## Friday, April 21, 2017

### Unizor - Indefinite Integrals - Problems 2

Notes to a video lecture on http://www.unizor.com

Indefinite Integral -
Problems 2

IMPORTANT:
All these problems require certain guessing based on known rules of integration (substitution and "by parts") and recollection of derivatives of known functions.
These problems are illustration of a notion of integration as a form of art more than a skill.

Problem 2.1:

d/ (cos(x)−sin(x))

One way to approach this problem was offered in a previous lecture and was based on the identity
cos(x)−sin(x) =
= (cos(x)·cos(π/4) −
− sin(x)·sin(π/4)) / (√2/2) =
= √2·cos(x+π/4)

The result of the integration using this method was
(√2/4)·[ln(1+sin(x+π/4)) −
− ln(1−sin(x+π/4))
] + C

Let's consider a different approach to "rationalize" this problem.
Recall that all trigonometric functions can be expressed as rational functions of a tangent of a half-angle.
In particular,
sin(x) =
= 2·tan(x/2) / (1+tan²(x/2))

cos(x) =
= (1−tan²(x/2)) / (1+tan²(x/2))

From these identities we derive
cos(x) − sin(x) =
= (1−tan²(x/2)−2·tan(x/2)) /
/ (1+tan²(x/2))

and
dx / (cos(x) − sin(x)) =
= (1+tan²(x/2))
dx /
/ (1−tan²(x/2)−2·tan(x/2))

Derivative of tangent can also be expressed in terms of tangent:
tanI(x) = 1+tan²(x)
Therefore,
tan(x/2) =
= (1/2)·(1+tan²(x/2))
dx

and
(1+tan²(x/2)) dx =
= 2·
d tan(x/2)

Therefore, our integral can be expressed in terms of an integral of some rational function of tangent as follows:
[2· d tan(x/2) /
/ (1−tan²(x/2)−2·tan(x/2))
]

Now let's substitute y=tan(x/2).

2· dy / (1−y²−2y) =
=−2·
dy/
[(y+1−√2)·(y+1+√2)] =
= (√2/2)·
[
dy / (y+1+√2) −
−
dy / (y+1−√2)
] =
= (√2/2)·
[ln(y+1+√2) − ln(y+1−√2)] + C

Going back through substitution, we have the result of integration:
(√2/2)·[ln(tan(x/2)+1+√2) − ln(tan(x/2)+1−√2)] + C

As we see, this result is quite different from the one we obtained in the previous lecture that we mentioned above, but it's still correct.

Let's check it by differentiation.

Dx {(√2/2)·[ln(tan(x/2)+1+√2) − ln(tan(x/2)+1−√2)] + C } =

= (√2/2)·
[1/(tan(x/2)+1+√2) −
− 1/(tan(x/2)+1−√2)
]·
·
Dx tan²(x/2) } =

= (√2/2)·
[1/(tan(x/2)+1+√2) −
− 1/(tan(x/2)+1−√2)
]·
·
[1+tan²(x/2)]/2 =

= −
[1+tan²(x/2)] /
/

[(tan(x/2)+1+√2
·(tan(x/2)+1−√2)
] =

= −
[1+tan²(x/2)] /
/

[(tan²(x/2)+2tan(x/2)−1]

The numerator equals to:
1/cos²(x/2).
The denominator equals to:
[sin²(x/2)+2sin(x/2)·cos(x/2)−
−cos²(x/2)
]/cos²(x/2)

The result of division of numerator by denominator, considering the minus sign in-front of the whole fraction, equals to:
1/[−sin²(x/2)−
−2sin(x/2)·cos(x/2)+
+cos²(x/2)
] =
= 1/
[cos(x)−sin(x)]

Since differentiation of our result proves that it's correct, it should differ from the result of the previous lecture for this integral by a constant.

Whoever suggests the trigonometric proof that the formula for a result from a previous lecture
(√2/4)·[ln(1+sin(x+π/4)) −
− ln(1−sin(x+π/4))
]

plus some constant equals to a new result
(√2/2)·[ln(tan(x/2)+1+√2) − ln(tan(x/2)+1−√2)]
will be mentioned in the notes for this lecture.

Problem 2.2:

d/ (cos(x)−sin(x))

Yes, still the same problem, but yet another approach to "rationalize" it.

Recall the wonderful Euler's formula that defines the exponential function of a complex argument:
eix = cos(x) + i·sin(x)
Using this formula, we can express both cos(x) and sin(x) as exponential functions of a complex argument as follows:
e−ix = ei·(−x) =
= cos(−x) + i·sin(−x) =
= cos(x) − i·sin(x)

Therefore,
cos(x) = (1/2)·(eix+e−ix)
sin(x) = (1/(2i))·(eix−e−ix)
Since i² = −1, we can replace 1/i with −i and the last expression for sin(x) would look like this:
sin(x) = −(i/2)·(eix−e−ix) =
= (i/2)·(e−ix−eix)

We are planning to use this representation to reduce our integral to something familiar. The problem is, of course, that we never explained what is a derivative or an integral when dealing with functions that take real arguments but take complex values, as is the case with functions like eix.
However, we will use the same techniques with these functions as we did with real functions. Strictly speaking, we have to prove that all these techniques are applicable (and they are!), but the procedures to prove all these properties are exactly the same as for real functions, So, we'll skip this and use all the available apparatus as if we have proven its applicability.

Let's convert the denominator first.
cos(x)−sin(x) =
= (1/2)·(eix+e−ix) −
− (i/2)·(e−ix−eix) =
= (1/2)·
[(1+i)·eix+(1−i)·e−ix] =
= (e−ix/2)
[(1+i)·e2ix+(1−i)] =
= (1−i)(e−ix/2)·
·
[e2ix·(1+i)/(1−i)+1]

Notice that
(1+i)/(1−i) = i
because
(1−i)·i = i−i² = 1+i
Also, e2ix = (eix
and e−ix = 1 / eix

That makes our denominator look like this:
cos(x)−sin(x) =
= (1−i)·
[i·(eix)² +1] / (2·eix)

As we see, all depends now on eix.
This prompts us to use substitution y=eix.
We also have to express dx in terms of dy as follows:
x = ln(y)/i = −i·ln(y)
Therefore,
dx = −i·d/ y
Now our integral equals to
2·y·(−i·dy/y)/{(1−i)·[i·y² +1]} =
[−2i/(1−i)]·
dy/[i·y² +1]

One more trivial substitution z=√i·y would result in the denominator within this integral to be equal to 1+z² familiar from differentiation of arctan(z).
In this case dy=dz/√i and our integral equals to
[−2√i/(1−i)]· dz/[z² +1]

The integral itself equals to arctan(z)+C.
As for a multiplier, as was shown in the chapter "Complex Numbers" (and trivially checked directly),
[(√2/2)·(i+1)]² = i
So, we replace i with
(1+i)/√2
Also, as we saw above,
(1+i)/(1−i) = i
That simplifies the result of integration to
−√2·i·arctan(z) + C

Reversing the substitutions, we get the following result of integration
d/ (cos(x)−sin(x)) =
=−√2·i·arctan((i+1)·eix/√2)+C

Notice that, regardless of the presence of an imaginary i, this expression should be a real function. To prove it, we have to get deeper into functions of complex arguments, which is beyond the scope of this course.

Check by differentiation:
[−√2·i·arctan((i+1)·eix/√2)]I=
= −√2·i·(i+1)·i·eix /
/ √2·[1+(1+i)²e2ix/2] =
= (1+i)·eix / (1+i·e2ix) =
= 2·eix /
[(1+i·e2ix)·(1−i)] =
= 2 /
[(1−i)·e−ix+(1+i)·eix)] =
= 1/
[cos(x)−sin(x)]

The end.