question (1 Viewer)

[pharmaceutical]

Differentiate tan-1(x/3)^2. hence find the exact value of 1/pi[integrate]from (root 3) to (0) {tan-1(x/3)}/(x^2 + 9)

 

[xclusv2bhung]

Differentiate tan-1(x/3)^2. hence find the exact value of 1/pi[integrate]from (root 3) to (0) {tan-1(x/3)}/(x^2 + 9)

lmao i like ur username ahahah.
pharmacy all the way !
lol n for the Q .. dont u jus apply the formula ? "

 

[watatank]

Differentiate tan-1(x/3)^2. hence find the exact value of 1/pi[integrate]from (root 3) to (0) {tan-1(x/3)}/(x^2 + 9)

uh, what?


did you mean

d/dx [ tan^-1(x/3)]²

= 2 * [ tan^-1(x/3)] * 1 / (9 + x²)

= 2[ tan^-1(x/3)] / (9 + x²)

im going to assume you want to find the integral of [ tan^-1(x/3)] / (9 + x²) from 0 to rt 3

I = 1/2 * integral 2[ tan^-1(x/3)] / (9 + x²)

= 1/2 * [ tan^-1(x/3)]²

b = rt3, a = 0

you can do the rest.

 

[xclusv2bhung]

oh .. didnt he want .. tan^-1 [ (x/3) ]^2 ?
thats what i thought anyway lmao

 

[watatank]

pft i duno its hard to read. i assumed it had to be that way since you cant do the second part.

 

[pharmaceutical]

lmao i like ur username ahahah.
pharmacy all the way !
lol n for the Q .. dont u jus apply the formula ? "

lol thanks yeah bro all the way... =DD
and for the Q dont worry i got it.

 

[xclusv2bhung]

rofl ahah nws.
im guna buddy ya.
no-one i know wants ter do pharmacy xD .. im like the only one ahahah.