not yet learned what tan x is the derivative of

Answer 3

\(\color{black}{\displaystyle \int \frac{\tan x}{\cos x}dx=\int \frac{\sin x/\cos x}{\cos x}dx=\int \frac{\sin x}{\cos^2x}dx}\)

Then, set. \(\color{black}{\displaystyle u=\cos x}\)

Answer 4

gone over u sub?

Answer 5

Nope

Answer 6

Ok, we will give it a quick glance:)

Answer 7

Basically, an integral of a form
\(\color{black}{\displaystyle \int f[g(x)]\times g'(x)~dx}\)

Can be reduced by setting
\(\color{black}{\displaystyle u=g(x)}\)
\(\color{black}{\displaystyle du=g'(x)~dx}\)

and replacing everything accordingly gives us:
\(\color{black}{\displaystyle \int f(u) du}\)

Answer 8

I will give you an example of such.

Answer 9

\(\color{black}{\displaystyle \int \frac{x}{\sqrt{x^2+3}}dx}\) or, alternatively, \(\color{black}{\displaystyle \int x\left[\frac{1}{\sqrt{x^2+3}}\right]dx}\).

If we set \(\color{black}{\displaystyle u=x^2+3}\),
then we will have \(\color{black}{\displaystyle du=2x~dx}\)

We don't have the 2x in there, so we will basically multiply times 2/2.

\(\color{black}{\displaystyle \frac{1}{2}\int\left[\frac{1}{\sqrt{x^2+3}}\right](2x)dx}\)

Now we can set the substitution:
\(\color{black}{\displaystyle u=x^2+3}\)
\(\color{black}{\displaystyle du=2x~dx}\)

and the integral becomes
\(\color{black}{\displaystyle \frac{1}{2}\int\frac{1}{\sqrt{u}} du=(1/2)\int u^{-1/2}du=(1/2)(2)u^{1/2}=\sqrt{u}=\sqrt{x^2+3}+C}\).

Answer 10

Another example ... while you read this one.

Answer 11

Is du the derivative of u?

Answer 12

The derivative of u (since u is a function of x that we substitute) is du/dx.
and, du/dx=u' ----> du=(u')dx.

Answer 13

So if our substitution was u=x^4, then

u=x^4
du/dx=4x^3 ---> du=4x^3 dx.

Answer 14

Okay, I think I'm getting it...

Answer 15

Ok, another example would be.
\(\color{black}{\displaystyle \int x^3(x^4+2)^{e}dx}\)

the derivative of x^4+2 is x^3, so we will multiply times 4/4, to get
\(\color{black}{\displaystyle \frac{1}{4}\int (x^4+2)^{e }(4x^3dx)}\)

and then set
\(\color{black}{\displaystyle z=x^4+2}\)
\(\color{black}{\displaystyle dz=4x^3~dx}\) (you can use other letters too)

our integral becomes
\(\color{black}{\displaystyle \frac{1}{4}\int z^{e } dz=\frac{1}{4(e+1)}z^{e+1}}\)

and substituting our original variable back, and with +C we get:
\(\color{black}{\displaystyle =\frac{(x^4+2)^{e+1}}{4e+4}+C}\)

Answer 16

\(e\) is just Euler's constant.

Answer 17

(oh excuse me, I meant to say the derivative of x^4+2 is \(\color{red}{4}\)x^3 ... mutliply times etc)

Answer 18

Ok, now, as you read I will repost about the problem.

Answer 19

This was our integral.
\(\color{black}{\color{black}{\displaystyle \int \frac{\tan x}{\cos x}dx=\int \frac{\sin x/\cos x}{\cos x}dx=\int \frac{\sin x}{\cos^2x}dx}}\)

and we simplified down to
\(\color{black}{\color{black}{\displaystyle \int \frac{\tan x}{\cos x}dx=\int \frac{\sin x/\cos x}{\cos x}dx=\color{blue}{\int \frac{\sin x}{(\cos x)^2}dx}}}\)
-----------------------------------------------
Then you know that the derivative of cos(x) is -sin(x), so multiplying times (-1)/(-1) (or, actually we can just multiply times -1 twice without changing anything, so I will do this.)

\(\color{black}{-\color{black}{\displaystyle \int \frac{-\sin x}{(\cos x)^2}dx}}\)

Now we can do a substitution.
\(\color{black}{p=\cos x}\)
\(\color{black}{dp=-\sin x~dx}\)

can you write the new integrall (in terms of p)

Answer 20

\(-\int \dfrac{-sinx}{p^2 dp}\)\)?

Answer 21

\(\large \color{black}{-\color{black}{\displaystyle \int \frac{\color{red}{-\sin x ~dx}}{(\color{blue}{\cos x})^2}}}\)

this is our integral?

Answer 22

right?

Answer 23

Yes

Answer 24

\(\large \color{black}{-\color{black}{\displaystyle \int \frac{\color{red}{-\sin x ~dx}}{(\color{blue}{\cos x})^2}}}\)
----------------------------
OK, then we substitute
\(\color{black}{\color{blue}{p=\cos x}}\)
\(\color{black}{\color{red}{dp=-\sin x ~dx}}\)

correct?

Answer 25

what part is replaced by dp, and what part is replaced by p?

Answer 26

(One sec, I'll be back, I need to carry my new tires)

Answer 27

So it would be \(\int dp/p^2\)

Answer 28

you forgot the negative in front of the integral

Answer 29

\(\color{black}{\displaystyle -\int p^{-2} dp}\)

Answer 30

Now we can put in p and dp?

Answer 31

integrate with respect to p

Answer 32

what do you get for that integral \(\color{black}{\displaystyle -\int p^{-2} dp}\)
?

Answer 33

(just apply the power rule)

Answer 34

\(\dfrac{(-cosx)^{-1}}{-1} * (sinx)\)

Answer 35

What is \(\color{black}{\displaystyle -\int x^{-2} dx}\) ?

Answer 36

\(\dfrac{-x^{-1}}{-1}dx\)?

Answer 37

Yes, exactly.

Answer 38

So, \(\color{black}{\displaystyle -\int p^{-2} dp}\) is?

Answer 39

(same thing, but different symbol)

Answer 40

oh, without the differential though

Answer 41

\(\dfrac{-x^{-1}}{-1}dp\)

Answer 42

Just the fraction?

Answer 43

Although, that would cancel out to just be \(x^{-1}\)

Answer 44

yes, just x^(-1).

Answer 45

and same way \(\color{black}{\displaystyle -\int p^{-2} dp=\frac{1}{p}+C}\)

Answer 46

\(\dfrac1{cosx} + c\)

Answer 47

YES !!

Answer 48

So basically the procedure is:

u = f(x)
du = f'(x) dx

Then, we solve the integral with respect to u (just like any other integral).
lastly, we substitute the original variable back.

Answer 49

I think I have the last question... just need you to check if that's okay?

Answer 50

sure.

Answer 51

Here it is

Answer 52

yes, you are correct!

Answer 53

Yay! :)

Answer 54

Yupss :)

Answer 55

Could you help with 1 or 2 definite integral problems? Just need to remember how to do them...

Answer 56

Sure