INTEGRAL CALCULUS dy ...

Question

INTEGRAL CALCULUS     	dy     	     	     	     	     	     	     	     	     	     	     	     	     	     	   Please use of the following as guide in solving the assignment.  	Antidifferentiation 	Differential Equation 	Chain Rule 	Logarithmic Funtion 	Exponential Function 	Integration by Parts 	Integrals Yielding Inverse Trigonometric Funtion 	Integrals Involving Hyperbolic Functions 	Inverse Hyperbolic Function 	Integration of Algebraic Functions by Trigonometric Substitution 	Trigonometric Integrals 	Integration of Rational Functions 	Integration by other Substitution Techniques 
	
Document Preview: INTEGRAL CALCULUSCreated with an evaluation copy of Aspose.Words. To discover the full versions of our APIs please visit: https://products.aspose.com/words/y3(1-2y4)5 dysec25xdx13-2xdxsin3tcos3t-1dt1+e2xexdxcos33x3sin3xdxtan5xsec3xdxx2x2+6dxdww2w2-72x-1x2-1dxdtt+22(t+1)82cos2x+1dxsin5?d?x2e4xdx24sinh5ydyCreated with an evaluation copy of Aspose.Words. To discover the full versions of our APIs please visit: https://products.aspose.com/words/Please use of the following as guide  in solving the assignment. AntidifferentiationDifferential EquationChain RuleLogarithmic FuntionExponential FunctionIntegration by PartsIntegrals Yielding Inverse Trigonometric FuntionIntegrals Involving Hyperbolic FunctionsInverse Hyperbolic FunctionIntegration of Algebraic Functions by Trigonometric SubstitutionTrigonometric IntegralsIntegration of Rational FunctionsIntegration by other Substitution Techniques

Robert · Accepted Answer

Sol: (1) 
 
3
5
41 2
y
dy
y
  
                 Substitute
4
3
1 2
8
u y
du y dy
 
 
 
Now,
   
 
 
 
   
3
5 5
4
3
5
5
4
3 5 1
5
4
3 4
5
4
3
5 4
4 4
1 1
81 2
1
81 2
1
8 5 11 2
1
8 41 2
1
1 2 32 1 2
y
dy du
uy
y
dy u du
y
y u
dy C
y
y u
dy C
y
y
dy C
y y

 

 

 

 
   
  
 
   
 
 
 
 
 




Sol: (2)   2sec 5x dx  
             2   int  sec 5 ,   5   5 :For the egrand x substitute u x and du dx 
   
 
 
2 21sec 5 sec
5
1
                    tan
5
1
                    tan 5
5
x dx u du
u C
x C

 
 
 
 
Sol: (3)  
1
3 2
dx
x
 
               
1
  int  ,   3 - 2   -2 :
3 2
For the egrand substitute u x and du dx
x
 

 
                                   
 
1 1 1
3 2 2
1
log
2
1
log 3 2
2
dx du
x u
u C
x C
 

  
   
 

Sol: (4)  
sin 3
cos3 1
t
dt
t 
 
sin 3
  int  ,   cos3 1  3sin 3 :
cos3 1
t
For the egrand substitute u t and du tdt
t
   

 
                     
 
sin 3 1 1
cos3 1 3
1
log
3
1
log cos3 1
3
t
dt dt
t u
u C
t C
 

  
   
 
 
Or 
                 
sin 3 2 3
log sin
cos3 1 3 2
t t
dt C
t
  
    
   

Sol: (5)  
21 x
x
e
dx
e


2 2
2
1 1
1
1
x x
x x x
x x
x x
x
x
x
x
e e
dx dx dx
e e e
e dx e dx
e e C
e C
e
e
C
e



 
 
   
   

 
  
 

Sol: (6) 
 
 
3
3
cos 3
sin 3
x
dx
x

 
 
      
 
3
3
2
3 3
2
cos 3
  int  ,
sin 3
        sin 3   sin 3 cos 3
cos 3
                    
x
For the egrand
x
substitute u x and du x x dx
x
du dx
u

 

 
 
 
    
 
 
 
 
     
23
3 3
6
2
6
7
2 8
2 8
3 3
cos 3 1 sin 3cos 3
sin 3 sin 3
1
1
2 8
sin 3 sin 3
2 8
x xx
dx dx
x x
u
u du
u
u udu
u u du
u u
C
x x
C




 
 
   
     
   
   
   
     
   
   
 




Sol: (7) 
5 3tan secx xdx                
5 3 2 2 For the integrand tan sec ,  use the triginometric identity tan sec 1:

INTEGRAL CALCULUS dy Please use of the following as guide in solving the assignment. Antidifferentiation Differential Equation Chain Rule Logarithmic Funtion Exponential Function Integration by Parts...

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