Integrate ?sec?^3 xdx

Asked by Thomas Albin | 21st Oct, 2012, 10:45: PM

Expert Answer:

Answer : Integral (sec3(x) dx)

Integral (sec(x) sec2(x) dx)

Now, use integration by parts.

Let u = sec(x). dv = sec2(x) dx
du = sec(x)tan(x). v = tan(x) 

Integral (sec3(x) dx) = sec(x)tan(x) - Integral (sec(x)tan2(x) dx)

Using tan2(x) = sec2(x) - 1.

Integral (sec3(x) dx) = sec(x)tan(x) - Integral (sec(x)[sec2(x)
- 1] dx)


Integral (sec3(x) dx) = sec(x)tan(x) - Integral( (sec3(x) - sec(x)) dx)

Now, separate into two integrals.

Integral (sec3(x) dx) = sec(x)tan(x) - [Integral (sec3(x) dx) -
Integral (sec(x) dx)] 


Integral (sec3(x) dx) = sec(x)tan(x) - Integral (sec3(x) dx) + 
Integral (sec(x) dx)
 
 
 
Now move - Integral (sec3(x) dx) to the LHS of equation

2 Integral (sec3(x) dx) = sec(x)tan(x) + Integral (sec(x) dx)

{integeral sec x = ln|sec(x) + tan(x)| }
2 Integral (sec3(x) dx) = sec(x)tan(x) + ln|sec(x) + tan(x)| 

now divide the complete equation by 2

Integral (sec3(x) dx) = (1/2)sec(x)tan(x) + (1/2)ln|sec(x) + tan(x)|


= (1/2)sec(x)tan(x) + (1/2)ln|sec(x) + tan(x)| + C Answer  { C is a constant } 

Answered by  | 23rd Oct, 2012, 11:44: PM

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