Unizor - Indefinite Integrals - Problems 1

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

Indefinite Integral -
Problems 1

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 1.1:

∫ sin(x)/cos³(x) dx

We know that (cos(x))^I = −sin(x).
Therefore, we can combine sin(x) in the numerator and dx to obtain −dcos(x).
Our integral is transformed into
−∫ dcos(x)/cos³(x)
This substitution allows to integrate as if we have a power function ∫ x^a dx = x^a+1/(a+1) where a=−3.
Since ∫ dx/x³ = −1/(2x²)+C,
our integral equals to
1/(2cos²(x))+C

Checking the answer:
(1/(2cos²(x)))^I = (1/2)(1/cos²(x))^I = (1/2)(−2/cos³(x))·(−sin(x)) = sin(x)/cos³(x).
The end.

Problem 1.2:

∫ ln(sin(x))·cot(x) dx

To find this integral, notice that
cot(x) = cos(x)/sin(x)
Since (sin(x))^I = cos(x),
we use sin(x) as an inner function:
ln(sin(x))·cot(x) = ln(sin(x))·(sin(x))^I/sin(x)
Next, notice that, since
(ln(x))^I = 1/x,
(ln(sin(x)))^I = (1/sin(x))·(sin(x))^I
Therefore,
ln(sin(x))·cot(x) = ln(sin(x))·(ln(sin(x)))^I


Using the above substitution, we obtain the answer:
∫ ln(sin(x))·cot(x) dx = ∫ ln(sin(x))·dln(sin(x)) = ln²(sin(x))/2 + C

Checking by differentiating the answer:
D_x (1/2)ln²(sin(x)) = ln(sin(x))·(1/sin(x))·cos(x) = ln(sin(x))·cot(x)
The end.

Problem 1.3:

∫ (ln(x)−3)/(x·√ln(x)) dx

To find this integral, break it in two integrals:
∫ ln(x)/(x·√ln(x)) dx −∫3/(x·√ln(x) dx) = ∫ √ln(x) dln(x) −∫3/√ln(x) dln(x) =

= (2/3)√ln³(x) − 6√ln(x) + C

Checking by differentiating the answer:
D_x ((2/3)√ln³(x) − 6√ln(x)) = (2/3)·(3/2)·√ln(x)·(1/x) − 6·(1/2)·(1/√ln(x)·(1/x) =
= (1/x)·(√ln(x) − 3/√ln(x)) = (ln(x)−3)/(x·√ln(x))
The end

Problem 1.4:

∫ x²·e^x/2 dx

Let's use the fact that exponential function does not change much with differentiation.
Transform our integral into
∫ 2x² de^x/2
Now we can use integration by parts twice getting
2x²·e^x/2 − 2∫ e^x/2 dx² = 2x²·e^x/2 − 2∫ e^x/2·2x dx = 2x²·e^x/2 − 2∫ 2x·2 de^x/2 =
= 2x²·e^x/2−8e^x/2·x+8∫ e^x/2 dx = 2e^x/2(x²−4x+8) + C

Checking by differentiating the answer:
D_x (2e^x/2(x²−4x+8)) = 2e^x/2·(1/2)·(x²−4x+8) + 2e^x/2·(2x−4) =
= e^x/2·(x²−4x+8+4x−8) = e^x/2·x²
The end

Problem 1.5:

∫ dx / √(−x²+4x+5)

Notice that
arcsin^I(x) = 1/√(1-x²)
The fact that a coefficient with x² is negative is very important. It prompts that we can try to transform our original function to be integrated into a form similar to a derivative of arcsin(x).
So, our task is to transform an expression under an integral into an expression that looks like the above derivative with linear transformation of variable x.
Basically, we have to express a quadratic polynomial under a square root as a full square expression.
−x²+4x+5 = 9−(x−2)² = 9·(1−(x−2)²/9) = 9·(1−((x−2)/3)²)

Let's substitute
y = (x−2)/3
Then x = 3y+2 and
dx = 3dy
Then our integral looks like this:
∫ 3dy/√(9·(1−y²)) = ∫ dy/√(1−y²) = arcsin(y) + C = arcsin((x−2)/3) + C

Checking by differentiating the answer:
D_x arcsin((x−2)/3) = 1/(3√(1−((x−2)/3))²)) = 1/√(9−x²+4x-4) = 1/√(−x²+4x+5)
The end

Problem 1.6:

∫ dx / (cos(x)−sin(x))

It's easy to deal with either cos(...) or sin(...).
To accomplish this, we can use the property
cos(π/4)=sin(π/4)=√2/2
and transform the denominator into the following form:
cos(x)−sin(x) = (cos(x)·cos(π/4) − sin(x)·sin(π/4)) / (√2/2) = √2·cos(x+π/4)

Substitute y = x+π/4.
Then dy = dx.
Our integral now looks like
∫ dy / (√2·cos(y)) = (√2/2)∫ dy / cos(y)

To find the above integral, we can multiply the nominator and denominator by sin^I(y) = cos(y) and express everything in terms of sin(y):
(√2/2)∫ dsin(y) / (1−sin²(y))

New substitution z = sin(y) leads to the following integral:
(√2/2)∫ dz / (1−z²)
Now we can use an obvious identity
1 / (1−z²) = (1/2)·((1/(1−z) + 1/(1+z))

Our integral equals to a sum of two integrals:
(√2/4)∫ dz / (1−z) + (√2/4)∫ dz / (1+z) = (√2/4)[−ln(1−z)+ln(1+z)] + C

Going back through substitutions used above, we obtain
(√2/4)·[ln(1+sin(x+π/4)) − ln(1−sin(x+π/4))]

Checking by differentiating the answer:
D_x {(√2/4)·[ln(1+sin(x+π/4)) − ln(1−sin(x+π/4))]}
Notice that
D_x ln(1+sin(x+π/4)) = [1/(1+sin(x+π/4))] · D_x(1+sin(x+π/4)) = cos(x+π/4)/(1+sin(x+π/4))
Similarly,
D_x ln(1−sin(x+π/4)) = [1/(1−sin(x+π/4))] · D_x(1−sin(x+π/4)) = −cos(x+π/4)/(1+sin(x+π/4))
Therefore, our derivative equals to the following:
(√2/4)·cos(x+π/4)·[1/(1+sin(x+π/4)) + 1/(1−sin(x+π/4))] =
= (√2/4)·cos(x+π/4)·2/(1−sin²(x+π/4)) = (√2/4)·cos(x+π/4)·2/(cos²(x+π/4)) =
= (√2/2)/cos(x+π/4) = 1/(cos(x)−sin(x))
The end