$\int (\tan x)^4 (\sec x)^2 xdx$

can you show me how it's done @amistre64 @Zarkon ? i'd love to learn that lol

Question

$\int (\tan x)^4 (\sec x)^2 xdx$

can you show me how it's done @amistre64 @Zarkon ? i'd love to learn that lol

anonymous · Answer

missed the other x outside, damn

lgbasallote · Answer

is it something like $\int u^4 \tan^{-1} u du$?

anonymous · Answer

Yes, now you can use by parts.

lgbasallote · Answer

haha we were doing this in another problem..it was just a typo of an xdx there but amistre and zarkon wanted to  do it so i typed it in :P

amistre64 · Answer

i was gonna do it by parts:

$$\int t^4s^2xdx=x\frac1{5}t^5-\frac{1}{5}\int t^5 dx $$

$$\int t^4s^2xdx=x\frac1{5}t^5-\frac{1}{5}\int t^3(s^2+1) dx $$

$$\int t^4s^2xdx=x\frac1{5}t^5-\frac{1}{5}\int t^3s^2+t^3 dx $$

$$\int t^4s^2xdx=x\frac1{5}t^5-\frac{1}{5}(\frac{1}{4}t^4+\int t^3 dx) $$

amistre64 · Answer

$$\int t^3dx=\int t(ss+1)dx$$
$$=\int (s)ts+t dx$$
$$=\frac{1}{2}s^2+\int t dx$$
$$=\frac{1}{2}s^2-ln(cos)+c$$

lgbasallote · Answer

huh..didnt know it was this complicated :P hahaha

amistre64 · Answer

not complicated, just drawn out :)

lgbasallote · Answer

would $\int u^4 \tan^{-1}udu$ work too?

amistre64 · Answer

you could give it a shot and see :)

amistre64 · Answer

down the tan-1 to a function without a trig

chrisplusian · Answer

my son calls me a math nerd for taking honors classes and being concerned that I am not at an A in the class right now. He says dad your a math nerd because a B isn't good enough. He should read this....... he would think I was normal:) keep up the good work you guys

amistre64 · Answer

:)  you have to take classes to learn this stuff?  lol

chrisplusian · Answer

Nope!

amistre64 · Answer

im saving my B for english class ;)

lgbasallote · Answer

hmm downing tan^-1 seems hard -__- i'll just take your method amistre

chrisplusian · Answer

I won't take a B in anything ..... well not yet. And I would definitly not take a B in a math

chrisplusian · Answer

class

amistre64 · Answer

D tan-1 = 1/(1+u^2)

u up to u^5/5

integrate u^5/5(1+u^2)

long divide it and see what we get

amistre64 · Answer

u^3 -u + u/(u^2+1)
             --------------------
u^2 + 1 ) u^5
               (u^5+u^3)
                      -u^3
                      (-u^3-u)
                                 u

lgbasallote · Answer

you used tan^-1 as u and u^4 as dv right?

amistre64 · Answer

yes

lgbasallote · Answer

and this one is the uv part right...

amistre64 · Answer

v du  yes

lgbasallote · Answer

oh vdu...so combined it's

$\LARGE (\tan^{-1} u)(\frac{u^5}{5}) - [u^3 - u + \frac{u}{(u^2 + 1)}]$ right?

amistre64 · Answer

yes, but finish integrateing the stuff after the - sign

lgbasallote · Answer

oh oh yeah it's integral...this is seriously complicated lol i see another u^2 + 1 so im guessing another arctan @_@ i'll just go with your method

amistre64 · Answer

$$...=tan^{-1}(u)\frac{u^5}{5}-\frac{1}{5}\int u^3-u+\frac{u}{u^2+1}dx$$

amistre64 · Answer

that actually lns up since we can create a du above it ....

lgbasallote · Answer

arggghhh this is a pain in the head i quit lol i'll use your method

amistre64 · Answer

$$...=tan^{-1}(u)\frac{u^5}{5}-\frac{1}{5}\left(\frac{u^4}{4}-\frac{u^2}{2}+\frac{1}{2}ln(u^2+1 )\right)$$

amistre64 · Answer

as long as i kept it all straight

lgbasallote · Answer

stop showing me these T_T i dont want to see anymore