Question

∫sin(2+3logx)dx/x

Answer 1

@jim_thompson5910

Answer 2

@UnkleRhaukus

Answer 3

Please help ..i just couldn't do it :(

Answer 4

let u = 2 + 3log(x)
du =

Answer 5

u mean d/dx (u)=d/dx(2+3logx)?

Answer 6

du/dx=0+3/x

Answer 7

yes du= 3/x dx

Answer 8

now sub these in and it should be easy from here

Answer 9

u mean ∫sinu du right?

Answer 10

it should be -cosu +c=-cos(2+3logx)+c?

Answer 11

@UnkleRhaukus

Answer 12

almost , dont forget that factor of 3

Answer 13

oh sorry...du/x=dx/x

Answer 14

du/3=dx/x

Answer 15

ok got it :)

Answer 16

thank you :)

Answer 17

so what'd you get as your final result?

Answer 18

its 1/3 -cos(2+3logx)+c

Answer 19

i think your mean the right thing , but i'd write it like this
-(1/3)cos(2+3logx)+c

Answer 20

YES yes :)

Answer 21

can u help for another problem please :)

Answer 22

ok, good work,
and welcome to openstudy!

Answer 23

thank you very much sir :)

Answer 24

I have done it like this ∫tan(sin^-1x)/root 1-x^2
let tan(sin^-1x)=z
=>d/dx tan(sin^-1x)=dz/dx
=>sec^2(sin^-1x).1/root 1-x^2=dz/dx

=>sec^2(sin^-1x)/dz=dx/root 1-x^2
=>1+tan^2(sin^-1x)/dz=dx/root 1-x^2
=>1+z^2/dz=dx/root 1-x^2 ?

Answer 25

@UnkleRhaukus

Answer 26

Is this right?

Answer 27

looks pretty good i think you have the right idea,

im not sure on the detail though, this sort of question with arctrig functions and sqrts gives me a headache, ,

Answer 28

but now i think i should let sin^-1x=z
=>1/root 1-x^2=dz/dx
=>dx/root1-x^2=dz

Answer 29

then integral tanz dz

Answer 30

its ln sec z +c

Answer 31

is this right? @UnkleRhaukus

Answer 32

@UnkleRhaukus U there sir?

Answer 33

@RadEn

Answer 34

@mukushla