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Integrating Trig functions (Namely tan^{2}x)

  • Thread starter aFk-Al
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19
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I know that
[tex]\int \tan^{2}x dx= \int \sec^{2}x-1\ dx = \tan x - x + C[/tex]
but i don't completely understand how this is derived. Because of this lack of comprehension, I have no idea what to do with [tex]\int \sec^{4}5x\ dx[/tex]. I went from there to get:
[tex]\int \sec^{4}5x\ dx=[/tex]
[tex]\int [\sec^{2}5x]^{2}\ dx=[/tex]
[tex]\int \tan^{2}5x\ dx = [/tex]
but I have no idea where to go from here. Any help would be appreciated. Thanks in advance.
 
Last edited:

Answers and Replies

ranger
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change from [text] to [tex]
 
radou
Homework Helper
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...and change from [\text] to [/tex].

Edit: boy, are we helpful. :biggrin:
 
19
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Ha there we go! Thank you. How about help with the problem now?
 
1,073
1
I know that

[tex]\int [\sec^{2}5x]^{2}\ dx=[/tex]
[tex]\int \tan^{2}5x\ dx = [/tex]
This is wrong, these two lines are not equivalent. What trigonometric identity do you know that relates secant and tangent?
 
19
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Doesn't [tex]\int \sec^{2}x\ dx=\int \tan x dx[/tex]?

Edit: I answered the question as I was asking it. It equals [tex] \tan x[/tex] not [tex] \int \tan x dx[/tex]. So what should I do instead? I would also still like to know how to integrate [tex] \tan^2 x[/tex]
 
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1,073
1
Doesn't [tex]\int \sec^{2}x\ dx=\int \tan x dx[/tex]?
No. What is the derivative of tan(x)?
 
19
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No. What is the derivative of tan(x)?

[tex]\sec^2 x[/tex]. roger. so what instead?
 
1,073
1
Doesn't [tex]\int \sec^{2}x\ dx=\int \tan x dx[/tex]?

Edit: I answered the question as I was asking it. It equals [tex] \tan x[/tex] not [tex] \int \tan x dx[/tex]. So what should I do instead? I would also still like to know how to integrate [tex] \tan^2 x[/tex]
Ok Starting with

[tex] \int \sec^{4} 5x dx[/tex]

split this up to

[tex] \int (\sec^{2} 5x)(\sec^{2} 5x) dx[/tex]

Then use an identity to simplify the first sec2(5x) and then it should be much simpler.
 
robphy
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Use complex numbers via
[tex]\cos(x)=\frac{1}{2}(e^{ix}+e^{-ix}})[/tex]
[tex]\sin(x)=\frac{1}{2i}(e^{ix}-e^{-ix}})[/tex]
where [tex]i^2=-1[/tex]
... and make use of
[tex]e^{ix}=\cos(x)+i\sin(x)[/tex] (to convert back)

[Integrating the exponential function is much easier.]
 
1,073
1
Use complex numbers via
[tex]\cos(x)=\frac{1}{2}(e^{ix}+e^{-ix}})[/tex]
[tex]\sin(x)=\frac{1}{2i}(e^{ix}-e^{-ix}})[/tex]
where [tex]i^2=-1[/tex]
... and make use of
[tex]e^{ix}=\cos(x)+i\sin(x)[/tex] (to convert back)

[Integrating the exponential function is much easier.]
With sines and cosines this is easy, but with secants and tangents i think this would actually complicate things.
 
mjsd
Homework Helper
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I know that
[tex]\int \tan^{2}x dx= \int \sec^{2}x-1\ dx = \tan x - x + C[/tex]
but i don't completely understand how this is derived. Because of this lack of comprehension,
FYI, the only trick you have used here is the fact that
[tex]\tan^2 x = \sec^2 x -1[/tex]
which is a trig identity derived from
[tex]\cos^2 x + \sin^2 x = 1[/tex]
btw, knowing this is inconsequential to how you may solve
[tex]\int \sec^4 (5x)\;dx[/tex]

as previous posters mentioned, split them up and use the identity once...your aim is to rewrite the expression in terms of things you can readily integrate like
[tex]\sec^2 x, \sec^2 x \tan^2 x [/tex]

hope I haven't given away too much :smile:
 
dextercioby
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[tex]\int \sec^{4}5x\ dx=[/tex]
[tex] =\frac{1}{5}\int \frac{du}{\cos^{4}u} [/tex]

where [itex] u=5x [/itex].

Now make the substitution

[tex] \tan u =y [/tex]

Daniel.
 
8
0
I know that
[tex]\int \tan^{2}x dx= \int \sec^{2}x-1\ dx = \tan x - x + C[/tex]
but i don't completely understand how this is derived. Because of this lack of comprehension, I have no idea what to do with [tex]\int \sec^{4}5x\ dx[/tex]. I went from there to get:
[tex]\int \sec^{4}5x\ dx=[/tex]
[tex]\int [\sec^{2}5x]^{2}\ dx=[/tex]
[tex]\int \tan^{2}5x\ dx = [/tex]
but I have no idea where to go from here. Any help would be appreciated. Thanks in advance.
The easiest way to go about this is

sec^4 (5x) dx
let 5x = y (which you can differentialte and substitue accordingly).

so lets integrate
I= sec^4 x dx.
= (sec^2x . sec^2x) dx
= ((1+tan^2x) .sec^2x)dx
= Sec^2 dx + sec^2x.tan^2x dx
= tan x + c + J

J = sec^2x.tan^2x dx

Now let tan x =t
therefore, sec^2xdx =dt
substituting
J = t^2 dt
J= t^3 / 3 + c

I=tan x + c + (t^3 / 3) + c
tan x +(tan^3x)/3 + constant of indefinate integration.
 
ashrafmod
that is good
 
19
0
This is great guys, thank you all. I just need to memorize all those stupid trig identities and I think this will be a lot easier for me. Thanks again everyone!

-Al
 
1,073
1
The easiest way to go about this is

sec^4 (5x) dx
let 5x = y (which you can differentialte and substitue accordingly).

so lets integrate
I= sec^4 x dx.
= (sec^2x . sec^2x) dx
= ((1+tan^2x) .sec^2x)dx
= Sec^2 dx + sec^2x.tan^2x dx
= tan x + c + J

J = sec^2x.tan^2x dx

Now let tan x =t
therefore, sec^2xdx =dt
substituting
J = t^2 dt
J= t^3 / 3 + c

I=tan x + c + (t^3 / 3) + c
tan x +(tan^3x)/3 + constant of indefinate integration.

Please don't post complete solutions to homework questions.
 
8
0
This is great guys, thank you all. I just need to memorize all those stupid trig identities and I think this will be a lot easier for me. Thanks again everyone!

-Al
You dont have to memorize the identitites.
I derive them when i need them

the basic ones are

Sin^2 x + cos^2 x= 1
Cos2x = Cos^2 - sin^2x
sin 2 x = 2. cosx. sinx

from the first one we can divide throught by cos and get values for
tan^2x +1=sec^x.
dividing by sin gives
1+cot^2x = csc^2x.

from the second one
you can substitute sin^2x or cos^2x as 1-cos^2x or 1-sin^2x
this is how how you integrate
 
19
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thanks for the advice, i'll work on that.
 
286
0
to add to what dickcruz said, the cos2x formulas (there are 3) are very useful in integration:

take cos2x= 1 - 2sin^2x and solve for sin^2x
(or take cos2x = 2cos^2x - 1 and solve for cos^2x
This gives you the power reducing formulas which you'll use when you have even powers.
i.e. to integrate (cos^2 x)
you would first change it to the integral of (1 + cos2x)/2; a relatively simple integral to take care of.
 
19
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Yes I know how to integrate that. Im sorry but I'm not following where to go after you have split up [tex]\sec^4 5x[/tex] to [tex]\sec^2 5x + \sec^2 5x * \tan^2 5x[/tex]. I understand that you can now integrate each part seperately. I'm pretty sure you can use the power reducing formula to break down [tex] \sec^2 5x[/tex] to 1/cos. As for the other half, I know how to integrate tanx * secx but it's different with sec^2*tan^2. Do I break it down to sin^2/cos^4?
 
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Gib Z
Homework Helper
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Dickcruz, plz use tex, makes an easier read.

Basically with integrals with secants and tangents, we convert of the the factors into the other trig function using an identity. eg tan^3 x= (sec^2 x-1)(tan x), then we make a substituion.
 

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