cant solve ?

Question

im in the 7th grade and have a problem with this integral

1 / [ 2 + cos x ] dx

Answer 1

2/3*sqrt(3) * arctan(1/3 * tan(1/2*x) * sqrt(3))

Answer 2

tan (2x) = 2.tan x /(1 - tan ² x)
Let t = tan (x/2)
dt/dx = (1/2).sec²(x/2)
dx = 2.dt / sec²(x / 2) = 2.dt / (t² + 1)
Use the facts that:
a) tan²(x/2) + 1 = sec²(x/2)
b) tan (x) = 2 tan (x/2) / (1 - tan²(x/2))
tan (x) = 2t / (1 - t²) and from this a triangle may be formed of sides 2t,(1 - t²) with hypotenuse (1 + t²)
From the triangle, cos x = (1 - t²) / (1 + t²)


I = ∫1/(2 + cos.x).dx
I = ∫ [1/ [2 + (1 - t ²) / (1 + t ²)] . 2.dt / (t ² + 1)
I = ∫ 2 dt / [ 2(t ² + 1) + 1 - t ² ]
I = 2 ∫ dt / (t ² + 3)
I = (2/√5) .tan^(-1)(t /√5) + C
I = (2/√5).tan^(-1) (tan(x/2)/√3) + C

Answer 3

let tanx/2=t, 1/2 sec^2(x/2)=d(t
now integral sec^2(x/2) d(x)/ 2+2tan^2(x/2) +1-tan^2(x/2)
[since cosx=1-tan^2x/2/(1+tan^2x/2)]
= 2 integral d(t)/3+t^2
= 2/sq.rt3 tan-1(t/sq.rt.3)
= 2/sq.rt.3 tan-1(tanx/2 / sq/.rt 3)

Answer 4

dnt get ppl 2 do homework 4 u

bt i like ur finkin i dnt no the ansa bt gud luk finding it

Answer 5

how dumb, that jason D can't even spell correctly. Anways, i think there are two unknown variable there. solve for D or X? No idea buddy, maybe someone will be better.