# What is the antiderivative of the function cos^2(ln x)?

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### 2 Answers

use t = ln(x) so dt = 1/x dx x = e^t, so e^t dt = dx so

integral(cos^2(ln x) dx) = integral(cos^2(t) e^t dt)

and cos^2(t) = 1/2cos(2t) + 1/2

integral(cos^2(ln(x)) dx) = 1/2 integral((cos(2t) + 1)e^t dt)

= 1/2 integral(e^t cos(2t) dt) + 1/2 integral(e^t dt)

Integrate(e^t cos(2t) dt) by parts using

u = e^t, du = e^t dt

dv =cos(2t) dt, v = sin(2t)/2

integral(cos(2t) e^t dt) = e^t sin(2t)/2 - integral(e^t sin(2t)/2 dt) =

= e^t sin(2t)/2 - 1/2 integrate(e^t sin(2t) dt)

integrate by parts again,

u = e^t, du = e^t dt and

dv = sin(2t) dt, v = -1/2 cos(2t)

to get

= e^t sin(2t)/2 - 1/2 (-1/2 e^t cos(2t) - integral(-1/2 cos(2t) e^t dt))

integral(cos(2t) e^t dt) = e^t sin(2t)/2 + 1/4 e^t cos(2t) - 1/4 integral(cos(2t) e^t) dt) so

Add 1/4 integral(cos(2t) e^t) dt) to both sides

5/4 integral(cos(2t) e^t dt) = e^t sin(2t)/2 + 1/4 e^t cos(2t))

Multiply both sides by 4/5 to get

integral(cos(2t) e^t dt) = 1/5e^t(2 sin(2t) + cos(2t))

integral(cos^2(ln(x)) dx) = 1/2 integral(e^t cos(2t) dt) + 1/2 integral(e^t dt)

= 1/2 (1/5 e^t(2 sin(2t) + cos(2t))) + 1/2 e^t + C

Substituting t = ln(x) we get

integral(cos^2(ln(x) dx) = 1/10 x (2 sin(2 ln(x)) + cos(2 ln(x)) + 1/2x + C

= 1/10 x (2 sin(2 ln(x)) + cos(2 ln(x)) + x/2) + C

One method to determine the primitive of the given function is to integrate by parts.

First, we'll apply the identity:

(cos t)^2 = (1+cos 2t)/2

According to this identity, we'll have:

[cos (ln x)]^2 = [1+cos (2ln x)]/2

We'll integrate both sides:

Int[cos (ln x)]^2 dx = Int [1+cos (2ln x)]dx/2

Int [1+cos (2ln x)]dx/2 = Int dx/2 + Int [cos (2ln x)]dx/2

Int [1+cos (2ln x)]dx/2 = x/2 + Int [cos (2ln x)]dx/2

We'll calculate the integral Int [cos (2ln x)]dx using parts:

Int udv = uv - Int vdu

Let u = [cos (2ln x)] => du = -2sin(2ln x)/x

dv = dx => v = x

Int [cos (2ln x)]dx = x*[cos (2ln x)] + Int x*2sin(2ln x) dx/x

Int [cos (2ln x)]dx = x*[cos (2ln x)] + 2Int sin(2ln x) dx

We'll integrate by parts again Int sin(2ln x) dx:

u = sin(2ln x) => du = 2cos (2ln x)/x

dv = dx => v = x

Int sin(2ln x) dx = x*sin(2ln x) - 2Int cos (2ln x)dx

Int [cos (2ln x)]dx = x*[cos (2ln x)] + 2[x*sin(2ln x) - 2Int cos (2ln x)dx]

Int [cos (2ln x)]dx = x*[cos (2ln x)] + 2[x*sin(2ln x)] - 4Int [cos (2ln x)dx]

Int [cos (2ln x)]dx + 4Int [cos (2ln x)dx] = x*[cos (2ln x)] + 2[x*sin(2ln x)]

5Int [cos (2ln x)]dx = x*[cos (2ln x)] + 2*sin(2ln x)] + C

Int [cos (2ln x)]dx = x*[cos (2ln x)] + 2*sin(2ln x)]/5 + C

Since we have determined the term Int [cos (2ln x)]dx, we can find out the requested antiderivative:

Int [1+cos (2ln x)]dx/2 = x/2 + x*[cos (2ln x)] + 2*sin(2ln x)]/10 + C

**The requested antiderivative of [cos (ln x)]^2 is Int [1+cos (2ln x)]dx/2 = x/2 + x*[cos (2ln x)] + 2*sin(2ln x)]/10 + C.**