Integrating 1/(5+3cosx).dx

could someone check this for me? the integral calculator online gives a strange answer!

int 1/5+3cosx. dx

use t=tanx/2 thus, cosx=1-t^2/1+t^2 and cos^(2)x/2=1/(1+t)^2

dt/dx=1/2.sec^(2).x/2
dt/dx=([1+t^2]/2)

By substitution, we eliminate cosx etc and get ...
int 1/5+3cosx. dx = int 2/(8-2t^2). dt
=int 1/(4-t^2).dt
=1/4.ln([2+t]/[2-t]) + C

is that correct? Thanks

Reply 1

elpaw

Hoofbeat

int 1/5+3cosx. dx = int 2/(8-2t^2). dt

i get int 2/(8+2t²) dt

then use t = 2 tany substitution

Reply 2

elpaw

I get 1/2 tan^-1 (1/2 tan x/2)

Im sure theres a way to simplify that.

Reply 3

Hoofbeat

elpaw

i get int 2/(8+2t²) dt

then use t = 2 tany substitution

Grrr where have my signs gone wrong?

elpaw

I get 1/2 tan^-1 (1/2 tan x/2)

Im sure theres a way to simplify that.

don't worry about simplifying as have to put boundary conditions in, so in a sense that's not too important. originally they were pi and 0 (which i think gives infinity and 0 when substitution occurs). the final answer is pi/4.