How do you find the integral of $\int {cot}^{n} (x)$ $int cot^n(x)$ if is an integer?

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How do you find the integral of $\int {cot}^{n} (x)$ $int cot^n(x)$ if is an integer?

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Answer 1 · 2015-10-12T23:48:33

See the explanation.

Explanation:

Let $n = 2 k + 1$ $n=2k+1$ , then:

$\int {cot}^{n} x d x = \int {cot}^{2 k + 1} x d x = \int {({cot}^{2} x)}^{k} cot x d x = I$ $int cot^nxdx = int cot^(2k+1)xdx = int (cot^2x)^k cotxdx = I$

${cot}^{2} x = t \Rightarrow 2 cot x (- {csc}^{2} x) d x = d t$ $cot^2x=t => 2cotx(-csc^2x) dx = dt$

$- 2 cot x (1 + {cot}^{2} x) d x = d t$ $-2cotx(1+cot^2x) dx = dt$

$cot x d x = - \frac{1}{2} \frac{d t}{t + 1}$ $cotxdx=-1/2 dt/(t+1)$

$I = - \frac{1}{2} \int \frac{t^{k}}{t + 1} d t$ $I = -1/2 int t^k/(t+1) dt$

$\frac{t^{k}}{t + 1} = k - 1 \sum i = 0 {(- 1)}^{i} t^{k - i - 1} + \frac{{(- 1)}^{k}}{t + 1}$ $t^k/(t+1) = sum_(i=0)^(k-1) (-1)^i t^(k-i-1) + (-1)^k/(t+1)$

$I = - \frac{1}{2} \int (k - 1 \sum i = 0 {(- 1)}^{i} t^{k - i - 1} + \frac{{(- 1)}^{k}}{t + 1}) d t$ $I = -1/2 int (sum_(i=0)^(k-1) (-1)^i t^(k-i-1) + (-1)^k/(t+1)) dt$

$I = - \frac{1}{2} (k - 1 \sum i = 0 {(- 1)}^{i} \int t^{k - i - 1} d t + {(- 1)}^{k} \int \frac{d t}{t + 1})$ $I = -1/2 (sum_(i=0)^(k-1) (-1)^i int t^(k-i-1)dt + (-1)^k int dt/(t+1))$

$I = - \frac{1}{2} (k - 1 \sum i = 0 {(- 1)}^{i} \frac{t^{k - i}}{k - i} + {(- 1)}^{k} ln | t + 1 |)$ $I = -1/2 (sum_(i=0)^(k-1) (-1)^i t^(k-i)/(k-i) + (-1)^k ln|t+1|)$

$I = - \frac{1}{2} ⎛ ⎝ k - 1 \sum i = 0 {(- 1)}^{i} \frac{{({cot}^{2} x)}^{k - i}}{k - i} + {(- 1)}^{k} ln ∣ ∣ {csc}^{2} x ∣ ∣ ⎞ ⎠$ $I = -1/2 (sum_(i=0)^(k-1) (-1)^i (cot^2x)^(k-i)/(k-i) + (-1)^k ln|csc^2x|)$

Let $n = 2 k$ $n=2k$ , then:

$\int {cot}^{n} x d x = \int {cot}^{2 k} x d x = \int {(cot x)}^{2 k} d x = I$ $int cot^nxdx = int cot^(2k)xdx = int (cotx)^(2k) dx = I$

$cot x = t \Rightarrow - {csc}^{2} x d x = d t \Rightarrow d x = - \frac{d t}{t^{2} + 1}$ $cotx=t => -csc^2xdx=dt =>dx=-dt/(t^2+1)$

$I = - \int \frac{t^{2 k}}{t^{2} + 1} d t$ $I = -int t^(2k)/(t^2+1)dt$

$\frac{t^{2 k}}{t^{2} + 1} = k - 1 \sum i = 0 {(- 1)}^{i} t^{2 (k - i - 1)} + \frac{{(- 1)}^{k}}{t^{2} + 1}$ $t^(2k)/(t^2+1) = sum_(i=0)^(k-1) (-1)^i t^(2(k-i-1)) + (-1)^k/(t^2+1)$

$I = - \int (k - 1 \sum i = 0 {(- 1)}^{i} t^{2 (k - i - 1)} + \frac{{(- 1)}^{k}}{t^{2} + 1}) d t$ $I = -int (sum_(i=0)^(k-1) (-1)^i t^(2(k-i-1)) + (-1)^k/(t^2+1)) dt$

$I = - (k - 1 \sum i = 0 {(- 1)}^{i} \int t^{2 (k - i - 1)} d t + {(- 1)}^{k} \int \frac{d t}{t^{2} + 1})$ $I = -(sum_(i=0)^(k-1) (-1)^i int t^(2(k-i-1)) dt + (-1)^k int dt/(t^2+1))$

$I = - (k - 1 \sum i = 0 {(- 1)}^{i} \frac{t^{2 (k - i - 1) + 1}}{2 (k - i - 1) + 1} + {(- 1)}^{k} arctan t)$ $I = -(sum_(i=0)^(k-1) (-1)^i t^(2(k-i-1)+1)/(2(k-i-1)+1) + (-1)^k arctant)$

$I = - (k - 1 \sum i = 0 {(- 1)}^{i} \frac{{(cot x)}^{2 (k - i) - 1}}{2 (k - i) - 1} + {(- 1)}^{k} arctan (cot x))$ $I = -(sum_(i=0)^(k-1) (-1)^i (cotx)^(2(k-i)-1)/(2(k-i)-1) + (-1)^k arctan(cotx))$

$I = - (k - 1 \sum i = 0 {(- 1)}^{i} \frac{{(cot x)}^{2 (k - i) - 1}}{2 (k - i) - 1} + {(- 1)}^{k} (x - \frac{π}{2}))$ $I = -(sum_(i=0)^(k-1) (-1)^i (cotx)^(2(k-i)-1)/(2(k-i)-1) + (-1)^k (x-pi/2))$

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