This is one of those integration by parts questions which loop around and you end up getting your answer. Before we start, it's worth remembering that Integral (sec(x) dx) = ln|sec(x) + tan(x)|.
Integral (sec^3(x) dx)
First, split sec^3(x) into sec(x) and sec^2(x).
Integral (sec(x) sec^2(x) dx)
Now, use integration by parts.
Let u = sec(x). dv = sec^2(x) dx
du = sec(x)tan(x). v = tan(x)
Integral (sec^3(x) dx) = sec(x)tan(x) - Integral (sec(x)tan^2(x) dx)
Use the identity tan^2(x) = sec^2(x) - 1.
Integral (sec^3(x) dx) = sec(x)tan(x) - Integral (sec(x)[sec^2(x)
- 1] dx)
Distribute the sec(x).
Integral (sec^3(x) dx) = sec(x)tan(x) - Integral( (sec^3(x) - sec(x)) dx)
Now, separate into two integrals.
Integral (sec^3(x) dx) = sec(x)tan(x) - [Integral (sec^3(x) dx) -
Integral (sec(x) dx)]
Distribute the minus over the brackets.
Integral (sec^3(x) dx) = sec(x)tan(x) - Integral (sec^3(x) dx) +
Integral (sec(x) dx)
Here's the part which gets tricky; we're going to move
- Integral (sec^3(x) dx) to the left hand side of our equation, resulting in TWO of them.
2 Integral (sec^3(x) dx) = sec(x)tan(x) + Integral (sec(x) dx)
And we know what the integral of sec(x) is (we stated it above).
2 Integral (sec^3(x) dx) = sec(x)tan(x) + ln|sec(x) + tan(x)|
All we have to do now is divide everything by 2, which is the same as multiplying everything by (1/2).
Integral (sec^3(x) dx) = (1/2)sec(x)tan(x) + (1/2)ln|sec(x) + tan(x)|
And don't forget to add the constant.
= (1/2)sec(x)tan(x) + (1/2)ln|sec(x) + tan(x)| + C
2007-03-02 19:09:45
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answer #1
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answered by Puggy 7
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Integral Of Sec 3x
2016-10-01 11:41:04
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answer #2
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answered by gabler 4
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This Site Might Help You.
RE:
How can you integrate sec^3[x]?
2015-08-14 00:19:13
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answer #3
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answered by Rolande 1
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Take the integral: integral 1/(x sqrt(x+1)) dx For the integrand 1/(x sqrt(x+1)), substitute u = x+1 and du = dx: = integral 1/((u-1) sqrt(u)) du For the integrand 1/((u-1) sqrt(u)), substitute s = sqrt(u) and ds = 1/(2 sqrt(u)) du: = 2 integral 1/(s^2-1) ds Factor -1 from the denominator: = -2 integral 1/(1-s^2) ds The integral of 1/(1-s^2) is tanh^(-1)(s): = -2 tanh^(-1)(s)+constant Substitute back for s = sqrt(u): = -2 tanh^(-1)(sqrt(u))+constant Substitute back for u = x+1: = -2 tanh^(-1)(sqrt(x+1))+constant Which is equivalent for restricted x values to: Answer: | | = log(1-sqrt(x+1))-log(sqrt(x+1)+1)+consta...
2016-04-08 21:03:02
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answer #4
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answered by Anonymous
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by partial integration.
sec^3 = cos^(-3)
2007-03-02 19:04:47
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answer #5
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answered by gjmb1960 7
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