∫x cos^2(log(x)) dx

Using integration by parts, we choose:

u = cos^2(log(x)) => du = -2cos(log(x))sin(log(x)) dx dv = x dx => v = (1/2)x^2

Now, apply the integration by parts formula:

∫u dv = uv - ∫v du

∫x cos^2(log(x)) dx = (1/2)x^2cos^2(log(x)) - ∫(1/2)x^2(-2cos(log(x))sin(log(x))) dx

Simplify the right-hand side:

∫x cos^2(log(x)) dx = (1/2)x^2cos^2(log(x)) + ∫x^2cos(log(x))sin(log(x)) dx

Now, let's consider the remaining integral, which is of the form:

∫x^2cos(log(x))sin(log(x)) dx

We can use a substitution method here. Let's set:

u = log(x)

Then, differentiate u with respect to x:

du = (1/x) dx

Rearrange for dx:

dx = x du

Now, substitute these values into our integral:

∫x^2cos(log(x))sin(log(x)) dx = ∫(x^2cos(u)sin(u)) x du = ∫x^3cos(u)sin(u) du

Now, integrate with respect to u:

∫x^3cos(u)sin(u) du = (1/4)x^3(sin(2u)) + C

Now, substitute back for u:

(1/4)x^3(sin(2log(x))) + C

Using the double angle identity for sine, sin(2u) = 2sin(u)cos(u), we can simplify further:

(1/4)x^3(2sin(u)cos(u)) + C (1/2)x^3sin(u)cos(u) + C

Substitute back u = log(x):

(1/2)x^3sin(log(x))cos(log(x)) + C

Now, we can substitute this back into our original equation:

∫x cos^2(log(x)) dx = (1/2)x^2cos^2(log(x)) + (1/2)x^3sin(log(x))cos(log(x)) + C

Factor out (1/2) and apply trigonometric identity for tan:

(1/2)[x^2cos^2(log(x)) + x^3sin(log(x))cos(log(x))] + C (1/2)x^2[cos^2(log(x)) + x sin(log(x))cos(log(x))] + C (1/2)x^2[cos^2(log(x)) + x sin(log(x))cos(log(x))] + C (1/2)x^2[cos^2(log(x)) + x sin(log(x))]cos(log(x)) + C (1/2)x^2[tan(log(x)) + C

So, we have successfully proven that:

∫x cos^2(log(x)) dx = (1/2)x^2tan(log(x)) + C