integrate 1/1+sec ax dx?

x-tan(ax/2)/a+ c

∫ 1 / (1+ sec ax) dx

Let u= ax
du = a dx
du = dt/a

∫ 1 / (1+ sec ax) dx = (1/a) ∫ du /(1+ sec u)
= (1/a) ∫ cos(u) du /(cos(u)+ 1)

Let t=tan(u/2)
dt = (1/2) sec^2(u/2) du
dt = (1/2) (1+tan^2(u/2)) du
dt = (1/2) (1+t^2) du
du = 2/(1+t^2) dt

Let t= tan(u/2)
cos(u) = cos^2(u/2) -sin^2(u/2)
= |cos^2(u/2) -sin^2(u/2)]/[cos^2(u/2)+sin^2(u/2)]
Multiply and divide the numerator and the denominator of the right side by cos^2(u/2)
cos(u) = (1-tan^2(u/2)) /(1+tan^2(u/2))
cos(u) = (1-t^2)/(1+t^2)

(1/a) ∫ cos(u) du /(cos(u)+ 1) = (1/a) ∫(2) (1-t^2) dt / (1+t^2)^2 ( (1-t^2)/(1+t^2) +1)
= (1/a) ∫ (1-t^2)/(1+t^2) dt

= (1/a) ∫ dt /(1+t^2) - (1/a) ∫ t^2 /(1+t^2) dt
= (1/a) tan^-1(t) - (1/a) ∫ t^2 /(1+t^2) dt

∫ t^2 /(1+t^2) dt = ∫ (1+t^2-1) /(1+t^2) dt = ∫ dt - ∫ dt/(1+t^2) = t - tan^1(t)
(1/a) tan^-1(t) - (1/a) ∫ t^2 /(1+t^2) dt = (1/a) tan^-1(t) - (1/a) ( t - tan^1(t) )
= (2/a) tan^-1(t) - (1/a) t

replace t by tan(u/2)
= (2/a) tan^-1(tan(u/2)) - (1/a) tan(u/2)

replace u by ax
= (2/a) tan^-1(tan(ax/2)) - (1/a) tan(ax/2) + C
= (2/a) (ax/2) - (1/a) tan(ax/2) + C
= x -(1/a) tan(ax/2) + C

Hello,


∫ {1 /[1 + sec(ax)]} dx =

let:

tan[(ax)/2] = u

hence:

(ax)/2 = arctan u

ax = 2arctan u

x = (2/a)arctan u

dx = (2/a)[1 /(1 + u²)] du

moreover let's apply the tangent half-angle formula cosθ = [1 - tan(θ/2)] /[1 + tan²(θ/2)] then secθ = 1 /cosθ =
[1 + tan²(θ/2)] /[1 - tan²(θ/2)] and therefore sec(ax) = {1 + tan²[(ax)/2]} /{1 - tan²[(ax)/2]} =
(being tan[(ax)/2] = u) =
(1 + u²) /(1 - u²)

yielding, by substitution:

∫ {1 /[1 + sec(ax)]} dx = ∫ {1 /{1 + [(1 + u²) /(1 - u²)]} } (2/a)[1 /(1 + u²)] du =

∫ {1 /{[(1 - u²) + (1 + u²)] /(1 - u²)} } (2/a)[1 /(1 + u²)] du =

∫ {1 /[(1 - u² + 1 + u²) /(1 - u²)]} (2/a)[1 /(1 + u²)] du =

∫ {1 /[2 /(1 - u²)]} (2/a)[1 /(1 + u²)] du =

∫ [(1 - u²) /2]} (2/a)[1 /(1 + u²)] du =

(simplifying)

∫ (1 - u²) (1/a)[1 /(1 + u²)] du =

∫ (1/a)[(1 - u²) /(1 + u²)] du =

being numerator and denominator of the same order, let's add and subtract 1 in the numerator:

∫ (1/a)[(1 + 1 - 1 - u²) /(1 + u²)] du =

∫ (1/a){[2 - (1 + u²)] /(1 + u²)} du =

(distributing and simplifying)

∫ (1/a){[2 /(1 + u²)] - [(1 + u²) /(1 + u²)]} du =

∫ (1/a){[2 /(1 + u²)] - 1} du =

(splitting into two integrals and factoring constants out)

(1/a)2 ∫ [1 /(1 + u²)] du - (1/a) ∫ du =

(2/a) ∫ [1 /(1 + u²)] du - (1/a)u =

(2/a)arctan u - (1/a)u + C

finally let's substitute back (ax)/2 for arctan u and tan[(ax)/2] for u:

(2/a)[(ax)/2] - (1/a)tan[(ax)/2] + C =

ending with:


x - (1/a)tan[(ax)/2] + C



I hope it's helpful

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Integrate 1/1+sec ax dx

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