Question

Can the trig identity be used this way?

Answer 1

\[\tan^2x+1=\sec^2x\]
Can it also be used like so...
\[\tan^4x+1=\sec^4x\]

Answer 2

If,\[\large\rm \sec^2x=1+\tan^2x\]Then,\[\large\rm (\sec^2x)^2=\sec^4x=(1+\tan^2x)^2\]

Answer 3

\[\large\rm =1+2\tan^2x+\tan^4x\]

Answer 4

Oh right... I see it. So while i was working on my problem. I got here.
\[\int\limits 4\tan^4x \sec^6x dx\] and while doing u-substition, I got this far... \[4 \int\limits u^4 \sec^4x dx\]

Answer 5

But how would i apply the my u-sub, that u=tanx, to \[1+2\tan^2x+\tan^4x\]

Answer 6

hmm first you have to use one of these identity 1+tan^2x=sec^2x
tan^4x is same as tan^2x times tan^2x :D

Answer 7

Oh..... so it would turn into \[(1+\tan^2x)^2\]

Answer 8

Oh okay. lol

Answer 9

So I got\[4 \int\limits u^4 seec^4x du = 4 \int\limits u^4 (1+\tan^2x)^2 du = 4 \int\limits u^4 (1+u^2)^2 du\]

Answer 10

But i cant distribute the u^4 unless i factor (1+u^2)^2 right?

Answer 11

\[\large\rm \int\limits_{}^{} \tan^4x \cdot \sec^2x \cdot \sec^2x \cdot \sec^2x dx \]
\[\large\rm \int\limits_{}^{} \tan^4x \cdot \sec^2x \cdot (1+tan^2x)\cdot (1+tan^2x) ~dx \]
let u =tanx

Answer 12

\[\large\rm 4 \int\limits_{}^{} u^4 \cdot (1+u^2)\cdot (1+u^2)\cdot \sec^2x~ dx \]
u =tan x
du =sec^2x dx
\[\large\rm 4 \int\limits_{}^{} u^4(1+u^2)(1+u^2) \cdot sec^2 x~~\cdot \frac{du}{sec^2x}\]
\[\large\rm 4 \int\limits_{}^{} u^4(1+u^2)(1+u^2) \cdot \cancel{sec^2 x}~~\cdot \frac{du}{ \cancel {sec^2x}}\]


cc

Answer 13

amiright? -:P

Answer 14

Yea thats where I am right now. :3

Answer 15

ohh yea :-=

Answer 16

\[4 \int\limits_{}^{} u^4 (1+u^2)(1+u^2) ~~dx = 4 \int\limits_{}^{} u^4(1+2u^2+u^4) ~~dx\]
distribute by u^4
and then \[\frac{ u^{n+1} }{ n+1 }\] that's it
g

Answer 17

Hm.. I got\[\frac{ 4\tan(x)^5 }{ 5 }+\frac{ 8\tan(x)^7 }{ 7 }+\frac{ 4\tan(x)^9 }{ 9 } +C\]

But it says that my answer is wrong

Answer 18

\[\frac{ 4 \tan^5(x)}{ 5 }+ \frac{8\tan^7(x)}{7}+\frac{4 \tan^9(x)}{9}+C\] hmm maybe type it like this ?? lel <_<

Answer 19

Try rewriting your\[\frac{ 4\tan(x)^5 }{ 5 }+\frac{ 8\tan(x)^7 }{ 7 }+\frac{ 4\tan(x)^9 }{ 9 } +C\]

Answer 20

Your exponents are in the wrong places. For example, your first term should be \[\frac{ 4\tan^5x }{ 5 }\]

Answer 21

Yea, i got it xD
Sorry about that.
Thanks though