Question

How do you find the integral of tan^4(x) dx? I assume you split it into sin^4(x)/cos^4(x), but after that, how do you integrate?

Answer 1

use by parts

Answer 2

integration by parts

Answer 3

convert tan into sec

Answer 4

tan^2*(sec^2-1)

Answer 5

u=tanx du=sec^2

Answer 6

Problem is, I don't really know how to deal with sec. Is there any other way I can integrate by parts?

Answer 7

is it mandatory to split tan into sin and cos?

Answer 8

it is easier just using tan into sec

Answer 9

tan^4=tan^2(sec^2-1)

Answer 10

tan^2sec^2-tan^2

Answer 11

integrate each one separately

Answer 12

tan^2sec^2 ----> u=tan du=sec^2, so... ----> \[\int\limits_{}u ^{2}du\]

Answer 13

and \[\int\limits_{}\tan ^{2}xdx=tanx-x\]

Answer 14

i've just solved it for you, did you understand?

Answer 15

one sec. Let me go through it :)

Answer 16

Yes, split them. \(\int\limits \tan^2(x)\tan^2(x)dx \). Then, you can convert one to a trig identity (which you should recognize). Then simple u-sub.

Answer 17

@SerikMB , I'm a little confused because my textbook says I should get 1/3(sinx/cosx) - 4/3 (sinx/cosx) + x + C ... I follow your process, I just don't see how they're the same.

Answer 18

then you need to find it by splitting

Answer 19

OK. Well thanks for walking me through the other way of doing it, though, it's good for me to know how! :)

Answer 20

\[\int\limits_{}\sin ^{2}/\cos ^{2}\]

Answer 21

then use integration by parts

Answer 22

no use just substition

Answer 23

OK. I think I can take it from there, just one question: which do I use as U, and which do I use as V'?

Answer 24

u=cos du=-sin

Answer 25

Ohok

Answer 26

do you know why?

Answer 27

because it makes it way simpler than doing it by parts?

Answer 28

It makes sense that cos would be the u because then du could be -sin which simplifies everything

Answer 29

ohh, sorry i messed up here. We need to use BY PARTS at final. No more messes promise)

Answer 30

Wait so no substitution? Why not?

Answer 31

because there is a square

Answer 32

i mean 4th degree

Answer 33

Right, and you can't deal with that because it won't clear sin^2(x) if you do substitution. OK, that makes sense.

Answer 34

so if it's by parts, what is U and what is V'?

Answer 35

wait a minute plz

Answer 36

sure thing

Answer 37

\[\int\limits_{}\sin ^{4}*\frac{ 1 }{ \cos ^{4} }\]

Answer 38

u=sin^4 and dv=sec^2

Answer 39

v=tanx and du=4sin^3cos

Answer 40

wait, i again faced up with different answer

Answer 41

i have to go now, but thanks for your help!! I think I'm getting it now :)