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# Best way to integrate ∫ (tan x)^2 (sec x)^3 dx ?

Thu 10 May, 2012 09:24 am
Hi.

Any ideas on the best way to integrate ∫ (tan x)^2 (sec x)^3 dx ?

(This is not homework.)

Cheers,
[The guy formerly known as "Oylok"]
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Type: Question • Score: 2 • Views: 32,512 • Replies: 13

View best answer, chosen by Kolyo
Tifinden

1
Thu 10 May, 2012 10:54 am
@Kolyo,
Dimensional Analysis of the axiomatic spectrum in which you function and then apply.
0 Replies

engineer

1
Thu 10 May, 2012 05:42 pm
@Kolyo,
When it's not homework I go to Wolfram.
raprap

1
Thu 10 May, 2012 05:59 pm
@Kolyo,
Try substitutinnd the identity (tanx)^2=(sec x)^2-1

Rap
engineer

1
Thu 10 May, 2012 10:29 pm
@engineer, Kolyo

1
Fri 11 May, 2012 09:27 am
@engineer,
Thanks,

I'll keep that in mind.

Still, I don't like the form of their answer:

Code:```Integrate[Tan[x]^2*Sec[x]^3, x] == (4*Log[Cos[x/2] - Sin[x/2]] - 4*Log[Cos[x/2] + Sin[x/2]] - Sec[x]^4*Sin[3*x] + 7*Sec[x]^3*Tan[x])/32 ```

I was hoping for an answer that was a linear combination of "sec^3 tan", "sec tan", and "ln |sec + tan|".
0 Replies

Kolyo

1
Fri 11 May, 2012 09:35 am
@raprap,
raprap wrote:

Try substitutinnd the identity (tanx)^2=(sec x)^2-1

Rap

Yes, that gives you (sec x)^5 - (sec x)^3 in the integrand.

At that point you have to use a reduction formula to change the integral of (sec x)^5 into an integral of (sec x)^3. Then you use the reduction formula again to reduce the integral of (sec x)^3 to an integral of (sec x), which you can then integrate.

I'm sorry: my original question was apparently a bit misleading, since it apparently gave the impression that I had no idea how to integrate this. I actually know the most popular (standard) method for integrating this. The reason I asked was really to find out whether anyone knew about the new method that a friend of mine invented.
uvosky Selected Answer

2
Fri 14 Sep, 2012 02:28 am
@Kolyo,
I would proceed by thus , let y = (sec x)^2
then dy = 2 secx secx tanx dx = 2 ( sec x)^2 tanx dx
so, 2 (tanx)^2 (sec x)^2 dx = secx tanx dy = (y(y-1) )^(1/2) dy , since
secx = y^(1/2) and tany = ( (secx)^2 - 1)^(1/2) = (y - 1)^(1/2) (by considering
positive square roots only ). Thus the substitution y = (sec x)^2 yields
2 ∫ ( tanx)^2 (secx)^3 dx = ∫ (y(y - 1) )^(1/2) dy , and this later form can be integrated easily by Euler substitution.

I am much eager to know what method your friend has invented.

Kolyo

1
Sat 29 Sep, 2012 01:55 am
@uvosky,
Very creative! The ribbon is yours!

My friend actually came up with a technique that works on everything of the form ∫ (tan x)^m (sec x)^n dx, where m is a non-negative integer and n is positive integer. (According to most textbooks for novices that we've read, there's no general approach, but he found one)

I'm a bit busy with work and the US elections right now, but I'll get back to you in November.
JohnFelgate

1
Sat 29 Sep, 2012 03:51 am
@Kolyo,
It may be different from your friend that want to invent so i excited to know about that
0 Replies

uvosky

1
Fri 5 Oct, 2012 02:42 am
@Kolyo,
I was out of touch from the forum for some time so noticed your reply lately; thanks very much for the ribbon , and about the general integral you mentioned I have some simple results for some of the special cases.
Letting y=secx we get dy= secx tanx and since obviously y^2 - 1=(tanx)^2 we get ∫ (secx)^(n+1) (tanx)^(2m+1) dx = ∫y^n (y^2 - 1)^m dy ; and this right hand side can be expanded in binomial of finite terms , since m is positive integer, and then be integrated term by term to get
(-1)^m y^(n+1) ∑ (mCk) (-1)^k (y^2k/(2k+n+1)) , where the summation is over k(from 0 to m).
Again by letting y=tanx we get dy= (secx)^2 dx and using 1+ y^2 = (secx)^2 we get ∫(secx)^(2n+2) (tanx)^m dx =∫(1+ y^2)^n y^m dy , and then as before we get y^(m+1) ∑ (nCk) (y^2k/(2k+m+1)) , where the summation is over k(from 0 to n) . So , we see that in variety of powers of tangent and secant we have completely considered
|tan: even ; sec: even | tan: odd ; sec : odd| tan : odd ; sec : even| so the only possible type of variety left is tan: even ; sec : odd , i.e. integrals of the form ∫(tanx)^(2m) (secx)^(2n+1) dx , this is the most difficult to handle and even though the original integral you posed is of this form and I successfully attacked it , the method can not be extended to get an answer in finite form and in elementary functions , for the general case . So , I"l be looking forward , eagerly, for your post on the general type of integral.
0 Replies

Kolyo

3
Wed 21 Nov, 2012 11:32 pm
@Kolyo,
So how do we integrate ∫ (tan x)^m (sec x)^n dx, where m >=0 and n > 0 are integers?

Preliminaries...

First, introduce two new functions:
u(x) = sec(x) + tan(x);
v(x) = sec(x) - tan(x).

Second, note that:
sec = (u + v) / 2, and
tan = (u - v) / 2.

Third, note that u(x) * v(x) = 1,
because
uv = (sec + tan)(sec - tan) = sec^2 - tan^2 = 1 = (tan^2 + 1) - tan^2 = 1.

Fourth, investigate properties of integration and differentiation:

(i)
u' = (sec + tan)' = sec' + tan' = sec*tan + sec^2 = (tan + sec) * sec = u * sec
v' = (sec - tan)' = sec' - tan' = sec*tan - sec^2 = (tan - sec) * sec = - (sec - tan) sec = - v * sec

(ii)
If n is an integer other than zero,
then
∫ sec(x) * u(x)^n dx
= ∫ u(x)^(n-1) * sec(x) * u(x) dx
= ∫ u(x)^(n-1) * u'(x) dx
= ∫ U^(n-1) * dU [where U = u(x)]
= 1/n * U^n + C = u(x)^n / n + C.

(iii)
If n is a positive integer,
then
∫ sec(x) * v(x)^n dx
= ∫ sec(x) * u(x)^(-n) dx [because v = 1/u]
= u(x)^(-n) / (-n) + C [because of (ii)]
= - v(x)^n / n + C.

(iv)
∫ sec(x) dx = ∫ u*sec(x)/u dx = ∫ u'/u dx = ∫ du/u = ln |u| + C

Main argument...

We can use all that to solve an integration problem like ∫ (tan x)^2 (sec x)^5 dx.

∫ (tan x)^2 (sec x)^5 dx
= ∫ sec(x) (tan x)^2 (sec x)^4 dx
= ∫ sec(x) [(u - v)/2]^2 [(u + v)/2]^4 dx
= 1/64 * ∫ sec(x) (u - v)^2 (u + v)^4 dx
= 1/64 * ∫ sec(x) (u^2 - 2uv + v^2) (u^4 + 4u^3 v + 6u^2 v^2 + 4uv^3 + v^4) dx
= 1/64 * ∫ sec(x) (u^2 - 2 + v^2) (u^4 + 4u^2 + 6 + 4v^2 + v^4) dx [using uv = 1]
= 1/64 * ∫ sec(x) (u^6 + 2u^4 - u^2 - 4 - v^2 + 2v^4 + v^6) dx [multiplying polynomials through and using uv = 1]
= 1/64 * [(1/6) u^6 + (1/2) u^4 - (1/2) u^2 - 4 ln |u| + (1/2) v^2 - (1/2) v^4 - (1/6) v^6] + C [using (ii), (iii), (iv) above]
= (1/384) u^6 + (1/128) u^4 - (1/128) u^2 - (1/16) ln |u| + (1/128) v^2 - (1/128) v^4 - (1/384) v^6 + C

I think that's done right, but I don't feel like checking the details. In any case, it takes a bit of work if you want to put it in a more familiar form, but at least the integration is done, and this approach works for integrating the general form of ∫ (tan x)^m (sec x)^n dx, where m >=0 and n > 0 are both integers.

Footnote:

Those same new functions, u(x) and v(x), can be used to perform a simpler, more intuitive version of the WeierstraB substitution.

Suppose you have any rational function of sin(x) and cos(x) that you want to integrate: F[sin x, cos x].

∫ F[sin x, cos x] dx
= ∫ F[tan/sec, 1/sec] dx
= ∫ F[(u - v)/(u + v), 2/(u+v)] (u*sec) dx / (u*sec)
= ∫ F[ (u^2 - 1) / (u^2 + 1), 2u / (u^2 + 1)] du / [u(u+v)/2]
= ∫ F[ (u^2 - 1) / (u^2 + 1), 2u / (u^2 + 1)] 2 du / (u^2 + 1)

So it reduces the integrand to a rational function in one variable just like the normal Weierstrass substitution does.

That's about all u(x) and v(x) do, as far we know...
uvosky

1
Thu 22 Nov, 2012 03:35 am
@Kolyo,
That's an wonderful method ! Nicely done , nicely done.
0 Replies

mark159159555

1
Wed 13 Jan, 2016 11:58 am
@Kolyo,
What an excellent method! You make those tedious integral clear and direct to calculate. Good job bro~
0 Replies

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