The indefinite integral with respect to x (assumed that, for future reference you should always list the variable of integration (dx, dy, dz, dq, dt, etc...)) of:
f(x)=1+sin2(x)csc(x)
csc(x) is the same as 1/sin(x), and sin2(x)/sin(x)=sin(x), so f(x) can be simplified to:
f(x)=1+sin(x)
the antiderivatives of both of the components of this function are easy and known. Since they are added together and not multiplied or divided, they can be and integrated individually and then added together in the end (capital-letter function names denote antiderivatives):
f(x)=h1(x)+h2(x)
h1(x)=1
h2(x)=sin(x)
H1(x)=x+C
H2(x)=-cos(x)+C
F(x)=H1(x)+H2(x)
F(x)=x-cos(x)+C
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arctan(x)
I will assume that this is sopposed to be integrated with respect to x. To make this problem easier, imagine that the integrand is x raised to the negative 3. The integral is 1/(-2x-2) plus some constant c.
The integral of -x2 is -1/3 x3 .
The integral of arcsin(x) dx is x arcsin(x) + (1-x2)1/2 + C.
The integral of 2-x = 2x - (1/2)x2 + C.