Integral Calculus: Integrating (secx)^2 - Learn How!

[Q.E.D]

It has been a while since I have had to do an integration such as this, it is probably quite simple lol. But could someone show me how to integrate (secx)^2, that is secx all squared. I know that the answer is tanx but i was wondering if someone could show me the method.

Thank you.

 

[Defennder]

Easiest way to do so is to differentiate tan x = sin x/ cos x by quotient rule, thus avoiding all the integration.

 

[Q.E.D]

That is a valid point thank you, however, I am specifically looking for the integration solution.

 

[BoundByAxioms]

Hmm, well I haven't taken many college mathematics classes yet, but I suspect that by taking an analysis class one could prove why $$\int$$sec²(x) dx=tan(x)+c. I think of it like this: I know (and can show) that the derivative of sin(x) is cos(x). Therefore, the antiderivative of cos(x) is sin(x)+c. Likewise, since I know that the derivative of tan(x) is sec²(x), the integral of sec²(x) must be tan(x)+c.

 

[snipez90]

Well it's easy if we're allowed to use the fact that the derivative of tan(x) is $$sec^2(x)$$. Just try a u-sub or use the pythagorean identity and then u-sub. But otherwise, I don't have any clever ideas.

 

[NoMoreExams]

Hmm, well I haven't taken many college mathematics classes yet, but I suspect that by taking an analysis class one could prove why $$\int$$sec²(x) dx=tan(x)+c. I think of it like this: I know (and can show) that the derivative of sin(x) is cos(x). Therefore, the antiderivative of cos(x) is sin(x)+c. Likewise, since I know that the derivative of tan(x) is sec²(x), the integral of sec²(x) must be tan(x)+c.

The OP isn't questioning that you could "prove" it by taking derivative of tan(x) + C. He just wants to know how to do the integral if we did not know what $$sec^{2}(x)$$ integrates to.

 

[kmikias]

Ok the way i know is like this.
sec^2(x) = tan(x) then sec^2(x) dx = tan(x) + c

 

[NoMoreExams]

Ok the way i know is like this.
sec^2(x) = tan(x) then sec^2(x) dx = tan(x) + c

that made no sense.

 

[snipez90]

Ok the way i know is like this.
sec^2(x) = tan(x) then sec^2(x) dx = tan(x) + c

Errr what? $$sec^2(x) = tan^2(x) + 1$$

If you substitute u = sec(x) and then perform a subsequent trig substitution, you could find the integral provided that you're okay with using the fact that the derivative of $$tan(x)$$ is $$sec^2(x)$$.

I think you would be hard-pressed to find a solution that doesn't use the fact that the derivative of $$tan(x)$$ is $$sec^2(x)$$ and though you might gain something from integration via substitution, it's kind of overkill since most people usually learn derivatives of trig functions first.

 

[Count Iblis]

Hmmm, let's see. Let's use the shorthand notation:

s = sin(x)

c = cos(x)

We want to integrate 1/c^2, in a "canonical way", i.e. using the usual rules of manipulation we always use and not using any ad hoc rules that only apply in this case, like recognizing that the answer is tan(x) and proving that by differentiation.

We can simply proceed in a similar way as how we would derive the reduction formulas for integration of 1/c^n, c^n, etc. etc:

1/c^2 = (c^2 + s^2)/c^2 = 1 + s^2/c^2

We now integrate s^2/c^2 using partial integration:

We integrate the factor s/c^2. This yields:

s/c - Integral of [1/c times derivative of s] dx = s/c - x

So, the integral of 1/c^2 is s/c

Note that if you replace 1/c^2 by 1/c^n, the above method yields a relation between the integral of 1/c^n and 1/c^(n-2).

 

[BoundByAxioms]

The OP isn't questioning that you could "prove" it by taking derivative of tan(x) + C. He just wants to know how to do the integral if we did not know what $$sec^{2}(x)$$ integrates to.

I know, I wasn't trying to offer a proof though. I was just demonstrating my reasoning. Am I correct in saying that his question could be answered in some sort of real analysis class though?

 

[HallsofIvy]

It has been a while since I have had to do an integration such as this, it is probably quite simple lol. But could someone show me how to integrate (secx)^2, that is secx all squared. I know that the answer is tanx but i was wondering if someone could show me the method.

Thank you.

Easiest way to do so is to differentiate tan x = sin x/ cos x by quotient rule, thus avoiding all the integration.

That is a valid point thank you, however, I am specifically looking for the integration solution.

It's not clear to me why you post here if you refuse to take the advice given. Defennder told you exactly how to find the integral.

 

[Marin]

Here's what I tried but I cannot find my mistake:

$$\displaystyle{\int}\frac{1}{\cos^2x} dx=\displaystyle{\int}\frac{1}{(Re\{e^{ix}\})^2} dx=Re \{\displaystyle{\int}\frac{1}{e^{2ix}} dx\}=Re\{\displaystyle{\int}e^{-2ix} dx \}=Re\{-\frac{1}{2i}e^{-2ix}\}$$

$$Re\{-\frac{1}{2i}e^{-2ix}\}=Re\{\frac{2i}{4}[\cos(-2x)+i\sin(-2x)]\}=Re\{\frac{2i}{4}[\cos(2x)-i\sin(2x)]\}=\frac{\sin2x}{2}=\sin x\cos x+C$$

My result is most obviously incorrect. Can anyone help me find the error? Thanks

 

[Count Iblis]

You cannot take the Re operator out of the integral sign from the denominator.

 

[Marin]

Ok, I see, but if it were in the numerator? I suppose it would be possible then, wouldn't it?

 

[Count Iblis]

Ok, I see, but if it were in the numerator? I suppose it would be possible then, wouldn't it?

In general, it only works if the Re sign is in front of the entire expression. If the denominator is a real function, then you can pull a Re sign from the numerator and place it in front of the expression.

You should try to prove this (e.g. using that the integral of a sum of two terms is the sum of the integrals of the two terms)

 

[HallsofIvy]

Whatever happened to Q.E.D.? The first response didn't do the problem for him so he went off to sulk?

 

[Gib Z]

Let $t=\tan \left( \frac{x}{2} \right)$, after simplifying the original integral directly to $$2\int \frac{1+t^2}{ (1-t^2)^2 } dt$$ use partial fractions and its quite straight forward from there.

 

[Defennder]

There's one problem here: You need to be able to replace dx with dt. But to do so you need differentiate arctan. How would you differentiate arctan without using d/dx tan x = sec^2 x ?

 

[Count Iblis]

There's one problem here: You need to be able to replace dx with dt. But to do so you need differentiate arctan. How would you differentiate arctan without using d/dx tan x = sec^2 x ?

Indeed. So, I gave the only correct solution earlier on in this thread: use the same reduction formula you use when you want to integrate 1/cos^n(x).

 

[Defennder]

1/c^2 = (c^2 + s^2)/c^2 = 1 + s^2/c^2

We now integrate s^2/c^2 using partial integration:

We integrate the factor s/c^2. This yields:

s/c - Integral of [1/c times derivative of s] dx = s/c - x

So, the integral of 1/c^2 is s/c

I don't follow this step. How does this follow from the previous? Could you be more explicit? c here isn't just x, it's cos x.

 

[Count Iblis]

I don't follow this step. How does this follow from the previous? Could you be more explicit? c here isn't just x, it's cos x.

derivative of s equals c. So 1/c times the derivative of c equals 1. The minus x you get upon integration then cancels the x you get when you integrate the right hand side of the identity

s/c = 1 + s^2/c^2

 

[Defennder]

Huh? (1/c)(d/dx cos x) = -tan x, why does that equal 1? And sin^2x/cos^2x + 1 = sec^2 x, not sinx/cosx.

 

[Count Iblis]

derivative of s equals c. So 1/c times the derivative of c equals 1. The minus x you get upon integration then cancels the x you get when you integrate the right hand side of the identity

s/c = 1 + s^2/c^2

Typo:

1/c^2 = 1 + s^2/c^2

 

[Count Iblis]

Huh? (1/c)(d/dx cos x) = -tan x, why does that equal 1? And sin^2x/cos^2x + 1 = sec^2 x, not sinx/cosx.

I made another typo: The derivative of s times 1/c equals 1. Anyway, I think the original post by me is free of typos. That is the standard way to derive the reduction formula.

 

[Defennder]

Ok, well I see how it works now, but in the end you still had to make use of the quotient rule to differentiate 1/c = s/c^2. But then again no one said it's disallowed. Except that if it isn't, one could just as well differentiate s/c directly to get the answer.

 

[Count Iblis]

Ok, well I see how it works now, but in the end you still had to make use of the quotient rule to differentiate 1/c = s/c^2. But then again no one said it's disallowed. Except that if it isn't, one could just as well differentiate s/c directly to get the answer.

Well, I did the reverse, used that the integral of s/c^2 is 1/c. I intepret this as a special case of the general rule of the integral of s/c^n being equal to 1/[(n-1) c^(n-1)]. The point here is that you can immediately see what the integral is because the s is the derivative of c and that is sitting in the numerator. So, it is part of a big class of integrals.

This is not like differentiating s/c to get the answer, because that's not a constructive way to obtain the answer. In that case you think the answer is s/c and you prove that by differentiation.

I think that the OP asks for the integral being done using some simple general algorithm that works for a large class of integrals. This is actually how we always want more complex integrals to be done. If you have some complicated function and want to integrate that, then you wouldn't be satisfied if someone just wrote down the answer and then said: "Well, this is the answer, you can prove that it is the answer by differentiating it". You would want to know how to obtain that answer using some method/algorithm that preferably works for some not too small class of cases.

Of course, in simple cases, like this one, this is not necessary. Also students are far better off knowing that the integral of 1/cos^2(x) equals tan(x) than having to derive it. But I can understand that someone wants to know if you could mechanically apply some common integration rules to obtain that result automatically.

 

[Defennder]

Well, I did the reverse, used that the integral of s/c^2 is 1/c. I intepret this as a special case of the general rule of the integral of s/c^n being equal to 1/[(n-1) c^(n-1)]. The point here is that you can immediately see what the integral is because the s is the derivative of c and that is sitting in the numerator. So, it is part of a big class of integrals.

I was not aware of that "general rule". It is also not very evident, I should say, certainly much less apparent to me than d/dx t = 1/c^2. It really depends on what you are given to start off with.

EDIT: I'll put it this way: Students learn differentiation before integration. And when they learn to differentiate, they'll inevitably come to the trigo functions and learn d/dx t = 1/c^2 by the quotient rule. Which is why the question of how to integrate 1/c^2 is moot. This is just one of the fundamental techniques we learn with which we use to differentiate and integrate more complex functions.