∫ (dx/(cot^3 x sin^7 x)) =?


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Question Number 178487 by cortano1 last updated on 17/Oct/22

$\int \frac{d x}{\cot^{3} x \sin^{7} x} = ?$
	Answered by Frix last updated on 18/Oct/22

	$\int \frac{d x}{\cot^{3} x \sin^{7} x} = \int \frac{d x}{\cos^{3} x \sin^{4} x} \overset{t = \sin x}{=}$ $= \int \frac{d t}{t^{4} {(t^{2} - 1)}^{2}} \underset{Method}{\overset{{Ostrogradski}^{'} s}{=}}$ $= - \frac{15 t^{4} - 10 t^{2} - 2}{6 t^{3} (t^{2} - 1)} + \frac{5}{4} \int (\frac{1}{t + 1} - \frac{1}{t + 1}) d t =$ $= - \frac{15 t^{4} - 10 t^{2} - 2}{6 t^{3} (t^{2} - 1)} + \frac{5 \ln \frac{t + 1}{t - 1}}{4} =$ $= \frac{{15 \sin}^{4} x - {10 \sin}^{2} x - 2}{{6 \cos}^{2} x \sin^{3} x} + \frac{5}{4} \ln \frac{1 + \sin x}{1 - \sin x} + C$
	Answered by Ar Brandon last updated on 17/Oct/22

	$I = \int \frac{d x}{\cot^{3} x \sin^{7} x} = \int \frac{d x}{\cos^{3} x \sin^{4} x} = \int \frac{\sec^{7} x}{\tan^{4} x} d x$ $= - \frac{\sec^{5} x}{{3 \tan}^{3} x} + \frac{5}{3} \int \frac{\sec^{5} x}{\tan^{2} x} d x = - \frac{\sec^{5} x}{{3 \tan}^{3} x} - \frac{{5 \sec}^{3} x}{3 \tan x} + 5 \int \sec^{3} x d x$ $= - \frac{\sec^{5} x}{3 \tan x} - \frac{{5 \sec}^{3} x}{3 \tan x} + 5 \int \frac{\cos x}{\cos^{4} x} d x = - \frac{\sec^{5} x}{3 \tan x} - \frac{{5 \sec}^{3} x}{3 \tan x} + 5 \int \frac{\cos x}{{(1 - \sin^{2} x)}^{2}} d x$ $= - \frac{\sec^{5} x}{3 \tan x} - \frac{{5 \sec}^{3} x}{3 \tan x} + \frac{5}{4} \int (\frac{1}{{(1 - \sin x)}^{2}} + \frac{2}{1 - \sin^{2} x} + \frac{1}{{(1 + \sin x)}^{2}}) d (\sin x)$ $= - \frac{\sec^{5} x}{3 \tan x} - \frac{{5 \sec}^{3} x}{3 \tan x} + \frac{5}{4} (\frac{1}{1 - \sin x} + 2 \tan x - \frac{1}{1 + \sin x}) + C$ $= - \frac{\sec^{5} x}{3 \tan x} - \frac{{5 \sec}^{3} x}{3 \tan x} + \frac{5}{2} \sec x \tan x + \frac{5}{2} \tan x + C$
	Answered by greougoury555 last updated on 18/Oct/22
	$let s=sin x I= ∫ (ds/((1−s^2 )^2 s^4 )) =∫ (ds/(((1−s^2 )s^2 )^2 )) Partial fractions (1/([(1−s^2 )s^2 ]^2 )) = ((1/(1−s^2 )) +(1/s^2 ))^2 = ((1/(1−s^2 )))^2 +(2/((1−s^2 )s^2 )) +(1/s^4 ) = [ (1/2)((1/(1−s)) +(1/(1+s)))]^2 +(2/(1−s^2 )) +(2/s^2 )+(1/s^4 ) = (1/(4(1−s)^2 )) +(5/(2(1−s^2 )))+(1/(4(1+s)^2 ))+(2/s^2 )+(1/s^4 ) I=∫ (ds/((1−s^2 )^2 s^4 )) = ∫ [ (1/(4(1−s)^2 )) +(5/(2(1−s^2 )))+(1/(4(1+s)^2 ))+(2/s^2 )+(1/s^4 ) ]ds = (1/(4(1−s)))+(5/4) ln ∣((1+s)/(1−s)) ∣−(1/(4(1+s)))−(2/s)−(1/(3s^3 )) +c = (1/(4(1−sin x))) +(5/4) ln ∣((1+sin x)/(1−sin x))∣−(1/(4(1+sin x)))−(2/(sin x))−(1/(3sin^3 x)) + c$
	$l e t s = \sin x$ $I = \int \frac{d s}{{(1 - s^{2})}^{2} s^{4}} = \int \frac{d s}{{((1 - s^{2}) s^{2})}^{2}}$ $P a r t i a l f r a c t i o n s$ $\frac{1}{{[(1 - s^{2}) s^{2}]}^{2}} = {(\frac{1}{1 - s^{2}} + \frac{1}{s^{2}})}^{2}$ $= {(\frac{1}{1 - s^{2}})}^{2} + \frac{2}{(1 - s^{2}) s^{2}} + \frac{1}{s^{4}}$ $= {[\frac{1}{2} (\frac{1}{1 - s} + \frac{1}{1 + s})]}^{2} + \frac{2}{1 - s^{2}} + \frac{2}{s^{2}} + \frac{1}{s^{4}}$ $= \frac{1}{4 {(1 - s)}^{2}} + \frac{5}{2 (1 - s^{2})} + \frac{1}{4 {(1 + s)}^{2}} + \frac{2}{s^{2}} + \frac{1}{s^{4}}$ $I = \int \frac{d s}{{(1 - s^{2})}^{2} s^{4}} = \int [\frac{1}{4 {(1 - s)}^{2}} + \frac{5}{2 (1 - s^{2})} + \frac{1}{4 {(1 + s)}^{2}} + \frac{2}{s^{2}} + \frac{1}{s^{4}}] d s$ $= \frac{1}{4 (1 - s)} + \frac{5}{4} \ln ∣ \frac{1 + s}{1 - s} ∣ - \frac{1}{4 (1 + s)} - \frac{2}{s} - \frac{1}{3 s^{3}} + c$ $= \frac{1}{4 (1 - \sin x)} + \frac{5}{4} \ln ∣ \frac{1 + \sin x}{1 - \sin x} ∣ - \frac{1}{4 (1 + \sin x)} - \frac{2}{\sin x} - \frac{1}{3 \sin^{3} x} + c$
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