Question

Can you check if I'm doing this right.

Answer 1

\[\int\limits_{}^{} \frac{ 1 }{ 4x^2+1 } dx\]

Answer 2

Oo this one is quite a bit different than the last one.
Not a simple sub.

You've learned trigonometric substitutions at this point?

Answer 3

Well I have learned them, just haven't memorized them yet.

Answer 4

Can I use the "u" for the denominator?

Answer 5

No u sub.
We'll have to make a trig sub here.
Let's make a minor adjustment first,
we want to include the 4 in that square with the x.

Answer 6

\[\large\rm 4x^2=2^2x^2=(2x)^2\]

Answer 7

So our denominator is of this form: \(\large\rm (stuff)^2+1\)

Answer 8

We want to look to our Pythagorean Identity which has addition in it.
\[\large\rm \cos^2\theta=1-\sin^2\theta\]\[\large\rm \sec^2\theta=\tan^2\theta+1\]

Answer 9

Looks like it's the one involving tangent, ya?

Answer 10

Do you see how if we replace the "stuff" with tangent,\[\large\rm (stuff)^2+1\quad=(\tan \theta)^2+1\quad=\sec^2 \theta\]We can apply our trig identity and it cleans things up nicely for us?

Answer 11

You are trying to create a tan^2theta+1 in your denominator,
to make use of your Pythagorean Identity.

So you want to make the substitution, \(\large\rm 2x=\tan\theta\)

Answer 12

What do you think?
Head explode?
Too much?

Answer 13

Gimme a sec to chew. o_o

Answer 14

hehe, ya trig subs are a lot to swallow ^^

Answer 15

How do you just instantly tell that you want the (stuff)^2 + 1 on the bottom?

Answer 16

Hmm I don't understand the question +_+

Answer 17

How do you figure out what basic integral form to get? Like, for example the last one you knew to get the ln|x| thingy.

Answer 18

Or maybe like, how can you tell that this question needs a \[\frac{ 1 }{ a }\tan^-1 (\frac{ u }{ a } + C\]

Answer 19

Well u-sub is very straight forward.
You're looking within your integral for `two things`:
~something
~and the derivative of that something

We had an x^2 in the denominator,
and an x in the top,
so a light should go off that maybe x^2 can be your u,
and it's derivative is in the numerator

Answer 20

Another light should go off for a problem like this.
You have a squared term and addition.
So you can make use of the form that your trig identity has.

You don't have an x showing up in the numerator, so no easy u-sub.

Answer 21

Something to note though:
In the last problem we could have also made the substitution x=tan theta.
It would have still given us the correct answer :)
Would've just been more work.

Answer 22

I know that this identity: \(\large\rm \sec^2\theta=\tan^2\theta+1\) allows me to `get rid of addition`.
That's why we would like to make use of it in this problem.

Answer 23

\[\large\rm \int\limits \frac{1}{(stuff)^2+1}\quad=\int\limits\frac{1}{\tan^2\theta+1}\quad=\int\limits\frac{1}{\sec^2\theta}\]Do you see how replacing the "stuff" that is being squared by tangent can help get rid of that pesky addition sign in the denominator?

Answer 24

To answer your question: There are a limited number of integral forms pertaining to the inverse trig function. You need to know them. Their integrals are the inverse sin of x, the inverse cosine of x and the inverse tan of x.

In the given problem you see 4x^2 in the denominator. Note how this resembles the form u^2+ a^2.

\[\int\limits_{?}^{?}\frac{ 1 }{ u^2+a^2 }du=\tan ^{-1}\frac{ u }{ a }+c\]

Answer 25

Okay. I see now.

Answer 26

This has to be memorized, or you must be able to derive it.

Then, the fact that the denominator of \[\frac{ 1 }{ 4x^2+1 }closely .resembles \frac{ 1 }{ u^2+a^2 }\]

Answer 27

tells us that the integral will be an inverse tangent.

Answer 28

Yes you can do that if you like the shortcut method :)
I don't really like memorizing integrals myself.

Answer 29

\[\large\rm \int\limits \frac{1}{4x^2+1}dx\quad=\int\limits\frac{1}{(\color{royalblue}{2x})^2+1}dx\]

\[\large\rm \color{royalblue}{2x=\tan \theta}\]Then take derivative to figure out what will take the place of our dx.

Answer 30

zepdrix is correct in saying "You have a squared term and addition.
So you can make use of the form that your trig identity has." Right: 4x^2 + 1 suggests u^2+a^2.

Answer 31

And U^2+a^2 in your denominator suggests the possibility that the integral has the inverse tangent form.

Answer 32

1/2sec^2 x

Answer 33

Is that the derivative?

Answer 34

\[\large\rm \color{royalblue}{2x=\tan \theta}\]\[\large\rm \color{orangered}{dx=\frac{1}{2}\sec^2\theta~d \theta}\]Yes, sounds right :)

We'll use this information to make our substitution,\[\large\rm \int\limits \frac{1}{(\color{royalblue}{2x})^2+1}\color{orangered}{dx}\]

Answer 35

I tried to color-code things so you can see how we'll make our replacements.

Answer 36

\[\large\rm \int\limits\limits \frac{1}{(\color{royalblue}{\tan \theta})^2+1}\color{orangered}{\frac{1}{2}\sec^2\theta~d \theta}\]The whole reason we chose that substitution was to make use of our Pythagorean Identity,
so let's apply it now, and also pull the 1/2 out front,\[\large\rm \frac{1}{2}\int\limits\frac{1}{\sec^2\theta}\sec^2\theta~d \theta\]

Answer 37

Ok with those few steps? :o

I gotta slow down sorry,
I get excited when doing integrals AND trig AT THE SAME TIME.
The two best things ever :D haha

Answer 38

Yeah I'm okay, I think.

Answer 39

So this integral is going to clean up really nicely,\[\large\rm \frac{1}{2}\int\limits d \theta\]

Answer 40

You can throw a 1 in there if you're more comfortable with that :)\[\large\rm \frac{1}{2}\int\limits\limits 1~d \theta\]What does that turn into when we integrate it?
Derivative of what gives you 1?

Answer 41

I got confused somewhere here.

Answer 42

Well let's at least wrap this one up,
and then we can summarize and see what went wrong :)

So if you took the derivative of x, you get 1, yes?
So then the integral of 1 is x+c

We're integrating in theta, not x, so we get theta+c from our 1.\[\large\rm \frac{1}{2}\int\limits d \theta\quad=\theta+c\]

Answer 43

...1/2 arctan (2x) + c. LORD HELP ME! Ahhhh! Okay okay.

Answer 44

Woops I forgot the 1/2, my bad.\[\large\rm \frac{1}{2}\int\limits\limits d \theta\quad=\frac{1}{2}\theta+c\]

Answer 45

Can I have a question to try?

Answer 46

We would like our final answer in terms of x, not theta.
So we have to undo our substitution.

Going back to our original sub,\[\large\rm 2x=\tan \theta\]Applying inverse gives us,\[\large\rm \arctan(2x)=\theta\]ya? :o

Answer 47

Like I make one up for you? :o

Answer 48

Yeah.

Answer 49

Like this one, or like the last one?

Answer 50

...Both? hehe.

Answer 51

\[\large\rm \int\limits\frac{(\ln x)^3}{x}dx\]Maybe try this one? :)

Answer 52

This is like the other problem we did,
nice simple u-substitution.

It might not jump out at you right away though.

Answer 53

We can stick with polynomial stuff if this is too tricky ^^

Answer 54

Yeah gimme a polynomial, then I'll try this one. x_x

Answer 55

\[\large\rm \int\limits\frac{3x}{5x^2-3}dx\]I made the numbers a little tricky :)
I'm so mean.

Answer 56

I think

Answer 57

\[\frac{ 3 }{ 10 }\ln(5x^2 - 3) + C\]

Answer 58

Is the second one...\[\frac{(\ln x)^4 }{ 4 }\]

Answer 59

Ooo nice c: Figured out that tricky log also huh? Awesome.

Answer 60

\[\large\rm \int\limits\frac{1}{1-x^2}dx\]Wanna try this one maybe?
Trig sub like this problem.

Answer 61

Haha in a second. food. xP

Answer 62

Alright.

Answer 63

I just skimmed through this, and I didn't understand anything.

Answer 64

lol

Answer 65

Am I supposed to use cosine and -sine here?

Answer 66

@zepdrix

Answer 67

Our identity:\[\large\rm 1-(\color{orangered}{\sin \theta})^2=\cos^2\theta\]And we have this showing up in our problem:\[\large\rm 1-\color{orangered}{x}^2\]

So yes, sine seems like an appropriate substitution for x.

Answer 68

Okay. Just making sure. hehe.

Answer 69

Ah darn it, I should've thought of a better problem :O(
This one is going to be a little more advanced.

Answer 70

Uhh. Am I supposed to have something that looks like this? \[\int\limits\frac{ \cos \theta }{ \cos^2 \theta }d \theta\]

Answer 71

Yesss good job \c:/

Answer 72

Now I don't know what to do hahaha.

Answer 73

Cancel out a cosine, and then apply an identity:\[\large\rm \int\limits \frac{1}{\cos \theta}d \theta\quad=\int\limits \sec \theta~d \theta\]

Answer 74

OH duh. -.- lol. I need to dye my hair blonde.

Answer 75

Have you seen the integral for secant yet?
It's a bit of a weird one.

Answer 76

\[\int\limits \sec \theta \space d \theta = \ln|\sec \theta - \tan \theta| + C\]

Answer 77

Is that it?

Answer 78

Oh. And I need to replace the stuff.

Answer 79

Um um, we want addition, but yes close,\[\large\rm \int\limits\limits \sec \theta \space d \theta \quad= \ln|\sec \theta + \tan \theta| + C\]

Answer 80

Yah, that's why I was saying this problem is a little more advanced,
we have to use the "triangle technique", can't just do a simple inverse trig to undo things.

Answer 81

Have you seen the triangle method or whatever its called? :)

Answer 82

Never heard of it. o_o

Answer 83

Our substitution was this:\[\large\rm \sin \theta=x\]Let's rewrite x as x/1 and then we'll draw out a triangle to illustrate this,\[\large\rm \sin \theta=\frac{opposite}{hypotenuse}=\frac{x}{1}\]