Two qn about integration (Understanding)

[qazxsw11111]

Hi everyone. I got two questions I want to ask.

1) I had just learning all the basic integration techniques (e.g. by parts, trigo etc) and just reading on my own about trigonometric substitution. I found this when I was having some difficulty integrating some basic trigo f(x) and was searching for some generic method when I stumbled upon this topic. I started reading but really couldn't understand it. From this passage, especially this part:

Trigonometric functions The six trigonometric functions of x may be expressed
in terms of cos x and sin x, so that the basic trigonometric polynomial integral takes
the form
R
sin^m x cosⁿ xdx. We can also allow m or n to be negative.
Case m odd We put u = cos x and du = -sin xdx and use sin2 x = 1 - u² on the
remaining even powers of sin x, to get a rational function of u.
Case n odd We put u = sin x and du = cos xdx and use cos2 x = 1 - u² on the
remaining even powers of cos x, to get a rational function of u.

From :http://www.math.jhu.edu/~jmb/note/methint.pdf

I just don't know the rationale for this. How can n be odd and then there is the "remaining even powers of cos x"? Sorry for being a newbie and all, but I really hope someone can explain the application. (or a simpler way of seeing it)

2)Also, on another different case,

in my lecture notes, integral of 1/(a ²-x²) is 1/2a ln abs ((a+x)/(a-x)) but integral of 1/ (x² -a²) is 1/2a ln abs ((x-a)/(x+a)). I thought it was just a matter of removing the minus sign out and putting it as a power of the integrated function which results in the numerator and denominator switching but this is clearly not the case here.

 

[smallphi]

1. When n is odd, you take ONE of the sine or cosine and together with dx it forms du. The rest is even powers of sine/cosine.

2. The two logarithms are indeed related by negative since their arguments are reciprocal:

abs ((a+x)/(a-x)) = 1 / abs ((x-a)/(x+a)) ( note that abs(x-a) = abs(a-x)) and there is a general formula

ln (something) = - ln (1/something)

 

[tiny-tim]

… Case m odd We put u = cos x and du = -sin xdx and use sin2 x = 1 - u² on the
remaining even powers of sin x, to get a rational function of u.
Case n odd We put u = sin x and du = cos xdx and use cos2 x = 1 - u² on the
remaining even powers of cos x, to get a rational function of u.
…
I just don't know the rationale for this. How can n be odd and then there is the "remaining even powers of cos x"? Sorry for being a newbie and all, but I really hope someone can explain the application. (or a simpler way of seeing it)

Hi qazxsw11111!

Suppose you have sin⁷xcosⁿx dx.

Using u = cos x and du = -sin xdx gives (1 - u²)³uⁿ du …

the left-hand part only has even powers of u (because one of the sinx's was absorbed into the du).

in my lecture notes, integral of 1/(a ²-x²) is 1/2a ln abs ((a+x)/(a-x)) but integral of 1/ (x² -a²) is 1/2a ln abs ((x-a)/(x+a)). I thought it was just a matter of removing the minus sign out and putting it as a power of the integrated function which results in the numerator and denominator switching but this is clearly not the case here.

erm … one answer is minus the other … log(p/q) = -log(q/p), and |x-a| = |a-x| … no problemo!

 

[qazxsw11111]

Ok...So let me repeat again what I think I understand. We always try to make one even power and one odd (since if it is both even we can use double angle formula)? Then apply the above formula? But one thing I don't understand is that how does the sinx gets absorbed into the dx?

Suppose you have sin7xcosnx dx.

Using u = cos x and du = -sin xdx gives (1 - u²)³un du …

the left-hand part only has even powers of u (because one of the sinx's was absorbed into the du).

From what I understand, the extra sine will be converted into du/dx dx which will 'times' together to become du. But since du/dx=-sinx, wouldn't there be a minus sign or is there a terrible misunderstanding on my part?

As for the second question, excellent responses! I just realized how blur I am lolx.

Thanks guys for your help!

 

[tiny-tim]

Ok...So let me repeat again what I think I understand. We always try to make one even power and one odd (since if it is both even we can use double angle formula)? Then apply the above formula?

Yes … if one is odd, we can use this u substitution … if both are even, we use the double angle formula until every term has one odd.

But one thing I don't understand is that how does the sinx gets absorbed into the dx?

From what I understand, the extra sine will be converted into du/dx dx which will 'times' together to become du. But since du/dx=-sinx, wouldn't there be a minus sign or is there a terrible misunderstanding on my part?

oops! yes, I forgot the minus!

 

[qazxsw11111]

Thanks for all your replies. I read on further and understood the trigo subs. But it mentions "If n is negative, the substitution u = tan x, du = sec2 xdx can be useful." What is the purpose of this? When is this applicable since I convert all to sin and cos? And what should I do if there is "For integrals of the form sinmxsin nxdx" (from the text)

At the double angled part I stumbled again. For example, if I have sin⁶x cos ⁸, then even I use double angle, it would result in a product of the two double angle formula with powers. Do I need to use substitution in this case? Since expanding them out would probably not do any good as it will result in sometiming like cos ⁿ2x. What is the easiest take? Thanks.

 

[tiny-tim]

… "If n is negative, the substitution u = tan x, du = sec2 xdx can be useful." What is the purpose of this? When is this applicable since I convert all to sin and cos?

For example, if you have n = -(m+2), then it's tan^mx sec²x, which is d/dx of tan^m+1x/(m+1) …

you can always use the previous method … but sometimes using sec and tan is obviously quicker!

And what should I do if there is "For integrals of the form sinmxsin nxdx" (from the text)

2sinmxsin nx = cos(m-n)x - cos(m+n)x …

you must memorise all your trigonometric identities !

At the double angled part I stumbled again. For example, if I have sin⁶x cos ⁸, then even I use double angle, it would result in a product of the two double angle formula with powers. Do I need to use substitution in this case? Since expanding them out would probably not do any good as it will result in sometiming like cos ⁿ2x. What is the easiest take?

That becomes (1 - cos2x)³(1 + cos2x)⁴/128 …

the odd powers of cos2x are then easy, and for the even powers, you have to use the double angle formula again.

 

[atyy]

2sinmxsin nx = cos(m-n)x - cos(m+n)x …

you must memorise all your trigonometric identities !

 

[qazxsw11111]

That becomes (1 - cos2x)3(1 + cos2x)4 /128 …

the odd powers of cos2x are then easy, and for the even powers, you have to use the double angle formula again.

Yes. I did until that step. Then I couldn't integrate it further as they are products. For products I only know the formula for f(x)^n f'(x) and by parts.

 

[tiny-tim]

Yes. I did until that step. Then I couldn't integrate it further as they are products. For products I only know the formula for f(x)^n f'(x) and by parts.

cos⁷2x is easy (7 is odd).

cos⁶2x = (1 + cos4x)³/8 …

which gives you a cos4x and a cos³4x, which are easy, and a (1 + cos8x).

 

[qazxsw11111]

cos⁷2x is easy (7 is odd).

cos⁶2x = (1 + cos4x)³/8 …

which gives you a cos4x and a cos³4x, which are easy, and a (1 + cos8x).

Im sorry, but how do you get these? Did you expand them out or something?

Thanks.

 

[tiny-tim]

cos²x = (1 + cos2x)/2

sin²x = (1 - cos2x)/2

learn them!

 

[qazxsw11111]

cos²x = (1 + cos2x)/2

sin²x = (1 - cos2x)/2

learn them!

Yeah, I know that. But what has it got to do with (1 - cos2x)³(1 + cos2x)⁴/128 to become cos⁷2x and cos ⁶2x

 

[tiny-tim]

'cos …

Yeah, I know that. But what has it got to do with (1 - cos2x)³(1 + cos2x)⁴/128 to become cos⁷2x and cos ⁶2x

'cos (1 - cos2x)³(1 + cos2x)⁴ has all the powers of cos2x up to the seventh,

and 'cos even powers of cos2x (or sin2x) can be written in terms of cos4x, even powers of cos4x can be written in terms of cos8x, etc.

 

[qazxsw11111]

You mean I need to use binomial to expand them out? Thats pretty tedious.

 

[tiny-tim]

You mean I need to use binomial to expand them out? Thats pretty tedious.

Yes, but you only need the x term … in the limit as x –> ∞, you can ignore x² terms and higher.

 

[qazxsw11111]

Yes, but you only need the x term … in the limit as x –> ∞, you can ignore x² terms and higher.

Im sorry. I don't get you. I don't see any x² terms when I expand them out. I can only see the expansion as a string of numbers with cos ⁿ 2x (with quite a number of terms by the way).

 

[tiny-tim]

oops!

Im sorry. I don't get you. I don't see any x² terms when I expand them out. I can only see the expansion as a string of numbers with cos ⁿ 2x (with quite a number of terms by the way).

oops! I replied to the wrong thread …

someone else was asking about the binomial theorem also …

please ignore last post!

sensible response …

You mean I need to use binomial to expand them out? Thats pretty tedious.

Yes, I agree!

Sometimes you just have to slog away …

 

[qazxsw11111]

Arg... Thats wickedly tiring! Is there any other method?

 

[HallsofIvy]

Ok...So let me repeat again what I think I understand. We always try to make one even power and one odd (since if it is both even we can use double angle formula)? Then apply the above formula? But one thing I don't understand is that how does the sinx gets absorbed into the dx?

It is not a matter of "making" one power odd. If either is odd, then you can factor out a single sine or cosine, leaving an even power so that you can use either sin²ⁿ(x)= (sin²(x))ⁿ= (1- cos²)ⁿ or cos²ⁿ(x)= (cos²(x))ⁿ= (1- sin²(x))ⁿ. In the first case, since you already have the "sin(x)dx" from the one you factored out, you can let u= cos(x) and have (1- u²)du, with, of course, additional powers of u from powers of cos(x).

But you cannot make a power of sine or cosine odd. Either it is or it isn't to start with. If you have an integral with only even powers of sine and cosine, say $\int sin^2(x)cos^2(x)dx$, you cannot use that method. In that case, you can use the "double angle formulas": sin²(x)= (1/2)(1- cos(2x)) and cos²(x)= (1/2)(1+ cos(2x)).

From what I understand, the extra sine will be converted into du/dx dx which will 'times' together to become du. But since du/dx=-sinx, wouldn't there be a minus sign or is there a terrible misunderstanding on my part?

As for the second question, excellent responses! I just realized how blur I am lolx.

Thanks guys for your help!

 

[qazxsw11111]

But you cannot make a power of sine or cosine odd. Either it is or it isn't to start with. If you have an integral with only even powers of sine and cosine, say $\int sin^2(x)cos^2(x)dx$, you cannot use that method. In that case, you can use the "double angle formulas": sin²(x)= (1/2)(1- cos(2x)) and cos²(x)= (1/2)(1+ cos(2x)).

But look at the question above. If it has a high power such as 6 or 8, expanding them out is absolutely crazy. Any althernative methods?