Question

how to integrate sec^3(X)

Answer 1

\[\int\limits \sec^3xdx\]

Answer 2

i think you should make it to int. sec^2(X)sec(X)

Answer 3

and then i am stuck in \[\int\limits \sec(x)+\int\limits \sec(x)\tan^2(x)\]

Answer 4

however, if its right. i do not know how to int. the second int. sec(x)tan^2(x)

Answer 5

what is \[\int\limits \sec(x)\tan^2(x)\]

Answer 6

could some body give me an explanation

Answer 7

the answer should be sec(x)tan(x)

Answer 8

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.

Answer 9

making progress?

Answer 10

Can anyone tell me where's the problem.... :(

Answer 11

in your posted solution there is a -- (double minus) on the last integral. The bottom line being you've shown 0=0.

Answer 12

Sorry.. i forgot the sign... But what i was trying to say is,... i can't find the 'answer' in that way..

Answer 13

\[ \int \sec^3 = \int \sec^2 . \sec = \tan . \sec - \int \sec \tan^2 \]
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.

Answer 14

That makes sense, and is probably similar to img_0001 above. Just the order of doing it is not the same :D

Answer 15

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