∫csc²(x)dx =
To integrate the function ∫csc²(x)dx, we can use the technique of trigonometric substitution
To integrate the function ∫csc²(x)dx, we can use the technique of trigonometric substitution.
First, let’s rewrite the integral using the reciprocal identity for csc(x):
∫csc²(x)dx = ∫(1/sin²(x))dx.
Now, to proceed with the substitution, we choose u = sin(x), which implies du = cos(x)dx. Rearranging this equation, we get dx = du/cos(x).
Substituting dx and sin(x) in terms of u in the integral, we have:
∫(1/sin²(x))dx = ∫(1/u²) * (du/cos(x)).
Notice that we also need to express cos(x) in terms of u. By using the Pythagorean identity, cos²(x) = 1 – sin²(x), we find cos(x) = √(1 – u²).
Substituting these expressions back into the integral, we have:
∫(1/sin²(x))dx = ∫(1/u²) * (du/√(1 – u²)).
Now, let’s simplify this further by multiplying the terms in the integral:
∫(1/u²) * (du/√(1 – u²)) = ∫(du/u²√(1 – u²)).
To continue, we can factor out 1/u² from the integrand:
∫(du/u²√(1 – u²)) = ∫(1/u²) * (1/√(1 – u²))du.
Now, let’s break this integral into two separate integrals:
∫(1/u²) * (1/√(1 – u²))du = ∫(1/u²)du * ∫(1/√(1 – u²))du.
The first integral, ∫(1/u²)du, is a standard integral that evaluates to -1/u.
The second integral, ∫(1/√(1 – u²))du, is also a standard integral that evaluates to arcsin(u).
Therefore, substituting these results back into the original integral:
∫csc²(x)dx = -1/u * arcsin(u).
Finally, we substitute back u = sin(x) to obtain the final result:
∫csc²(x)dx = -1/sin(x) * arcsin(sin(x)) = -cot(x) * arcsin(sin(x)).
So, the integral of csc²(x)dx is given by -cot(x) * arcsin(sin(x)).
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