Question

Evaluate the integral. 5 sec^4 x dx

Answer 1

next step is... \[5 \int\limits \sec^4 x \]

Answer 2

then i am lost.

Answer 3

\[\large \int (\sec^2 x)((\sec^2 x)dx\]
have you tried thaT?

Answer 4

@lgbasallote so then that becomes 1/3 tan x (sec^2 x)

Answer 5

wait is that the answer?

Answer 6

i dont know the answer. i just dont know what to do after that?

Answer 7

\[\large \int (\sec^2 x)(\sec^2 x)dx \implies \int (\tan^2 x + 1)(\sec^2 x)dx\]
do you remember that?

Answer 8

ugleh!

Answer 9

huh?

Answer 10

what rule is that?

Answer 11

it's part of the rules of integrating sec and tan.. uploading for you.

Answer 12

it's a trig identity
\[\large \tan^2 x + 1 = \sec^2 x\]

Answer 13

gotcha!

Answer 14

so we have \[\huge \int (\tan^2 x + 1)\sec^2 x dx\]
next step is to distribute sec^2 x
\[\huge \int (\tan^2 x \sec^2 x + \sec^2 x)dx\]

Answer 15

I have circled what's important for this section.

Answer 16

so how do you integrate that?

Answer 17

you distribute the integral symbol
\[\int \tan^2 x \sec^2 xdx + \int \sec^2 x dx\]
the second term is easy to integrate..integrate that first

Answer 18

that would become....tan x?

Answer 19

yup

Answer 20

so the whole answer would be 5 tan x +5/x tan^3 x +c

Answer 21

\[ \tan^3 x \over 3 \]

Answer 22

then the first term is u-sub

Answer 23

wait what did you do?

Answer 24

well the whole problem was 5 sec^4 x dx

Answer 25

we were just doing the sec^4 x

Answer 26

part

Answer 27

lol i know i just cant understand your notation sorry
\[\Large 5[\int \tan^2 x \sec^2 x dx + \int sec^2 x dx]\]
\[\Large 5[\frac{\tan^3 x}{3} + \tan x]\]
\[\Large \frac{5\tan^3 x}{3} + 5\tan x\]
is this what you meant?

Answer 28

yes

Answer 29

well then you're right! :D