## ∫cscxcotxdx

### To solve the integral ∫csc(x)cot(x)dx, we will use a technique called u-substitution

To solve the integral ∫csc(x)cot(x)dx, we will use a technique called u-substitution.

First, let’s simplify the integrand. Recall that csc(x) is equal to 1/sin(x), and cot(x) is equal to cos(x)/sin(x). Therefore, we can rewrite the integrand as (1/sin(x))(cos(x)/sin(x)):

∫(1/sin(x))(cos(x)/sin(x)) dx

Next, let’s introduce a new variable u to simplify the expression. We will let u = sin(x), meaning du/dx = cos(x). To find dx in terms of du, we can rearrange the equation to get dx = du/cos(x). Substituting these values into the original integral, we have:

∫(1/u)(cos(x)/u) (du/cos(x))

Simplifying further, we can cancel out the cos(x) terms:

∫(1/u)(1/u) du

Now, we can simplify the integral. Multiplying the fractions together, we have:

∫1/u^2 du

This integral is straightforward to evaluate. Using the power rule, we know that the integral of u^n du is equal to (u^(n+1))/(n+1) + C. In this case, n = -2, so:

∫1/u^2 du = (-1/u^(-1+1))/(-1+1) + C

Simplifying:

∫1/u^2 du = -1/(u^0) + C

Since u = sin(x), this becomes:

∫1/u^2 du = -1/(sin(x)^0) + C

Recall that any term raised to the power of 0 is equal to 1. Therefore, this simplifies to:

∫1/u^2 du = -1 + C

So, the final answer to ∫csc(x)cot(x)dx is -1 + C, where C is the constant of integration.

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