Find ∫cscx(cotx+cscx)dx .

Answer 1

To solve this integral, we can use the trigonometric identity:

csc(x) = 1/sin(x)
cot(x) = 1/tan(x) = cos(x)/sin(x)

Now let's substitute these expressions into the given integral:

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

Now, let's make a substitution to simplify this integral. Let u = sin(x), then du = cos(x) dx.

∫ (cos(x) + 1) / sin^2(x) dx
= ∫ (1/u^2 + 1) du
= ∫ (1/u^2) du + ∫ du
= -1/u + u + C
= -1/sin(x) + sin(x) + C

Therefore, ∫ csc(x)(cot(x) + csc(x)) dx = -1/sin(x) + sin(x) + C.