Evaluating the Integral of csc^2(x)dx using Trigonometric Substitution

∫csc^2(x)dx

To evaluate the integral of csc^2(x)dx, we can use a trigonometric identity

To evaluate the integral of csc^2(x)dx, we can use a trigonometric identity. Recall that the identity csc^2(x) = 1 + cot^2(x) holds true for all real values of x where cos(x) ≠ 0.

Using this identity, we can rewrite the integral as:

∫csc^2(x)dx = ∫(1 + cot^2(x))dx

Now, let’s break this integral into two separate integrals:

∫(1 + cot^2(x))dx = ∫1dx + ∫cot^2(x)dx

The integral of a constant term is simply the product of the constant and the variable of integration, so ∫1dx is just x.

For the second integral, we can use a trigonometric substitution to simplify it. Let’s substitute u = cot(x), then du = -csc^2(x)dx. Rearranging this equation, we can express dx in terms of du and csc^2(x):

dx = -du / csc^2(x)

Substituting this back into the integral:

∫cot^2(x)dx = ∫cot^2(x) (-du / csc^2(x)) = -∫u du

Now, we integrate the u term:

-∫u du = – (u^2 / 2) + C

Recall that we made a substitution: u = cot(x). Substituting back:

– (u^2 / 2) + C = – (cot^2(x) / 2) + C

Finally, summing up the two integrals:

∫csc^2(x)dx = x – (cot^2(x) / 2) + C

Therefore, the integral of csc^2(x)dx is given by x – (cot^2(x) / 2) plus an arbitrary constant C.

More Answers: