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 }
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)
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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