Question

evaluate the integral :

Answer 1

\[\int\limits_{}^{} \frac{ \sec^6(x) }{ \tan^2(x)}\]

Answer 2

do i convert the sec to 1/cos^2(x) ?

Answer 3

hmm have to use this identity 1+tan^2x =sec^2x
\[\huge\rm \int\limits_{ }^{ } \frac{ \frac{1}{\cos^6x }}{ \frac{\sin^2x}{\cos^2x} } \]
\[\int\limits_{ }^{ } \frac{1}{\cos^6x} \cdot \frac{\cos^2x}{\sin^2x}\]
this looks mssy hmm

Answer 4

sec=1/cosx btw

Answer 5

can we cross out the 2 signs and we will be left with

\[\frac{ 1 }{ \cos^4(x) *\sin^2(x) }\]

Answer 6

rewrite it

\[\frac{ \sec^2(x)\sec^2(x)\sec^2(x) }{ \tan^2(x)}\]

Remember the identity: \[\tan^2(x)=\sec^2(x)-1\]

Answer 7

that's not gonna work bec if you use sin^2x=(1-cos(2x))/2 identity it would not help you to cancel cos(x) or let u=cos x....

Answer 8

https://www.youtube.com/watch?v=xFrO9gUYdB4

Answer 9

im lost @_@

Answer 10

\[\int\limits_{ }^{ } \frac{\sec^6x}{\tan^2x}\]
we can use this identity 1+tan^2x=sec^2
but first we need to rewrite to get sec^2x or tanx
\[\int\limits_{ }^{ } \frac{\sec^6x}{ tanx \cdot tanx}\]
if i split tan^2x i wouldn't be apply to use that identity
therefore \[\frac{ \sec^4x \cdot \sec^2x }{ \tan^2x }\]
or just like legomyego mentioned
\[\frac{ \sec^2x \cdot \sec^2x \cdot sec^2x }{ \tan^2x }\]
sec^2x=tan^2x +1 \[\int\limits_{ }^{} \frac{ (1+\tan^2x)(1+\tan^2x) \cdot \sec^2x }{ \tan^2x }\]
let u =tanx
hope that would work hmm

Answer 11

alternatively , you can make this \[\int(\tan^2(x)+1)^2(\cot^2(x)+1)dx\] although how you would recognize this is anyone's guess

Answer 12

u sub ?

Answer 13

\[I=\int\limits \frac{ \sec ^4x \sec ^2x }{ \tan ^2x }dx=\int\limits \frac{ \left( 1+\tan ^2x \right)^2\sec ^2x }{ \tan ^2x }dx\]
\[=\int\limits \frac{ 1-2\tan ^2x+\tan ^4x }{ \tan ^2x }\sec ^2x dx\]
put tan x=t
\[\sec ^2x dx=dt\]
\[I=\int\limits \frac{ 1-2t^2+t^4 }{ t^2 }dt\]
?

Answer 14

hmm i kind of did what satelite did but i think it makes it more complicated ?

or is there an easier way in doing these

Answer 15

\[I=\int\limits t ^{-2 }dt-2\int\limits dt+\int\limits t^2 dt+c\]

Answer 16

that is the easy one
when i see sec^nx and tan^mx i think abt 1+tan^2x=sec^2x :D

Answer 17

correction use + 2tan^2 x in place of -2 tan^2x

Answer 18

@marcelie this is trig powers, its one of the hardest things you have to do in calc 2. Theres no easy way around them really. Luckily they dont really show up in any other class I hear except this one.

Answer 19

lool yeah but are there like rules for these like when sin is odd and cos is odd for their powers?

Answer 20

\[I=\int\limits \frac{ \left( 1+2 \tan ^2x+\tan ^4x \right) }{ \tan ^2x }\sec ^2x dx\]

Answer 21

\[\int\limits_{ }^{} \sin^ax ~~\cos^b x \]
when a or b = odd
apply this identity \[\sin^2x= 1-\cos^2x ~~~or ~~\cos^2x= 1-\sin^2x\]
if a or b = even \[\sin^2(x) =\frac{ 1-\cos(2x) }{ 2}~~~~\cos^2(x)=\frac{1+\cos(2x)}{2}\]

Answer 22

in the end
\[I=\int\limits t ^{-2}dt+2 \int\limits dt+\int\limits t^2 dt+c\]

Answer 23

i think it would be easy if you sub u for tan\[\int\limits_{ }^{} \frac{ (1+u^2)(1+u^2) }{ u^2 } \sec^2x\]

Answer 24

= \[(1+2u^2+u^4)/ u^2\]

Answer 25

\[(u ^{-2})(1+2u^2+u^4)\]

Answer 26

looks good \[\int\limits_{ }^{} \frac{ 1+2u^2+u^4 }{ u^2 } \cancel{\sec^2x }\cdot \frac{du}{\cancel{\sec^2}}\]
\[\int\limits_{ }^{} \frac{1}{u^2}+\frac{2u^2}{u^2}+\frac{u^4}{u^2} du \]

Answer 27

\[\frac{ 1 }{ u^2 }+2+u^2\]

Answer 28

yes^^

Answer 29

now you can rewrite 1/u^2 = u^{-2} would be easy to integrate :D

Answer 30

okay so then i got this

\[-\frac{ 1 }{ u }+2u+\frac{ 1 }{ 3 }u^3\]

Answer 31

looks good and +Chocolates :)

Answer 32

lool okie so then


final answer:

\[\frac{ 1 }{ 3 }\tan^3(x)+2\tan(x)-\frac{ 1 }{ \tan(x) }+c\]

Answer 33

so that 1/tanx converts to cot (x)

Answer 34

hmm you don't have to but ofc u can .... i would leave it like that

Answer 35

oh okayy :D

Answer 36

i love these type of questions <3 funn!! :D

Answer 37

loool D: but question are there rules how to format these ? when its sec or tan

Answer 38

hmm use this identity 1+tan^2x=sec^2x
doesn't matter if the exponents are odd or even
but first rewrite the function to get sec^2x `or` tanx
i always let u equal to base of highest exponent
here is an example \[\large\rm \int\limits_{ }^{ } \sec^5x \cdot \tan^3x ~dx\]
i will split tan^3x = tan^2x * tanx \[\large\rm \int\limits_{ }^{ } \sec^5x \cdot \tan^2x \cdot \color{REd}{tan x} ~dx\] apply the identity
\[\large\rm \int\limits_{ }^{ } \sec^5x \cdot (sec^2x-1 ) \cdot \color{REd}{tan x} ~dx\]
now let u = secx
the reason why we need tan x or sec^2x because when we let u= something
the derivative of that *something* should already in the function
\[\large\rm \int\limits_{ }^{ } \sec^5x \cdot (1+sec^2x) \cdot \color{REd}{tan x} ~dx\]
u= secx ( base of highest exponent )
du = secx tan x dx
dx=du/(secx tanx)
\[\int\limits_{ }^{ }\color{REd}{u^5}(u^2 -1) \tan x \cdot \frac{du}{secx ~tanx}\]
u=sec
u^5 is same as u^2 secx
\[\int\limits_{ }^{ }\color{blue}{ u^4}(u^2 -1) \tan x \cdot \color{REd}{secx} \cdot \frac{du}{secx ~tanx}\]

Answer 39

not good at explaining stuff
but hopefully that makes sense

Answer 40

oh okay ill look at it right now

Answer 41

correction
u=sec
u^5 is same as `u^2 secx `

typoo it should be u^4 secx