Integrating tan^3 x: Tips and Tricks for Solving ∫tan3x dx

[Cali920]

Homework Statement

∫tan³x dx


2. The attempt at a solution
∫tan x + ∫tan²x

∫tan x (sec²x - 1) dx

∫(tan x (sec²x - tan x) dx

∫tan x sec²x dx - ∫tan x dx

u = sec x
du = sec x tan x dx

∫tan x sec x sec x - ∫tan x dx

Now I'm stuck..

∫ du * u - ∫ tan x dx ?

 

[Panphobia]

For the second part, tan(x) = sin(x)/cos(x), and a simple u substitution should be fine for that.

 

[Cali920]

For the second part, tan(x) = sin(x)/cos(x), and a simple u substitution should be fine for that.

so ∫ du * u - (sin x /cos x)? What do I do with the du * u? I'm confused...

Thanks for the response! I appreciate it!

 

[Panphobia]

it is ∫ u * du- ∫(sin x /cos x)dx and maybe you should learn the basics of integration before you even use substitution. The integral of x is (x^2)/2, so ∫ u * du = (u^2)/2.

 

[HallsofIvy]

I presume it was a typo, but of course, "\int tan^3(x) dx" is NOT "\int tan(x) dx+ \int tan(x) dx".

Personally, I would have written this integral as \int \frac{sin^3(x)}{cos^3(x)} dx and used the standard "odd power of sine or cosine" technique: factor out one of the sines to use with the dx:
\int \frac{sin^2(x)}{cos^3(x)} sin(x)dx= \int \frac{1- cos^2(x)}{cos^3(x)} (sin(x)dx)
and let u= cos(x), du= - sin(x)dx, to get
\int \frac{1- u^2}{u^3} (- du)= -\int (u^{-3}- u^{-1})du= \int (u^{-1}- u^{-3})du

 

[Mark44]

Homework Statement

∫tan³x dx2. The attempt at a solution
∫tan x + ∫tan²x

I presume it was a typo, but of course, "\int tan^3(x) dx" is NOT "\int tan(x) dx+ \int tan(x) dx".

Just to set the record straight, ∫tan³x dx ≠ ∫tan x dx + ∫tan^2[/color]x dx
I think that's what HallsOfIvy was attempting to say, but omitted the exponent on the second integral.

 

[STEMucator]

If you prefer:

$\int tan^3(x) dx = \int (sec^2(x) - 1) tanx dx = \int sec^2(x) tan(x) dx - \int tan(x) dx$

Both the latter integrals can be integrated quite easily. For the integral on the left, set $u = sec(x) \Rightarrow du = sec(x)tan(x)dx$.

 

[Panphobia]

Yea, I assumed that was a typo since the rest of the work was correct.

 

[Cali920]

AH! I didn't even realize I did that.

= ∫ tan x * ∫ tan²x
= ∫ tan x (sec²x - 1) dx
= ∫ (tan x sec²x) dx - ∫ (tan x) dx
u = tan x
du = sec²x dx

= ∫ u * du - ∫ u

Is this correct so far?Or can you do two u substitutions?:
AH! I didn't even realize I did that.

= ∫ tan x * ∫ tan²x
= ∫ tan x (sec²x - 1) dx
= ∫ (tan x sec²x) dx - ∫ (tan x) dx
u = tan x & u = cos x
du = sec²x dx & du = -sin x dx?

 

[Mark44]

AH! I didn't even realize I did that.

= ∫ tan x * ∫ tan²x

Right off the bat you have a mistake. ∫tan³(x)dx ≠ ∫ tan x dx * ∫ tan²x dx

= ∫ tan x (sec²x - 1) dx
= ∫ (tan x sec²x) dx - ∫ (tan x) dx
u = tan x
du = sec²x dx

= ∫ u * du - ∫ u

Is this correct so far?

The first integral looks OK, but not the second. It's missing the du factor. In addition, the same substitution you're using in the first integral won't work in the second.

 

[STEMucator]

I believe the OP is confused about the integral operation commuting over multiplication.

Please note:

$\int f(x) g(x) dx ≠ \int f(x) dx \int g(x) dx$

So for example $\int tan^3(x) dx = \int tan^2(x)tan(x) dx ≠ \int tan^2(x) dx \int tan(x) dx$

 

[lurflurf]

you need sec(x)^2 as a factor

$$\tan^3(x)=\frac{\tan^3(x)}{\tan^2(x)+1}\sec^2(x)=\left( \tan(x) - \frac{\tan(x)}{\tan^2(x)+1} \right) \sec^2(x)$$

 

[Cali920]

I believe the OP is confused about the integral operation commuting over multiplication.

Please note:

$\int f(x) g(x) dx ≠ \int f(x) dx \int g(x) dx$

So for example $\int tan^3(x) dx = \int tan^2(x)tan(x) dx ≠ \int tan^2(x) dx \int tan(x) dx$

Oh okay. So then what can I do if I can't do ∫ tan x dx * ∫ tan²x dx?

 

[STEMucator]

Oh okay. So then what can I do if I can't do ∫ tan x dx * ∫ tan²x dx?

Look at post #7, I already gave you most of the work really.

Just note that $sec^2(x) - tan^2(x) = 1$. Much like $sin^2(x) + cos^2(x) = 1$. If you want to get your hands dirty, $csc^2(x) - cot^2(x) = 1$ can also be useful for these trig integrals.

 

[Cali920]

If you prefer:

$\int tan^3(x) dx = \int (sec^2(x) - 1) tanx dx = \int sec^2(x) tan(x) dx - \int tan(x) dx$

Both the latter integrals can be integrated quite easily. For the integral on the left, set $u = sec(x) \Rightarrow du = sec(x)tan(x)dx$.

Look at post #7, I already gave you most of the work really.

Just note that $sec^2(x) - tan^2(x) = 1$. Much like $sin^2(x) + cos^2(x) = 1$. If you want to get your hands dirty, $csc^2(x) - cot^2(x) = 1$ can also be useful for these trig integrals.

I think I started off doing that originally then got stuck which is why I tried a different u substitution but ->(from my first post):

=∫tan x (sec²x - 1) dx

∫(tan x (sec²x - tan x) dx

∫tan x sec²x dx - ∫tan x dx

u = sec x
du = sec x tan x dx

∫tan x sec x sec x - ∫tan x dx

∫udu - ∫tan x dx ?

 

[STEMucator]

I think I started off doing that originally then got stuck (from my first post):

=∫tan x (sec²x - 1) dx

∫(tan x (sec²x - tan x) dx

∫tan x sec²x dx - ∫tan x dx

u = sec x
du = sec x tan x dx

∫tan x sec x sec x - ∫tan x dx

∫udu - ∫tan x dx ?

Watch the second line there. You have an extra bracket you don't need.

So far that looks fine. I'm sure you can integrate udu, the second integral can be easily solved by letting $tan(x) = \frac{sin(x)}{cos(x)}$ and making a simple substitution.

 

[Mark44]

Oh okay. So then what can I do if I can't do ∫ tan x dx * ∫ tan²x dx?

Continue what you were doing earlier.

AH! I didn't even realize I did that.
∫ tan³ x dx[/color]
[STRIKE]= ∫ tan x * ∫ tan²x
[/STRIKE]= ∫ tan x (sec²x - 1) dx
= ∫ (tan x sec²x) dx - ∫ (tan x) dx
u = tan x
du = sec²x dx

= ∫ u * du - ∫ u

I've already replied about the first integral being OK, but the second one is not OK.

 

[Cali920]

Watch the second line there. You have an extra bracket you don't need.

So far that looks fine. I'm sure you can integrate udu, the second integral can be easily solved by letting $tan(x) = \frac{sin(x)}{cos(x)}$ and making a simple substitution.

whoops! THANK YOU!

=∫tan x (sec2x - 1) dx

∫(tan x sec2x - tan x) dx

∫tan x sec2x dx - ∫tan x dx

u = sec x
du = sec x tan x dx

∫tan x sec x sec x - ∫tan x dx

∫udu - ∫tan x dx ?

= 1/2 u^2 + C -∫ (sin x / cos x) dx?

so u = cos x?

 

[STEMucator]

whoops! THANK YOU!

=∫tan x (sec2x - 1) dx

∫(tan x sec2x - tan x) dx

∫tan x sec2x dx - ∫tan x dx

u = sec x
du = sec x tan x dx

∫tan x sec x sec x - ∫tan x dx

∫udu - ∫tan x dx ?

= 1/2 u^2 + C -∫ sin x / cos x dx?

so u = cos x?

While I'm sure you meant $sec^2(x)$ on those lines, you should make it explicit or someone might confuse it for $sec(2x)$.

Yes $u = cos(x)$ so that $-du = sin(x) dx$.

 

[Cali920]

While I'm sure you meant $sec^2(x)$ on those lines, you should make it explicit or someone might confuse it for $sec(2x)$.

Yes $u = cos(x)$ so that $-du = sin(x) dx$.

Just didn't copy and paste right. But thanks again! I really appreciate it.

=∫tan x (sec²x - 1) dx

∫(tan x sec²x - tan x) dx

∫tan x sec²x dx - ∫tan x dx

u = sec x
du = sec x tan x dx

∫tan x sec x sec x - ∫tan x dx

∫udu - ∫tan x dx ?

= u²/2 + C -∫ (sin x / cos x) dx

u = cos x
du = - sin x dx → - du = sin x dx

so then u²/2 + C + du/u?

 

[STEMucator]

Just didn't copy and paste right. But thanks again! I really appreciate it.

=∫tan x (sec²x - 1) dx

∫(tan x sec²x - tan x) dx

∫tan x sec²x dx - ∫tan x dx

u = sec x
du = sec x tan x dx

∫tan x sec x sec x - ∫tan x dx

∫udu - ∫tan x dx ?

= u²/2 + C -∫ (sin x / cos x) dx

u = cos x
du = - sin x dx → - du = sin x dx

so then u²/2 + C + du/u?

Close, but watch your negative sign, it doesn't just 'disappear' because you integrated. The integrand itself looks fine though.

 

[Cali920]

Close, but watch your negative sign, it doesn't just 'disappear' because you integrated. The integrand itself looks fine though.

I made it positive I thought? because it was - - du/u?

 

[Mark44]

so then u²/2 + C + du/u?

How did ∫du/u become just du/u?

The integral sign shouldn't just disappear.

 

[STEMucator]

I made it positive I thought? because it was - - du/u?

Looks like you knew the answer already then, I should've read that more carefully. I just knew that $\int tan(x) dx = - ln|cos(x)| + c = ln|sec(x)| + c$. So I jumped a bit there by accident and forgot the negative myself.

 

[Cali920]

How did ∫du/u become just du/u?

The integral sign shouldn't just disappear.

Looks like you knew the answer already then, I should've read that more carefully. I just knew that $\int tan(x) dx = - ln|cos(x)| + c = ln|sec(x)| + c$. So I jumped a bit there by accident and forgot the negative myself.

That's okay! Thank you! & I forgot to put the sign in there! THANK YOU Mark44!

u = cos x
du = - sin x dx → - du = sin x dx

so then u²/2 + C + ∫ du/u

Okay, but I'm not sure where to go from here or how to integrate du/u?

 

[lurflurf]

$$\int \! \frac{\mathrm{d}u}{u}=\log(|u|)+\mathrm{Constant}$$

 

[Cali920]

$$\int \! \frac{\mathrm{d}u}{u}=\log(|u|)+\mathrm{Constant}$$

Thank you! Is that a rule?

u = cos x
du = - sin x dx → - du = sin x dx

= u²/2 + C + ∫ du/u
= u²/2 + C + log (|u|) + C

then plug back in u?

= u²/2 + C + log (|cos x|) + C?

 

[Cali920]

wait I think there is a sign mishap somewhere if we put the integral sign back in...oh no.

 

[Cali920]

u = cos x
du = - sin x dx → -du = sin x dx

= u²/2 + C - ∫ -du/u
= u²/2 + C - (- [log (|u|) + C])?

 

[STEMucator]

wait I think there is a sign mishap somewhere if we put the integral sign back in...oh no.

No that looks fine. Except you have another $u$ sitting around that you need to sub back in for. You can also combine both constant terms into one constant term. Call it $C$.

Yes it is also a rule that $\int \frac{1}{x} dx = ln|x|$.

$\int x^n dx =$

$\frac{x^{n+1}}{n+1}$ if $n ≠ -1$

$\ln|x|$ if $n = -1$