Question

how do you integrate tan^6(x)dx

Answer 1

\[\int\limits \tan ^{6}dx\]

Answer 2

\[\tan^6x=\tan^2x\cdot\tan^4x=\tan^2x(\sec^2x-1)^2\]expand and u-sub

Answer 3

Go TuringTest!!!!! :D

Answer 4

\[\tan^6x=\tan^4x\cdot\tan^2x=\tan^4x(\sec^2x-1)=\tan^4x\sec^2x-\tan^4x\]\[=\tan^4x\sec^2x-\tan^2(\sec^2-1)=\tan^4x\sec^2x-\tan^2\sec^2+\tan^2x\]\[=\tan^4x\sec^2x-\tan^2\sec^2+\sec^2x-1\]\[u=\tan x\to du=\sec^2xdx\]I bet there is a faster way though...

Answer 5

@UTLEO got it?
are we good?

Answer 6

I am trying to understand it =]

Answer 7

UTLEO, be careful, he wants a medal...:D haha....

Answer 8

LOL

Answer 9

say it was\[\int\tan^4xdx\]I'm utilizing the identity\[\tan^2x=1-\sin^2x\]watch what happens:\[\int\tan^4xdx=\int\tan^2x\cdot\tan^2xdx=\int\tan^2x(\sec^2x-1)dx\]\[=\int\tan^2x\sec^2xdx-\int\tan^2xdx\]now the first integral can be done with a u-sub\[u=\tan x\to du=\sec^2xdx\]so the first integral becomes\[\int\tan^2x\sec^2xdx=\int u^2du=\frac13\tan^3x+C\]right?

Answer 10

Great answer!!

Answer 11

for the other integral we can use the same identity\[\int\tan^2xdx=\int\sec^2x-1dx=\tan x-x+C\] so the whole thing is\[\int\tan^4xdx= \frac13\tan^3x-\tan x+x+C\]right?

now look at what I did above for tan^6
I used the exact same procedure, but had to repeat it one extra time..

Answer 12

glad to help, see you around :D

Answer 13

I have one more "tricky" question to me
please help me for this one \[\int\limits \cos pixcos(\pi x/2)dx\]

Answer 14

\[\int\cos(\pi x)\cos(\frac{\pi x}2)dx\]?

Answer 15

yea

Answer 16

there is a simple trig identity that fixes so many of these types of integrals...

Answer 17

\[\cos\alpha\cos\beta=\frac12[\cos(\alpha-\beta)+\cos(\alpha+\beta)]\]now the integral should be no problem ;)

Answer 18

haven't thought of that, thanks for reminding!

Answer 19

those
sina*cosb
sina*sinb
cosa*cosb
ones always get people
don't be one of them, remember those identities!

Answer 20

definitely!