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

[aFk-Al]

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

 

[ranger]

change from [text] to

 

[radou]

...and change from [\text] to [/tex].

Edit: boy, are we helpful.

 

[aFk-Al]

Ha there we go! Thank you. How about help with the problem now?

 

[d_leet]

I know that

\int [\sec^{2}5x]^{2}\ dx=
\int \tan^{2}5x\ dx =

This is wrong, these two lines are not equivalent. What trigonometric identity do you know that relates secant and tangent?

 

[aFk-Al]

Doesn't \int \sec^{2}x\ dx=\int \tan x dx?

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

 

[d_leet]

Doesn't \int \sec^{2}x\ dx=\int \tan x dx?

No. What is the derivative of tan(x)?

 

[aFk-Al]

No. What is the derivative of tan(x)?

\sec^2 x. roger. so what instead?

 

[d_leet]

Doesn't \int \sec^{2}x\ dx=\int \tan x dx?

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

Ok Starting with

\int \sec^{4} 5x dx

split this up to

\int (\sec^{2} 5x)(\sec^{2} 5x) dx

Then use an identity to simplify the first sec²(5x) and then it should be much simpler.

 

[robphy]

Use complex numbers via
\cos(x)=\frac{1}{2}(e^{ix}+e^{-ix}})
\sin(x)=\frac{1}{2i}(e^{ix}-e^{-ix}})
where i^2=-1
... and make use of
e^{ix}=\cos(x)+i\sin(x) (to convert back)

[Integrating the exponential function is much easier.]

 

[d_leet]

Use complex numbers via
\cos(x)=\frac{1}{2}(e^{ix}+e^{-ix}})
\sin(x)=\frac{1}{2i}(e^{ix}-e^{-ix}})
where i^2=-1
... and make use of
e^{ix}=\cos(x)+i\sin(x) (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]

I know that
\int \tan^{2}x dx= \int \sec^{2}x-1\ dx = \tan x - x + C
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
\tan^2 x = \sec^2 x -1
which is a trig identity derived from
\cos^2 x + \sin^2 x = 1
btw, knowing this is inconsequential to how you may solve
\int \sec^4 (5x)\;dx

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
\sec^2 x, \sec^2 x \tan^2 x

hope I haven't given away too much

 

[dextercioby]

\int \sec^{4}5x\ dx=

=\frac{1}{5}\int \frac{du}{\cos^{4}u}

where u=5x.

Now make the substitution

\tan u =y

Daniel.

 

[dickcruz]

I know that
\int \tan^{2}x dx= \int \sec^{2}x-1\ dx = \tan x - x + C
but i don't completely understand how this is derived. Because of this lack of comprehension, I have no idea what to do with \int \sec^{4}5x\ dx. I went from there to get:
\int \sec^{4}5x\ dx=
\int [\sec^{2}5x]^{2}\ dx=
\int \tan^{2}5x\ dx =
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 let's 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

 

[aFk-Al]

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

 

[d_leet]

The easiest way to go about this is

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

so let's 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.

 

[dickcruz]

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 don't 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

 

[aFk-Al]

thanks for the advice, i'll work on that.

 

[drpizza]

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.

 

[aFk-Al]

Yes I know how to integrate that. I am sorry but I'm not following where to go after you have split up \sec^4 5x to \sec^2 5x + \sec^2 5x * \tan^2 5x. I understand that you can now integrate each part seperately. I'm pretty sure you can use the power reducing formula to break down \sec^2 5x 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?

 

[Gib Z]

Dickcruz, please 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.