/sci/ - Science & Math

>>9542487
Doesn't* you third world shitnigger

>>9542487
[math]\int{sin^{5}x/cos^{8}x }dx[/math]

did i help?

>>9542487
Is this on your calc II test?
It's ez m8, start with sec^2x(sec^2 -1)^2secxtanx and go from there.

>>9542493
No. That's not equal to sec(sec(sec x)) * tan(tan(tan(tan(tan x)))).

>>9542497
How is that not equal? Sec^3 is equal to 1/cos*1/cos*1/cos

[math]\int{sec^{3}(x)tan^{5}(x)} dx[/math]
= [math]\int{sec^{2}(x)tan^{4}(x)sec(x)tan(x)} dx[/math]
u=sec(x) du = sec(x)tan(x)dx
[math]\int{u^{2}(u^{2}-1)^{2}} du[/math]
=[math]\int{u^{6}-2u^{4}+u^{2}} du[/math]
=[math]\frac{u^{7}}{7}-2\frac{u^{5}}{5}+\frac{u^{3}}{3}[/math]
=[math]\frac{sec^{7}}{7}-2\frac{sec^{5}}{5}+\frac{sec^{3}}{3}[/math]

OP here, I only made this post to show someone how to use LaTeX

>>9542513
>I only made this post to show someone how to use LaTeX
Tell him you're the worst [math]\rm\LaTeX[/math] user I've seen today.

>>9542513
use \displaystyle

>>9542504
he's doing a shit meme about sin^2(x) = sin(sin(x)), iterated trig functions

>>9542513
The tans and secs shouldn't be italicized. Use \tan etc.

>>9542513
that's a dumb way to solve it

simply use identity of sec^2 then just chain rule

no need to thank me, nerd

>>9542487
cant even evaluate that expression

>>9542513
You forgot the +C brainlet.

>>9542846
or dfrac instead of frac

>>9542869
[math]{\displaystyle \int{\sec{x}(\tan^2{x}+1)\tan^5{x}}dx
= \int{\sec{x}\tan^7{x}+\tan^5{x}}dx}

I don't see how this is easier than my way

>>9543857
[math]{\displaystyle \int{\sec{x}(\tan^2{x}+1)\tan^5{x}}dx
= \int{\sec{x}\tan^7{x}+\tan^5{x}}dx}[/math]

I don't see how this is easier than my way*

>>9543861
[math]{\displaystyle \int{\sec{(x)}(\tan^7{(x)}+\tan^5{(x)})}dx}[/math]

FFS