بسم الله الرحمن الرحيم
السلام عليكم ورحمة الله تعالى وبركاته

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Here's how I'd solve it. Hold on tight, because this is going to be a bumpy ride...

First step for any complex integral involving trig is to use the tangent half-angle formula. See first link.

Denote t = tan(x/2). Then:

1/(tan²(x)+sin(x))
= 1/((2t/(1-t²))² + 2t/(1+t²))
= (t^6 - t^4 + t^3 - t^2 + 1)/(2t^5 + 4t^4 - 4t^3 + 4t^2 + 2t)
= 1/2 [ (t^6 - t^4 + t^3 - t^2 + 1)/(t^5 + 2t^4 - 2t^3 + 2t^2 + t) ]
= 1/2 t + 1/2 (-2t^5 + t^4 - t^3 - 2t^2 + 1)/(t^5 + 2t^4 - 2t^3 + 2t^2 + t)

The first term is straightforward, since tan(x) is a standard integral:

1/2 ∫ tan(x/2) dx
= -log(cos(x/2))

The second term is trickier. We're going to attack this using a generalisation of partial fraction expansion.

Recall that integration by partial fractions attempts to convert an expression of the form p(x)/q(x) where p and x are polynomials to the form:

A/(x-a) + B/(x-b) + &ldots; + C/(x-c)

the integral of which is:

A log (x-a) + B log (x-b) + &ldots; + C log (x-c)

What we're going to do is the same thing, only working with t instead of x. The denominator is a quintic, so we expect that the integral will be of the form:

A log (t-a) + B log (t-b) + C log (t-c) + D log (t-d) + E log (t-e)

where a, b, c, d and e are the roots of the denominator. That is:

∫ (-2t^5 + t^4 - t^3 - 2t^2 + 1)/(t^5 + 2t^4 - 2t^3 + 2t^2 + t) dx = A log (t-a) + B log (t-b) + C log (t-c) + D log (t-d) + E log (t-e)

Taking the derivative of both sides, and using the fact that d/dx t = 1/2 (1 + t²) gives:

(-2t^5 + t^4 - t^3 - 2t^2 + 1)/(t^5 + 2t^4 - 2t^3 + 2t^2 + t)
= 1/2 [ A (1 + t²)/(t-a) + B (1 + t²)/(t-b) + C (1 + t²)/(t-c) + D (1 + t²)/(t-d) + E (1 + t²)/(t-e) ]

We can then cross-multiply and solve for A, B, C, D and E, and then we have our integral.

I'll leave you to find the roots of the denominator. There are some that are complex, so you'll need to introduce i temporarily, but you will be able to remove them from the final integral using the following identities:

arctanh(x) = 1/2 log [ (1+x)/(1-x) ]
arctan(x) = i/2 log [ (1-ix)/(1+ix) ]