Question

How to solve integral tan^3 x in terms of sin and cos

Answer 1

Do you mean, you need to convert to sines and cosines BEFORE solving the integral? Or you just need the final answer to be sines cosines?

Answer 2

the answer key does it in terms of sec and tan, but I did it using sin and cos and got it wrong, but i think it should work out too..so i think i 've made some mistakes ...and so I want to know if it is possible to solve it by converting tan into sin and cos

Answer 3

Hmm, it's much easier to do with tan and secant :O having trouble doing it that way?

Answer 4

didnt realize u can use sec and tan >.> i tried it for 1 min but thought couldnt solve using that so went on with sin and cos

Answer 5

\[\large 1+\tan^2x=\sec^2x \qquad \rightarrow \qquad \tan^2x=\sec^2x-1\]Using this identity gives us,
\[\int\limits \tan^3x dx \qquad =\qquad \int\limits \tan x(\tan^2x)dx\]
\[ \int\limits\limits \tan x(\sec^2x-1) dx \qquad = \qquad \int\limits \tan x \sec^2x dx - \int\limits \tan x dx\]

Answer 6

so basically when I see tan, always try solving it using tan and sec right

Answer 7

Ummmmm

Answer 8

Yes, unless the problem involves any other trig functions. Because that might mean you can simplify things down by converting everything to sines and cosines :D

Answer 9

Surprisingly, the level of difficulty of these integrals often comes down to the exponent on the trig function, whether it's even or odd.

Answer 10

yea, i guess if the question already have sin and cos present, then it might be more helpful to convert it to sin and cos

Answer 11

you always change the odd one right

Answer 12

so it will be (x^2)(x^2) (x) < the x becomes the derivative

Answer 13

Yes very good c:

Answer 14

Odd is the one we like ^^

Answer 15

just wondering what if they are both even

Answer 16

err actually, EVEN is the one we want for Tangent i guess :)
Odd for sine and cosine.

Answer 17

i usuall just pick anyone lol, it should work out ..@@

Answer 18

why even for tangent?

Answer 19

they have similar trig identities though
tan^x + 1 = sec ^2 x
cos^x + sin^x = 1

Answer 20

They don't have similar derivatives though. tangent produces a square. Sines and cosines don't :O

Answer 21

sin^2 x + cos^2 x = 1 < is this one right..
they are all involving squares..

Answer 22

the derivative of sine does not produce cos^2 right? :D

Answer 23

oh..u mean the derivative , true :D

Answer 24

then i guess i will search for sec/tan if it pops up an even function. and i'm pretty sure my prof will make it like an undoable question and i will just get stuck ..

Answer 25

From where we broke down the tangent into 2 integrals, you understand how to do the first one? It's a pretty straightforward U-sub.

Answer 26

yes i do. Can u help me with another hard integral question? just need a guide

Answer 27

usuall its always deciding what to start with is the hardest part..

Answer 28

Yesss so true! :D

Answer 29

bring it :O I'll try to help XD

Answer 30

thanks!

integral of 1/ x^3 sqrt (x^2- 1)

Answer 31

All in denominator?

Answer 32

yes

Answer 33

I think has to use trig sub in here?

Answer 34

Hmm yah it looks like it :O

Answer 35

if i set x= sec theta, i will get a random tan..

Answer 36

random? XD lol

Answer 37

and cant even cancel out the x^3..

Answer 38

lol..not random haha, something that i dont want it to appear in my equation lol

Answer 39

So what you'll do is, Let,
x=secQ
dx=secQ tanQ dQ

(those are thetas +_+)

And then just plug everything in, and after much cancellations, convert leftovers to sines and cosines if you need to :O

It shouldn't be too bad hmm

Answer 40

Imma try it on paper real quick :D to make sure I'm not talkin out my butt

Answer 41

haha thanks!

Answer 42

yah it worked out really nice :O

Need to see some steps? :D

Answer 43

yes plz. want ot know what to do with the x^3

Answer 44

x^3 = sec^3x?

Answer 45

Yes :O true story.

Answer 46

then do u have 1/ sec^2 theta somewhere ?

Answer 47

1/sec^2? Hmm

Answer 48

\[\large x=\sec \theta \qquad dx=\sec \theta \cdot \tan \theta \;d \theta\]

\[\int\limits\limits\limits \frac{(dx)}{x^3 \sqrt{x^2-1}} \quad \rightarrow \quad \int\limits\limits \frac{(\sec \theta \cdot \tan \theta d \theta)}{\sec^3\theta \sqrt{\sec^2\theta-1}}\]

Answer 49

i got \[\frac{ 1 }{ 2 }\sec ^-1 x + 2\frac{ \sqrt{1-x} }{ x }\] ....

Answer 50

for what part? plugging everything in?

Answer 51

Or that's what your integral gave you? I'm confused :D

Answer 52

yea..final ans..ok..just ignore me then, lol i'm wrong

Answer 53

Oh sorry I'm a little behind :D Was making sure you got the setup.

Yes that looks close to the right answer, hmm

Answer 54

Or is that from the book? :D

Answer 55

no..I did it!

Answer 56

xD

Answer 57

Ew you draw your triangles toward the left! You're one of those!! lolol

Answer 58

LOLLL

Answer 59

The only part I'm a little worried about - You didn't apply the half-angle formula correctly :O it should have changed the 1/2 (1+cos2Q)

Answer 60

do u?

Answer 61

oh opps ..@@

Answer 62

ahh..did it too fast..

Answer 63

sin 2 theta / 2 .. can i still bring out the 1/2

Answer 64

you mean write it as (1/2)sin2Q ... ? :o

Answer 65

cuz ur equation is now 1/2 (theta + sin 2 theta / 2)
to bring out the 1/2, dont u need to times theta by 2

Answer 66

From here you need to remember your Double Angle Formula for Sine :O

Answer 67

1/2 ( 1 - cos2x)!

Answer 68

Integrating should have given you something like this,
\[\large \frac{1}{2}\left(\theta+\frac{1}{2}\sin2\theta\right)\]

Answer 69

yes , i am here

Answer 70

i think i should use 2sinxcosx here tho?

Answer 71

Yessss

Answer 72

That's the double angle formula silly :O

Answer 73

You were thikning of the half angle formula, with the 1/2(1-cos2x)

Answer 74

I have no idea what those formulas called...LOL
it is pain enuff to remember the formulas @@ .we are not given any formulas on the exam
no calculators..

Answer 75

No calc? ooo nice :D I wish i could take a class like that haha

Answer 76

............I wish u can write it for me :(
this is my last exam.. and i'm really not prepared.. i still have 12 more ques to go before finishing the first package

Answer 77

poor guy :U