anonymous
  • anonymous
how to integrate sec^3(X)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[\int\limits \sec^3xdx\]
anonymous
  • anonymous
i think you should make it to int. sec^2(X)sec(X)
anonymous
  • anonymous
and then i am stuck in \[\int\limits \sec(x)+\int\limits \sec(x)\tan^2(x)\]

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anonymous
  • anonymous
however, if its right. i do not know how to int. the second int. sec(x)tan^2(x)
anonymous
  • anonymous
what is \[\int\limits \sec(x)\tan^2(x)\]
anonymous
  • anonymous
could some body give me an explanation
anonymous
  • anonymous
the answer should be sec(x)tan(x)
Callisto
  • Callisto
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JamesJ
  • JamesJ
The best thing to do is to start with sec x. sec^2 x, and integral first by parts. Remember that the derivative of tan x is sec^2 x Follow that path and see where it gets you.
JamesJ
  • JamesJ
making progress?
Callisto
  • Callisto
Can anyone tell me where's the problem.... :(
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JamesJ
  • JamesJ
in your posted solution there is a -- (double minus) on the last integral. The bottom line being you've shown 0=0.
Callisto
  • Callisto
Sorry.. i forgot the sign... But what i was trying to say is,... i can't find the 'answer' in that way..
JamesJ
  • JamesJ
\[ \int \sec^3 = \int \sec^2 . \sec = \tan . \sec - \int \sec \tan^2 \] as (sec)' = tan.sec. Now tan^2 = sec^2 - 1, hence the right hand side of that last expression is equal to \[ \tan . \sec - \int \sec^3 + \int \sec \] therefore \[ 2 \int \sec^3 = \tan . \sec - \int \sec \] Now you're almost home.
Callisto
  • Callisto
That makes sense, and is probably similar to img_0001 above. Just the order of doing it is not the same :D
anonymous
  • anonymous
let u = sec x dv = sec^2 x then du = sec x tan x v = tan x so int(sec^3 x dx) = sec x tan x - int(sec x tan^2 x dx) = sec x tan x - int(sec x (sec^2 x - 1) dx) = sec x tan x - int(sec^3 x dx) + int(sec x dx) = sec x tan x - int(sec^3 x dx) + ln (sec x + tan x) + C so now we have int(sec^3 x dx) = sec x tan x - int(sec^3 x dx) + ln (sec x + tan x) + C solve for int(sec^3 x dx): 2 int(sec^3 x dx) = sec x tan x + ln (sec x + tan x) + C int(sec^3 x dx) = 1/2 (sec x tan x + ln (sec x + tan x)) + C note that the end result C is different from the above C, but since it's an arbitrary constant it doesn't matter or 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 hope this helps u !! :D

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