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Calculate-9-x-2-x-6-dx-




Question Number 76892 by john santu last updated on 31/Dec/19
Calculate ∫ ((√(9−x^2 ))/x^6 ) dx .
Calculate9x2x6dx.
Answered by jagoll last updated on 31/Dec/19
Answered by MJS last updated on 01/Jan/20
∫((√(9−x^2 ))/x^6 )dx=       [t=(x/( (√(9−x^2 )))) → dx=(((9−x^2 )^(3/2) )/9)dt]  =(1/(81))∫(dt/t^4 )+(1/(81))∫(dt/t^6 )=−(1/(243t^3 ))−(1/(405t^5 ))=  =−(((9−x^2 )^(3/2) )/(243x^3 ))−(((9−x^2 )^(5/2) )/(405x^5 ))=−(((2x^2 +27)(9−x^2 )^(3/2) )/(1215x^5 ))+C
9x2x6dx=[t=x9x2dx=(9x2)329dt]=181dtt4+181dtt6=1243t31405t5==(9x2)32243x3(9x2)52405x5=(2x2+27)(9x2)321215x5+C
Commented by jagoll last updated on 01/Jan/20
sorry sir. the equation ∫ ((√(9−x^2 ))/x^6 ) dx .
sorrysir.theequation9x2x6dx.
Commented by MJS last updated on 01/Jan/20
thank you, it′s just a typo, I corrected it
thankyou,itsjustatypo,Icorrectedit
Commented by jagoll last updated on 01/Jan/20
thanks you sir
thanksyousir
Answered by petrochengula last updated on 03/Jan/20
let x=3sinθ⇒dx=3cosθdθ  =∫((√(9−9sin^2 θ))/(3^6 sin^6 θ))3cosθdθ=(1/3^4 )∫((cos^2 θ)/(sin^6 θ))dθ  =(1/(81))∫cot^2 θcosec^4 θdθ  =(1/(81))∫cot^2 θ(1+cot^2 θ)cosec^2 θdθ  let t=cotθ⇒−dt=cosec^2 θdθ  =−(1/(81))∫t^2 (1+t^2 )dt  =−(t^3 /(243))−(t^5 /(405))+c  =−((cot^3 θ)/(243))−((cot^5 θ)/(405))+c  =−(((cot(sin^(−1) ((x/3))))^3 )/(243))−(((cot(sin^(−1) ((x/3))))^5 )/(405))+c
letx=3sinθdx=3cosθdθ=99sin2θ36sin6θ3cosθdθ=134cos2θsin6θdθ=181cot2θcosec4θdθ=181cot2θ(1+cot2θ)cosec2θdθlett=cotθdt=cosec2θdθ=181t2(1+t2)dt=t3243t5405+c=cot3θ243cot5θ405+c=(cot(sin1(x3)))3243(cot(sin1(x3)))5405+c

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