Question

∫ (secxtanx)/(4+sec^2x) dx
Please show steps on what to do, consider inverse trigonometric functions.
Should I make a u-substitution?

Answer 1

if you put u= sec x ,
what will be du= ... ?

Answer 2

secxtanx would be the du?

Answer 3

yes, that correct.
so substitute these in your integral, what will your new integral in terms of u and du ?

Answer 4

Give me a moment to write this out

Answer 5

and actually, du = sec x tan x dx

Answer 6

sure, take your time :)

Answer 7

Wouldn't having the u=4+sec^2x be easier? and then it would just be du=2secxtanx?

Answer 8

That way it'll cancel with the first secxtanx

Answer 9

u= 4+sec^2x
du = 2 sec x (sec x tan x) dx
not easier.

Answer 10

So my dx= 1/2 du secxtanx?

Answer 11

Alright so I have ∫ (secxtanx) (4)+ (u)^2 dx where u=secx?
Or does that first secxtanx get cancelled out somehow..

Answer 12

lol, don't confuse me
\(\large \int \dfrac{(\sec x \tan xdx)}{4+\sec^2x}=\int \dfrac{du}{4+u^2}\)
because numerator = sec x tan x dx = du

Answer 13

Lol I had the 4+sec^2x up top so it would be (4+sec^2x)^-1

Answer 14

Don't I have to solve for dx?

Answer 15

means ? do what you think needs to be done, and i'll check/verify.
also, didn't you get the step i did ?

Answer 16

Ok so I have the∫ du/(4+(u)^2)
Would I just anti-derive it to be 1/4 * ln(u)^2?

Answer 17

no...
antiderivative of 1/(x^2+a^2) is 1/a tan^{-1} (x/a) +c
heard of it ?

Answer 18

No, no I havent.

Answer 19

now you have :)

Answer 20

I'm still so very lost... :/ I'm stuck at ∫ du/ 4+u^2 x still

Answer 21

Where is this x^2 coming from in relation to my problem?

Answer 22

doesn't these 2 look similar ?
\(\huge \int \dfrac{1}{x^2+a^2}dx= \dfrac{1}{2}\tan^{-1}(\dfrac{x}{a})+c\)
and
\(\huge \int \dfrac{1}{u^2+4}du=....?\)
you must be knowing, 4= ...^2

Answer 23

No I never learned that. My teacher never showed us that formula.

Answer 24

So the final answer would be 1/2 (tan^-1) (sec^2)/(4) +C?

Answer 25

a= 4
or 2 ?

Answer 26

and u = sec x only.

Answer 27

2

Answer 28

and sorry, the formula is
\(\huge \int \dfrac{1}{x^2+a^2}dx= \dfrac{1}{a}\tan^{-1}(\dfrac{x}{a})+c\)

Answer 29

I'm just trying to work this out in a way my teacher would deem fit since we haven't learned that formula

Answer 30

ok, so lets prove it using a trigonometric substitution
u =2tan y
du = sec^2 y dy
\(\huge \int \dfrac{1}{u^2+4}du=\int \dfrac{\sec^2 ydy}{4\tan^2 y+4}\)
can you simplify the denominator ??

Answer 31

There is no 2tanx in the problem though so how can that be u?

Answer 32

hmmm.. ?
i am making a new substitution in 1/ (u^2+4) du
u= 2 tan y
didn't get what you doubt is ?

Answer 33

Alright let's back track here and etch a sketch the thing.
Tell me if this is just totally wrong b/c this is the way my brain wants to work.
∫ (secxtanx)(4+sec^2x)^-1
u=4+sec^2x
du=2secx+secxtanx dx

Integrating it
∫ (2secx)/(u)
(2secx)/(4+sec^2x) final answer.

That is how my teacher taught us but I feel like I'm going wrong somewhere in there or what.

Answer 34

The first secxtanx cancel out b/c of the secxtanx in the du

Answer 35

du=2secx+secxtanx dx <----incorrect.

Answer 36

What is the du then, just secxtanx? I though I had to use the chain rule for that.

Answer 37

du = 2 sec x (sec x tan x ) dx
by chain rule.
this du is of no use here.

Answer 38

if u= sec x
then du = sec x tan x dx
which is actually the numerator!

Answer 39

My poor eraser, okay let me figure it out that way.

Answer 40

\[\large \color{orangered}{u=\sec x}\]\[\large \color{royalblue}{du=\sec x \tan x dx}\]

\[\large \int\limits\limits \frac{\color{royalblue}{\sec x \tan x dx}}{4+(\color{orangered}{\sec x})^2} \qquad \rightarrow \qquad \int\limits \frac{\color{royalblue}{du}}{4+(\color{orangered}{u})^2}\]

Maybe colors will help sort things out.
I always find that helps me at least :D

Answer 41

∫ du/4+(u)2 dx
okay, and how do I find the anti derivative of that or is that just it?

Answer 42

You have two options, you can either:
~Make a trigonometric substitution

Or

~Just memorize the formula,

depending on which method you have learned at this point.

Answer 43

Yes that's where I'm stuck, the second part there.

Answer 44

do the trigonometric substitution please

Answer 45

The point of making a trig sub is that it allows us to take advantage of our trig identities. We can simplify the denominator to one term, getting rid of the addition.

Since it's of the form \(\large a^2+x^2\) we'll make the substitution \(\large x=a\tan\theta\)
Here is what that will do for us.

\[\large a^2+x^2 \qquad \rightarrow \qquad a^2+(a \tan \theta)^2 \qquad \rightarrow \qquad a^2(1+\tan^2\theta)\]
From here, we revisit our Trig Identity, \(\large \color{royalblue}{1+\tan^2\theta=\sec^2\theta}\)

Which gives us, \(\large a^2\sec^2\theta\)

See how that gets rid of the addition?
This term that it leaves us with is much easier to integrate.
So let's rewrite our integral a sec, so we have an \(\large a^2\) in the bottom.

Answer 46

I never learned that formula though

Answer 47

the a^2+x^2 thing

Answer 48

You didn't? Then why are you working on this problem?
Have you learned this formula then?

\[\large \int\limits \frac{1}{a^2+x^2}dx \qquad = \qquad \frac{1}{a}\arctan\left(\dfrac{x}{a}\right)\]

Answer 49

I have not -_- I'm just going to say screw it and learn it anyhow. Apparently my teacher left something out or I missed something. I'll just go with it. Okay, so re-writing the problem with the a^2(sec^2θ).. continue

Answer 50

1/2 (arctan)*sec/2 ?

Answer 51

We first want to get the proper form in the denominator.\[\large \int\limits \frac{1}{4+u^2}du \qquad = \qquad \int\limits \frac{1}{2^2+u^2}du\]
Ok now we can properly see our \(\large a\) value is \(\large 2\).
We'll make the substitution,\[\large u=a \tan \theta \qquad \rightarrow \qquad \color{royalblue}{u=2\tan \theta}\]As with any substitution, we need to take a derivative, so we can replace the differential \(\large du\).
\(\large \color{orangered}{du=2\sec^2\theta}\)


Plugging in all the pieces,\[\large \int\limits\limits \frac{\color{orangered}{du}}{2^2+\color{royalblue}{u}^2} \qquad =\qquad \int\limits\limits \frac{\color{orangered}{\color{orangered}{2\sec^2\theta}}}{2^2+(\color{royalblue}{2\tan \theta})^2} \]

Answer 52

That last orange term should have a \(\large d\theta\), my bad.

So this will simplify to,\[\large \frac{2}{4}\int\limits \frac{\sec^2\theta d \theta}{1+\tan^2\theta}\]

Applying our Trig Identity to the denominator gives us,\[\large \frac{1}{2}\int\limits\limits \frac{\sec^2\theta\; d \theta}{\sec^2\theta} \qquad =\qquad \frac{1}{2}\int\limits d \theta\]

Answer 53

Trig Substitution can be a little tricky to get used to.
With U-substitutions, you're replacing a big chunk of stuff with a single variable.

With Trig Substitutions, you're replacing something with a `function` of a single variable. The function being the Trig function.

So it might seem a bit strange at first.

Answer 54

FINALLY integrating gives us,\[\large \frac{1}{2}\int\limits d \theta \qquad = \qquad \frac{1}{2}\theta+C\]

The final steps will be rewriting our answer in terms of \(\large x\).
To do so, we have to back-substitute into \(\large u\).
From there, back-substitute again into \(\large x\).

Answer 55

Alright, I had someone else solve the problem to equal (1/2)arctan(sec(x)/2) + C. Does this seem correct or no?

Answer 56

Yah that looks right :)
Too confusing I guess? Heh.

Answer 57

Oh boy. Now I have no idea where to go lol

Answer 58

Wanna see the last steps or not so much? :O Too much to take in?

Answer 59

I'm not sure which one is the correct answer now..

Answer 60

What do you think?

Answer 61

I don't understand what you're asking.
The answer you posted is correct.

I hadn't finish the problem yet...
Where are you seeing conflicting answers...?

Answer 62

Where did dθ come from in yours

Answer 63

Is that in place of du?

Answer 64

\(\large 2\tan\theta\) replaced \(\large u\).

\(\large 2\sec^2\theta \;d\theta\) replaced \(\large du\).

Answer 65

mkay i'm trying to get this all into this tiny space give me one moment to catch up

Answer 66

I don't get why she gave us this tiny space to fit this huge problem in, alright continue from 1/2∫ dθ= 1/2 θ+C

Answer 67

So we have to look back at the substitution we made, and try to solve it for \(\large \theta\).

\[\large u=2\tan \theta \qquad \rightarrow \qquad \frac{u}{2}=\tan \theta \qquad \rightarrow \qquad \color{orangered}{\arctan\left(\frac{u}{2}\right)=\theta}\]There, we've successfully solved for theta.

\[\large \frac{1}{2}\color{orangered}{\theta}+C \qquad = \qquad\frac{1}{2}\color{orangered}{\arctan\left(\frac{u}{2}\right)}+C\]

Answer 68

Now we need to go from \(\large u\) to \(\large x\).
Looking back at our substitution, \(\large \color{royalblue}{u=\sec x}\),

\[\large \frac{1}{2}\arctan\left(\frac{\color{royalblue}{u}}{2}\right)+C \qquad = \qquad \frac{1}{2}\arctan\left(\frac{\color{royalblue}{\sec x}}{2}\right)+C\]

Answer 69

If your teacher didn't give you much room, then she probably just wanted you to memorize the simple formula at this point :3 that would be my guess.

Answer 70

I know we weren't taught this. Augh. It's a lot to take in. You have no idea how much help you are.

Answer 71

If I could give more medals I would. thank you thank you thank you!

Answer 72

heh np :3