Simplifying and Expanding Integration Using the Chain Rule

[stephen8686]

Homework Statement

∫ sec² (7-3θ) dθ

Homework Equations

∫ cos(kx+b)dx= ((sin(kx+b))/k)+C
∫ sin (kx+b)dx=-cos "
(k and b are constants)

The Attempt at a Solution

The answer in the book is: ∫ -1/3(tan(7-3x))+C. I assume by x they meant θ.
The book only gave the two formulas above for solving this, but my question is can these by simplified and expanded by saying ∫ f(g(x)) dx = F(g(x))/g'(x)? Because I looked online for a rule like this and found nothing. If this involves U-substitution or something of the sort, I couldn't find that in the textbook. Also, if you have any really good textbooks or sources for learning integration I don't like my textbook.

 

[SammyS]

Homework Statement

∫ sec² (7-3θ) dθ

Homework Equations

∫ cos(kx+b)dx= ((sin(kx+b))/k)+C
∫ sin (kx+b)dx=-cos "
(k and b are constants)

The Attempt at a Solution

The answer in the book is: ∫ -1/3(tan(7-3x))+C. I assume by x they meant θ.
The book only gave the two formulas above for solving this, but my question is can these by simplified and expanded by saying ∫ f(g(x)) dx = F(g(x))/g'(x)? Because I looked online for a rule like this and found nothing. If this involves U-substitution or something of the sort, I couldn't find that in the textbook. Also, if you have any really good textbooks or sources for learning integration I don't like my textbook.

What is the derivative of ##\ \tan(7-3\theta) \ ## ?

 

[SteamKing]

Homework Statement

∫ sec² (7-3θ) dθ

Homework Equations

∫ cos(kx+b)dx= ((sin(kx+b))/k)+C
∫ sin (kx+b)dx=-cos "
(k and b are constants)

The Attempt at a Solution

The answer in the book is: ∫ -1/3(tan(7-3x))+C. I assume by x they meant θ.
The book only gave the two formulas above for solving this, but my question is can these by simplified and expanded by saying ∫ f(g(x)) dx = F(g(x))/g'(x)? Because I looked online for a rule like this and found nothing. If this involves U-substitution or something of the sort, I couldn't find that in the textbook. Also, if you have any really good textbooks or sources for learning integration I don't like my textbook.

Instead of going off on wild goose chases and inventing spurious integration rules, refresh your knowledge of the derivatives of trig functions:

http://www.sosmath.com/calculus/diff/der03/der03.html

 

[MidgetDwarf]

Two ways to this problem.

You can prove the integral of the basic sec^2(x) or you can use the following fact. the derivative of tan(x) is sec^2(x). This should be in your book, right after they prove it.

 

[stephen8686]

∫ f(g(x)) dx = F(g(x))/g'(x)
In this case f(x)=sec^2 (x) and g(x)= 7-3x (or θ, doesn't matter)
So F(g(x))=tan(7-3x) because ∫sec^2 (x) dx = tan x
That would be divided by only g'. f is not in the denominator, it is just d/dx (7-3x) = -3. That is on the bottom. I don't use derivatives of trig functions

I'm just saying that this: ∫ f(g(x)) dx = F(g(x))/g'(x), gave me the right answer here, but I don't want to use it unless I know that it will always work when integrating composite functions.

 

[stephen8686]

I'm not trying to prove trig identities. I am taking ∫ sec^2(x)=tan x to be a given. The thing is that there is a function inside of the sec^2(x) function, which I have named g. Putting this function inside of the trig function creates a composite function which can't be integrated by only integrating the outer function (the trig identity).

 

[SteamKing]

I'm not trying to prove trig identities. I am taking ∫ sec^2(x)=tan x to be a given. The thing is that there is a function inside of the sec^2(x) function, which I have named g. Putting this function inside of the trig function creates a composite function which can't be integrated by only integrating the outer function (the trig identity).

You're making this too complicated.

The basic integral is ∫ sec²(u) du = tan (u) + C

Now, if your particular integral does not have a simple u as the argument of the secant function, you can use u-substitution to simplify things.

Your particular argument is u = (7 - 3x). You should be able to calculate du very easily, do a little algebra, and obtain the correct result.

 

[stephen8686]

thank you, and I found exactly what I was looking for on wikipedia under "leibniz's notation":

Would have made things so much simpler to put this in the textbook instead of arbitrary sin and cos integrals.

 

[Ray Vickson]

Homework Statement

∫ sec² (7-3θ) dθ

Homework Equations

∫ cos(kx+b)dx= ((sin(kx+b))/k)+C
∫ sin (kx+b)dx=-cos "
(k and b are constants)

The Attempt at a Solution

The answer in the book is: ∫ -1/3(tan(7-3x))+C. I assume by x they meant θ.
The book only gave the two formulas above for solving this, but my question is can these by simplified and expanded by saying ∫ f(g(x)) dx = F(g(x))/g'(x)? Because I looked online for a rule like this and found nothing. If this involves U-substitution or something of the sort, I couldn't find that in the textbook. Also, if you have any really good textbooks or sources for learning integration I don't like my textbook.

[tex] \int f(g(x) \, dx = \frac{F(g(x))}{g'(x)} \Longleftarrow \text{FALSE!} [/tex]
at least if you mean that ##F## is the antiderivative of ##f##. For one thing, differentiation of ##F(g(x))/g'(x)## does not give back ##f(g(x))##. For another thing it is easy to give counterexamples. For example, take ##f(y) = \sqrt{y}## and ##g(x) = x^2 + 1##., so that ##f(g(x)) = \sqrt{x^2+1}##. Then, we have ##F(y) = (2/3) y^{3/2}##, so
[tex] F(g(x))/g'(x) = \frac{1}{3x} (x^2+1)^{3/2}, [/tex]
but
[tex] \int \sqrt{x^2+1} \, dx = \frac{1}{2} x \sqrt{x^2+1} + \frac{1}{2} \, \text{arcsinh}(x) + \text{const.} [/tex]

You have already been advised to not try making up you own, personal, integration rules, because if you are not an expert you are almost guaranteed to get it wrong.

You say you do not like your textbook. Well, buy a cheap, used copy of another, or go to the library and check out an alternative. Or, do the modern equivalent of going to the library, and download one of the many free calculus textbooks available on-line.

 

[2nafish117]

Homework Statement

∫ sec² (7-3θ) dθ

Homework Equations

∫ cos(kx+b)dx= ((sin(kx+b))/k)+C
∫ sin (kx+b)dx=-cos "
(k and b are constants)

The Attempt at a Solution

The answer in the book is: ∫ -1/3(tan(7-3x))+C. I assume by x they meant θ.
The book only gave the two formulas above for solving this, but my question is can these by simplified and expanded by saying ∫ f(g(x)) dx = F(g(x))/g'(x)? Because I looked online for a rule like this and found nothing. If this involves U-substitution or something of the sort, I couldn't find that in the textbook. Also, if you have any really good textbooks or sources for learning integration I don't like my textbook.

you are kind of right
∫ f(g(x))dx = ∫ { f(u)du/g'(x) } taking g(x)=u; [note:g'(x)is still inside the integral to be evaluated]
now this integral is equal to F(g(x))/g'(x) (assuming ∫f(x)dx=F(x) ) only if g(x) is a linear function ie g(x)=ax+b a≠0;
now g'(x)=a (a constant which can be taken out of the integral woohoo!)
this gives us ∫ { f(u)du/g'(x) } =(1/a)*∫f(u)du= F(g(x))/g'(x) [i now remind you again that g must be a linear function only]
now try out this integral just for fun
∫x sin(3x^2)dx (hint take 3x^2 as some other variable t)

2nafish117

 

[Thewindyfan]

You could also use u-substitution to better simplify the integral for yourself before arriving at the answer.

 

[stephen8686]

you are kind of right
∫ f(g(x))dx = ∫ { f(u)du/g'(x) } taking g(x)=u; [note:g'(x)is still inside the integral to be evaluated]
now this integral is equal to F(g(x))/g'(x) (assuming ∫f(x)dx=F(x) ) only if g(x) is a linear function ie g(x)=ax+b a≠0;
now g'(x)=a (a constant which can be taken out of the integral woohoo!)
this gives us ∫ { f(u)du/g'(x) } =(1/a)*∫f(u)du= F(g(x))/g'(x) [i now remind you again that g must be a linear function only]
now try out this integral just for fun
∫x sin(3x^2)dx (hint take 3x^2 as some other variable t)

2nafish117

Thank you so much. This really helps.I took ∫x sin(3x^2)dx and got -cos(3x^2)/6+C. I checked this on an online integral calculator. Thanks again!