Mastering Trigonometric Substitution: How to Integrate ∫csc²(x)dx

∫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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