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

__Indefinite Integral -__

Problems 1

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)**d**x**We know that

**(cos(x))**^{I}= −sin(x)Therefore, we can combine

*in the numerator and*

**sin(x)***d*to obtain −

**x***d*.

**cos(x)**Our integral is transformed into

**−∫**d**cos(x)/cos³(x)**This substitution allows to integrate as if we have a power function

**∫ x**d^{a}**x = x**^{a+1}/(a+1)*.*

**a=−3**Since

*,*

**∫**d**x/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)**d**x**To find this integral, notice that

**cot(x) = cos(x)/sin(x)**Since

*,*

**(sin(x))**^{I}= cos(x)we use

*as an inner function:*

**sin(x)**

**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)**d**x = ∫ ln(sin(x))·**d**ln(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))**d**x**To find this integral, break it in two integrals:

**∫ ln(x)/(x·√ln(x))**d**x −∫3/(x·√ln(x)**d**x) = ∫ √ln(x)**d**ln(x) −∫3/√ln(x)**d**ln(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))= (1/x)·(√ln(x) − 3/√ln(x)) = (ln(x)−3)/(x·√ln(x))

The end

*Problem 1.4*:

**∫ x²·e**d^{x/2}**x**Let's use the fact that exponential function does not change much with differentiation.

Transform our integral into

**∫ 2x²**d**e**^{x/2}Now we can use integration by parts twice getting

**2x²·e**d^{x/2}− 2∫ e^{x/2}**x² = 2x²·e**d^{x/2}− 2∫ e^{x/2}·2x**x = 2x²·e**d^{x/2}− 2∫ 2x·2**e**

= 2x²·ed^{x/2}== 2x²·e

^{x/2}−8e^{x/2}·x+8∫ e^{x/2}**x = 2e**^{x/2}(x²−4x+8) + C*Checking by differentiating the answer*:

*D*

_{x}**(2e**

= e^{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}·x²The end

*Problem 1.5*:

**∫**d**x / √(−x²+4x+5)**Notice that

**arcsin**^{I}(x) = 1/√(1-x²)The fact that a coefficient with

*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*

**x²***.*

**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

*and*

**x = 3y+2***d*

**x = 3**d**y**Then our integral looks like this:

**∫ 3**d**y/√(9·(1−y²)) = ∫**d**y/√(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*:

**∫**d**x / (cos(x)−sin(x))**It's easy to deal with either

*or*

**cos(...)***.*

**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

*d*.

**y =**d**x**Our integral now looks like

**∫**d**y / (√2·cos(y)) = (√2/2)∫**d**y / cos(y)**To find the above integral, we can multiply the nominator and denominator by

**sin**^{I}(y) = cos(y)*:*

**sin(y)**

**(√2/2)∫**d**sin(y) / (1−sin²(y))**New substitution

**leads to the following integral:**

*z = sin(y)*

**(√2/2)∫**d**z / (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)∫**d**z / (1−z) + (√2/4)∫**d**z / (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))= (√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

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