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dx-cot-3-x-sin-7-x-

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Question Number 178487 by cortano1 last updated on 17/Oct/22
∫ (dx/(cot^3 x sin^7 x)) =?
∫dxcot3xsin7x=?
Answered by Frix last updated on 18/Oct/22
∫(dx/(cot^3 x sin^7 x))=∫(dx/(cos^3 x sin^4 x))=^(t=sin x) =∫(dt/(t^4 (t^2 −1)^2 ))=_(Method) ^(Ostrogradski′s) =−((15t^4 −10t^2 −2)/(6t^3 (t^2 −1)))+(5/4)∫((1/(t+1))−(1/(t+1)))dt= =−((15t^4 −10t^2 −2)/(6t^3 (t^2 −1)))+((5ln ((t+1)/(t−1)))/4)= =((15sin^4 x −10sin^2 x −2)/(6cos^2 x sin^3 x))+(5/4)ln ((1+sin x)/(1−sin x)) +C
∫dxcot3xsin7x=∫dxcos3xsin4x=t=sinx=∫dtt4(t2−1)2=Ostrogradski′sMethod=−15t4−10t2−26t3(t2−1)+54∫(1t+1−1t+1)dt==−15t4−10t2−26t3(t2−1)+5lnt+1t−14==15sin4x−10sin2x−26cos2xsin3x+54ln1+sinx1−sinx+C
Answered by Ar Brandon last updated on 17/Oct/22
I=∫(dx/(cot^3 xsin^7 x))=∫(dx/(cos^3 xsin^4 x))=∫((sec^7 x)/(tan^4 x))dx =−((sec^5 x)/(3tan^3 x))+(5/3)∫((sec^5 x)/(tan^2 x))dx=−((sec^5 x)/(3tan^3 x))−((5sec^3 x)/(3tanx))+5∫sec^3 xdx =−((sec^5 x)/(3tanx))−((5sec^3 x)/(3tanx))+5∫((cosx)/(cos^4 x))dx=−((sec^5 x)/(3tanx))−((5sec^3 x)/(3tanx))+5∫((cosx)/((1−sin^2 x)^2 ))dx =−((sec^5 x)/(3tanx))−((5sec^3 x)/(3tanx))+(5/4)∫((1/((1−sinx)^2 ))+(2/(1−sin^2 x))+(1/((1+sinx)^2 )))d(sinx) =−((sec^5 x)/(3tanx))−((5sec^3 x)/(3tanx))+(5/4)((1/(1−sinx))+2tanx−(1/(1+sinx)))+C =−((sec^5 x)/(3tanx))−((5sec^3 x)/(3tanx))+(5/2)secxtanx+(5/2)tanx+C
I=∫dxcot3xsin7x=∫dxcos3xsin4x=∫sec7xtan4xdx=−sec5x3tan3x+53∫sec5xtan2xdx=−sec5x3tan3x−5sec3x3tanx+5∫sec3xdx=−sec5x3tanx−5sec3x3tanx+5∫cosxcos4xdx=−sec5x3tanx−5sec3x3tanx+5∫cosx(1−sin2x)2dx=−sec5x3tanx−5sec3x3tanx+54∫(1(1−sinx)2+21−sin2x+1(1+sinx)2)d(sinx)=−sec5x3tanx−5sec3x3tanx+54(11−sinx+2tanx−11+sinx)+C=−sec5x3tanx−5sec3x3tanx+52secxtanx+52tanx+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
lets=sinxI=∫ds(1−s2)2s4=∫ds((1−s2)s2)2Partialfractions1[(1−s2)s2]2=(11−s2+1s2)2=(11−s2)2+2(1−s2)s2+1s4=[12(11−s+11+s)]2+21−s2+2s2+1s4=14(1−s)2+52(1−s2)+14(1+s)2+2s2+1s4I=∫ds(1−s2)2s4=∫[14(1−s)2+52(1−s2)+14(1+s)2+2s2+1s4]ds=14(1−s)+54ln∣1+s1−s∣−14(1+s)−2s−13s3+c=14(1−sinx)+54ln∣1+sinx1−sinx∣−14(1+sinx)−2sinx−13sin3x+c

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