How do I solve the integral ∫1/√(1+cosx) dx using substitution?

[Motivanka]

Hey guys I can't solve one integral ∫1/√(1+cosx) dx I have started like ∫1/√(1+(cos^2 x/2 -sin^2 x/2)) dx = ∫1/√(cos^2 x/2 + cos^2 x/2) dx = ∫1/√(2cos^2 x/2) dx = 1/√2 ∫1/(cos x/2) dx = { substitution t= x/2 dx= 2dt } = 2/√2 ∫ 1/cost dt= 2/√2 ∫1/ ( cos^2 t/2 - sin^2 t/2) dt = 2/√2 ∫1/(cos^2 t/2 (1-(sin^2 t/2 /cos^2 t/2))) dt = 2/√2 ∫ 1/ cos^2 t/2(1-tg^2 t/2)) dt = { substitution t/2=u dt=2du} = 4/√2 ∫1/ cos^2u(1-tg^2 u) du = after this I don't know what to do can someone help me please.

 

[sk1105]

Your equations are hard to interpret in this form. The best thing to do is read up on how to use Latex to type mathematical expressions in PF (click 'latex preview' then 'latex guide'), but just for now it would be easier if your steps were each on their own line.

 

[PeroK]

Somewhere in the middle of that muddle you got the integral down to 1/cos(t), which is sec(t). Why not google for "integral sec" and see what comes up?

 

[Motivanka]

Your equations are hard to interpret in this form. The best thing to do is read up on how to use Latex to type mathematical expressions in PF (click 'latex preview' then 'latex guide'), but just for now it would be easier if your steps were each on their own line.

Thanks :) I am new so I didnt know that

 

[Motivanka]

Somewhere in the middle of that muddle you got the integral down to 1/cos(t), which is sec(t). Why not google for "integral sec" and see what comes up?

THANK YOUUU A LOT you gave me an idea :D