Question

Find the reduction formula of ʃcot5xdx.

Answer 1

Answer on Question #53133 – Math – Integral Calculus

Find the reduction formula of $\mathrm{fcot}^{\wedge}5(\mathrm{x})\mathrm{dx}$.

Solution

$$\begin{array}{l} I _ {n} = \int \cot^ {n} x d x = \int \cot^ {n - 2} x \cot^ {2} x d x = \int \cot^ {n - 2} x (\csc^ {2} x - 1) d x \\ = - \int \cot^ {n - 2} x d x + \int \cot^ {n - 2} x \csc^ {2} x d x. \end{array}$$ $$\csc^ {2} x = - (\cot x) ^ {\prime}.$$ $$\int \cot^ {n - 2} x \csc^ {2} x d x = \int \cot^ {n - 2} x d (\cot x) = - \frac {\cot^ {n - 1} x}{n - 1} + c,$$

where $c$ is an arbitrary real constant.

The reduction formula of $I_{n}$:

$$I _ {n} = - I _ {n - 2} - \frac {\cot^ {n - 1} x}{n - 1}.$$

The reduction formula of $I_{5}$:

$$I _ {5} = - I _ {3} - \frac {\cot^ {3} x}{3}.$$ $$I _ {3} = - I _ {1} - \cot x.$$ $$I _ {1} = \ln | \sin x | + c,$$

where $c$ is an arbitrary real constant.

Thus

$$\int \cot^ {5} x d x = - (- \ln | \sin x | - \cot x) - \frac {\cot^ {3} x}{3} + c = \ln | \sin x | + \cot x - \frac {\cot^ {3} x}{3} + c,$$

where $c$ is an arbitrary real constant.

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