Training creative thinking for high school students through solving some Algebra - Analysis exercises in many ways

3. Topics on antiderivatives and integrals

Example 1 : (University – A2013)

Integral calculation:

2 x 2  1

I 2 1

ln xdx

Method 1:

2 x 2  1

2 2 1

I  x 2 ln xdx   ln xdx   (  x 2 )ln xdx

1 1 1

We consider:

2 12

1 1 2 2 1

1 1 1

I 2   (  x 2 ) ln xdx   ln xd ( x )  x ln x 1  x 2 dx

 1 ln x 2  1 2  1 ln 2  1

x 1 x 1 2 2

I 1   ln xdx  2ln 2  1

So I  I  I

 2ln 2  1  1ln 2  1  1 (5ln 2  3 )

1 2

Method 2:

2 2 2

2 x 2  1 2 1 21

I  x 2 ln xdx   (1  x 2 ) ln xdx   ln xd ( x  x )

1 1 1



1 2 2 1 1

 ln x ( x  )  ( x 

x 1 1

) dx

 ln x ( x  1 ) 2  ( x  1 ) 2

x 1 x 1

 1 (5ln 2  3)

So I  1 (5ln 2  3)

Method 3:

2 x 2  1 21

I  x 2 ln xdx   (1  x 2 )ln xdx

1 1

Let t  ln x  x  e t and dx e t dt ; x  1  t  0; x  2  t  ln 2

2 1 2 2

tt  tt  t

I   (1 

2 t ) t . e dt  ( e  e ) t dt   td ( e  e )

1 1

 t ( e t  e  t )

ln 2

  ( e t e  t ) dt

 t ( e t  e  t ) ln 2  ( e t  e  t ) 2

1 1

 ln 2( e ln 2  e  ln 2 )  ( e ln 2  e  ln 2 )

 ln 2(2  1 )  (2  1 )  1 (5ln 2  3)

2 2 2

I  1 (5 ln 2  3)

Example 2 : Calculate the integral

3 3

x 2 dx

0 1

Method 1 : Use variable conversion by trigonometric method

Put

x  tan t  dx   1

 tan 2 t  dt

Change perspective

x  0  t  0; x =

t 



3 3 3 3

I= tan 3 t dt  tan t  tan t  1  1  dt  tan t (tan t 2  1) dt  tan tdt

0 0 0 0



3 3

 d ( cos t ) 3

 tan d  tant 

0 0

c ost

 ln 2 2

Method 2 : Use integration by parts

 x 2 u3

Set 

I  1 x 2ln  x 2 1 3 -

x ln  x 2 1  dx

 dv xdx2 0 

 x 2 10

 3ln 2 1

ln( x 2 11) d  x 2 1   3ln 2  J

With J=

ln( x 2 1) d  x 2 1 

Put

  ln( x 2 1) u







 du 



d ( x 2  1)

x 2  1

  dv d ( x 2 1)

  v x 2 1

I = 3ln2 -

[  x 2 1  ln  x 2 1 

  d  x 2 1  ]

= 3ln2 -

1  8ln 2  3  3  ln 2

2 2

Method 3 : Divide the numerator by the denominator to separate into 2 integrals.

3 x 3 3 x

1 3 d ( x 2  1) 3

I= ( x  x 2  ) dx  xdx  

x 2  1 dx 

2

 ln 2

0 1 0 0

2 0 x

 1 2

x 2

So it is easy to see that method 3 is much simpler.

Exercise :

Integral calculation

0 x 2 dx 𝜋

 1

I = ( x 2  1) 3 dx

Answer I = 32

Example 3 : Calculate the integral

I =  x 3(3  2 x 4) 3dx

Method 1 : If we think in the usual way, we can expand it into the sum of integrals as follows

1 1

I =  x 3 27  54 x 4 36 x 8 8 x 1 2  dx   27 x 3 54 x 7 36 x 1 1  8 x 1 5 ) dx

0 0

= 27 x 4 54 x 8 36 x 12 

8 x 16 1  5

4 8 12 16 0 2

Method 2 : Transform

Let t = 3  2 x 4  dt   8 x 3 dx

Change perspective

x  0 t  3,

x  1 t  1

I = 

1 1

3

t dt 

8 3

1 t 4

8 4

3  5

1 2

Method 3 : Using differential

Notice that  1 d (3  2 x 4 )  x 3 dx we have the following solution



1 1 3

1 (3  2 x 4 ) 4 1 5



I = 

 3  2 x 4 

8 3

d 3  2 x 4   

8 4 3 2

This problem may seem quite easy to some of you, but notice that if the degree of  3  2 x 4  is higher than degree 3, then method 1 will be too difficult to do.

but if you use method 2 or 3 it is easy to do. Here using method 3 is the easiest and most effective

Example 4 : Calculate the integral

I = 1 x 2  2

x 4  2 x 3  5 x 2  4 x  4 dx

Method 1 : We analyze the denominator of the expression under the integral sign.

x 4  2 x 3  5 x 2  4 x  4 = ( x 4  2 x 3  x 2 )  4( x 2  x )  4

= ( x 2  x  2) 2

Using the integral separation method

x 2  2

x 4  2 x 3  5 x 2  4 x  4

 x 2  2

( x 2  x  2) 2

 Ax  B  Cx  D x 2  x  2 ( x 2  x  2) 2

Then using the coefficient identification method we find ABCD. However, we see that although we can still reach the solution, following this method is quite long and complicated.

Method 2 : Divide both the numerator and denominator by x 2

I = 4

1x 2 

1 



x 2

  2( x 

2 )  5

2 x 2 x

Let t = x +

2  dt   1  2  dx

x x 2 

Change perspective



x  1  t  9 , x  1  t  3 2 2

3 dt

2 d ( t  1) 1

3

I = t 2 2 t  1   ( t  1) 2

9 3



t  1

2 44

Method 3 : If the student is better at putting expressions into differential equations, he can do it as follows:

I  

1  2

x 2

4  2

dx =



 x    1

2

 1   3

1x

2 

 2   2( x 

x 

)  5

x 2

Example 5 : Calculate the integral

xx 2  1

I dx

Method 1 : Transform

x 2 1

Put

tx2 t 2  1 xdx tdt

Change the limit x  2  t  ; x   t  1

2 dx

3 tdt

3 dt

xx 2  1

1 1

I  

 t  t 2 1  ( t 2 1)

Let t  tan u  dt =

travel

cos 2 u

Change the limit t  1  u  , t 

u 

I  

travel



		travel		
3		cos 2 u		3
	tan 2 u 		1	
4				4

Maybe you are interested!

Method 2 : Change variable

Put

1 t dx  dt xx 2

Change perspective

x  2 so

t 1

; x so t = 1

xx 2  1

I dx



1 1  t 2

1  t 2

dt

Let t = sin u, then dt = cos udu

Change the t 

1  u  ; t  1  u  π 2 4 2 6



4 c os udu 4 π

I 

 du 12

1  sin 2 u



6 6

Method 3 : Set x 

c ost

infer

dx  sin tdt

cos 2 t

Change the limit x = 2 to deduce

t  ; x  2

infer

t 



 sin tdt  





cost

1  cos 2t

sin t



I 

 3 sin tdt

 dt 12



4 cos 2 t 4 4

All three of the above methods are essentially methods of transformation.

Exercise

Lesson 1: (university-college 2003-block A)

xx 2  4

2 3 dx 1 5

Calculate the integral I =

Result I = 4 ln 3

2 dx 𝜋

x x 2

I  Result I =



I sin xc os xdx

sin x  cos x

Result I

 1  – c os x  sin x  1 ln  c

4 2

sin  x    1

4 

sin  x    1



4 





4.Trigonometry topic

Example 1: Solve the equation

3cos 4 x  8cos 6 x  2cos 2 x  3  0

Comment: From the appearance of the 4x arc , we can bring it back to the x arc using the doubling formula as follows:

cos 4 x  2cos 22 x  1  2  2cos 2x  1   1 = 8 cos 4x  8cos 2x  1

Method 1 : Equation

4cos 6 x  12cos 4 x  11cos 2 x  3  0

Let t = 𝑐𝑜𝑠 2 𝑥 ; 0≤ t ≤1

Then we have

4 t 3  12 t 2  11 t  3  0

 t  0.5

  t  1



From there we get the solution of the equation as

x = k k , k Z

4 2

Comment 2: From the appearance of even powers of cos, we can convert to the 2 x arc by the reduction formula and from the 4 x arc we convert to the 2 x arc by the doubling formula.

Method 2 :

3  cos 22 x  1  

1  cos 2 x ) 3 2  1  cos 2 x 3  0