Question

can someone help me integrate these two please?

Answer 1

bottom one is a u sub

Answer 2

@Psymon will do the top one

Answer 3

that's what I tried. let me try again though

Answer 4

\(u=x+3\) \(x=u-3\) and so \(2(u-3)-1=2u-7\)

Answer 5

For the first one, let u = cot(x) :P
if u = cotx, then du would be -csc^2(x)dx. Solving for dx gives us
dx = -du/csc^2(x). Now we can substitute this dx value into our integral:
\[\int\limits_{}^{}\frac{\csc^{2}x}{\cot^{3}x}dx = \int\limits_{}^{}\frac{ \csc^{2}x }{ u^{3} }*-\frac{ du }{ \csc^{2}x }\]
pretty straightforward from there?

Answer 6

or use \(\frac{\csc^2(x)}{\cot^3(x)}=\tan(x)\sec^2(x)\)

Answer 7

I went with what I saw first, lol x_x

Answer 8

@satellite73 ok i see what you did there. My teacher taught the dx/du part differently. and yea i tried using tan(x)sec2(x) but I guess I messed it up

Answer 9

i see what you did @Psymon though. Thank you guys!!

Answer 10

\(\tan(x)\sec^2(x)\) is a function sitting right next to its derivative right?

Answer 11

Yeah, I suppose its how ive always done it. Hopefully it doesnt confuse you from what youre used to.

Answer 12

you do it in your head
answer is almost \(\tan^2(x)\) since the derivative of \(\tan^2(x)\) is \(2\tan(x)\sec^2(x)\) you are only off by a factor of 2

Answer 13

nah it won't, a little easier actually so thank you!
and yes I understand that. Thanks(:

Answer 14

Alrighty, awesome then ^_^

Answer 15

My brain just refuses to ever think of tan as 1/cot and vice versa. I have noidea why x_x

Answer 16

I wasnt trying to yell at ya or anything >.<