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# How do you integrate ( sec x)^3

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How do you integrate    ( sec x)^3

Please explain ...

Please you Latex if possible as some ot the non-latex answers are just too hard to follow.

Thanks in advance :)

Oct 30, 2016

### Best Answer

#8
+10

$$\displaystyle \int\sec^{3}x\,dx=\int\sec x\sec^{2}x\,dx$$,

integrate by parts,

$$\displaystyle = \sec x\tan x - \int\tan^{2}x\sec x\,dx$$,

$$\displaystyle = \sec x\tan x - \int(\sec ^{2}x -1)\sec x \,dx$$,

split the iintegral on the rhs into two and switch the first one to the lhs (doubling up with the original),

$$\displaystyle 2\int\sec^{3}x\,dx = \sec x\tan x +\int\sec x\,dx$$.

The integral of sec(x) has been deal with in an earlier post, multiply top and bottom by (sec(x) + tan(x)), and, (use a substitution if you must), it's of the standard form ' the thing on the top is the derivative of the thing on the bottom ' so it integrates directly to the natural log of the thing on the bottom, sec(x) + tan(x).

Finally divide throughout by 2.

Tiggsy.

Oct 31, 2016

### 11+0 Answers

#1
0

I do not use LaTex, but here is your answer:

Answer: | = 1/2 (tan(x) sec(x)-log(cos(x/2)-sin(x/2))+log(sin(x/2)+cos(x/2)))+constant

Oct 30, 2016
#2
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Thanks but I don't want the answer I want a full understandable explanation.

Oct 30, 2016
#3
+5

substitute   u = tan^2(x)      so du = -2tan(x)sec^2(x)dx     (chain rule and quotient rules for differentiation used here)

and use the the identity  sec^2(x) =  1 + tan^2(x) so you want to integrate

sec(x){1+tan^2(x)} dx          using sec^3(x)  = sec(x) {1+tan^2(x)}

From the u  - substitution,  sec(x) = sqrt(1+u)       , 1+ tan^2(x)=  (1 + u)  and

dx= -du/2tan(x)sec^2(x) = -du/2sqrt(u)(1+u).

If you now set up this integral you should get the integral of

-1/2(sqrt(1+u)/u du  which you can now integrate by parts.

Oct 30, 2016
#4
+5

Here is your solution step-by-step in non-LaTex everyday language!.

Take the integral:
integral sec^3(x) dx

Use the reduction formula, integral sec^m(x) dx = (sin(x) sec^(m-1)(x))/(m-1) + (m-2)/(m-1) integral sec^(-2+m)(x) dx, where m = 3:
= 1/2 tan(x) sec(x)+1/2 integral sec(x) dx

Multiply numerator and denominator of sec(x) by tan(x)+sec(x):
= 1/2 tan(x) sec(x)+1/2 integral(sec^2(x)+tan(x) sec(x))/(tan(x)+sec(x)) dx

For the integrand (sec^2(x)+tan(x) sec(x))/(tan(x)+sec(x)), substitute u = tan(x)+sec(x) and du = (sec^2(x)+tan(x) sec(x)) dx:
= 1/2 tan(x) sec(x)+1/2 integral1/u du

The integral of 1/u is log(u):
= (log(u))/2+1/2 tan(x) sec(x)+constant

Substitute back for u = tan(x)+sec(x):
= 1/2 tan(x) sec(x)+1/2 log(tan(x)+sec(x))+constant

Factor the answer a different way:
= 1/2 (tan(x) sec(x)+log(tan(x)+sec(x)))+constant
Which is equivalent for restricted x values to:

Answer: |= 1/2 (tan(x) sec(x)-log(cos(x/2)-sin(x/2))+log(sin(x/2)+cos(x/2)))+constant

Oct 30, 2016
#5
+105
0

Thanks very much.  :)

Are you the same guest or 3 different guests?

Anyway I'll take a proper look in the morning.

I don't know the reduction formula but maybe I can follow the rest://

Oct 30, 2016
#6
+5

Try this online calculator and see if it helps you in solving your Integral:

http://www.integral-calculator.com/

Oct 30, 2016
#7
+5

No,we are different guests,but essentially we have both given you the same method/idea using the 'u substitution' and integration by parts..(I put the first reply on.)   The Reduction Formula method is just an adapation of integration by parts. Good luck.

Oct 30, 2016
#8
+10
Best Answer

$$\displaystyle \int\sec^{3}x\,dx=\int\sec x\sec^{2}x\,dx$$,

integrate by parts,

$$\displaystyle = \sec x\tan x - \int\tan^{2}x\sec x\,dx$$,

$$\displaystyle = \sec x\tan x - \int(\sec ^{2}x -1)\sec x \,dx$$,

split the iintegral on the rhs into two and switch the first one to the lhs (doubling up with the original),

$$\displaystyle 2\int\sec^{3}x\,dx = \sec x\tan x +\int\sec x\,dx$$.

The integral of sec(x) has been deal with in an earlier post, multiply top and bottom by (sec(x) + tan(x)), and, (use a substitution if you must), it's of the standard form ' the thing on the top is the derivative of the thing on the bottom ' so it integrates directly to the natural log of the thing on the bottom, sec(x) + tan(x).

Finally divide throughout by 2.

Tiggsy.

Guest Oct 31, 2016
#9
+105
0

Thanks everyone, you all helped :)

Tiggsy, I can follow what you have done and I can see that early on I/you had already determined that

d/dx secx = tanx secx

I suppose that the bit that is still troubling me is multiplying by (secx+tanx)/(secx+tanx)

I mean I can see that it works but I would not think of it ... is there something I am missing, I mean a hint of what to do or something?  Is this just a 'trick' that I must commit to memory for questions such as this?

Is there some extra working that can be thrown in to make it more obvious?

Oct 31, 2016
#10
+5

The derivative of secx being tanx secx would be regarded as a standard result, commit it to memory or look it up. Alternatively, if you want to go back a step, differentiate (cosx)^(-1) using function of a function.

The trick of multiplying secx top and bottom by secx + tanx is just that, a trick that happens to work. Store it as a technique at the back of your mind in case something similar occurs.

Tiggsy.

Oct 31, 2016
#11
+105
0

Ok I'll try

Thanks Tiggsy :/

Nov 1, 2016