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

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.

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ranger
Gold Member
change from [text] to $$radou Homework Helper ...and change from [\text] to$$.

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

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?

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 $$\tan^2 x$$

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Doesn't $$\int \sec^{2}x\ dx=\int \tan x dx$$?
No. What is the derivative of tan(x)?

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

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

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
Homework Helper
Gold Member
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.]

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
Homework Helper
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
Homework Helper
$$\int \sec^{4}5x\ dx=$$
$$=\frac{1}{5}\int \frac{du}{\cos^{4}u}$$

where $u=5x$.

Now make the substitution

$$\tan u =y$$

Daniel.

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.

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

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

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.

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

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

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.

Yes I know how to integrate that. Im 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?

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Gib Z
Homework Helper
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.