Question

evaluate the following integral.\\(\int \cot^{3}x \csc^{2}x \\, dx\\)\\(\int \cot^{3}x \csc^{2}x \\, dx = \square\\)

Explanation:

Step1: Substitution

Let \( u = \cot x \). Then, the derivative of \( u \) with respect to \( x \) is \( du = -\csc^{2}x \, dx \), which implies \( -du = \csc^{2}x \, dx \).
The integral \( \int \cot^{3}x \csc^{2}x \, dx \) can be rewritten in terms of \( u \) by substituting \( u = \cot x \) and \( -du = \csc^{2}x \, dx \). So the integral becomes \( \int u^{3} (-du) \).

Step2: Integrate with respect to u

Simplify the integral: \( \int u^{3} (-du) = - \int u^{3} \, du \).
Using the power rule for integration, \( \int u^{n} \, du = \frac{u^{n + 1}}{n+1} + C \) (where \( n
eq - 1 \)), for \( n = 3 \), we have \( - \int u^{3} \, du=- \frac{u^{4}}{4}+C \).

Step3: Substitute back

Substitute \( u=\cot x \) back into the expression. So we get \( - \frac{\cot^{4}x}{4}+C \).

Answer:

\( -\frac{\cot^{4}x}{4} + C \)