How do you find the indefinite integral of ∫ dx/ (√5-x^2) ^3 ?

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Answer 1 · 2018-04-15T10:00:49

#=1/5*x/sqrt(5-x^2)+C#

#I=int1/(sqrt(5-x^2))^3dx#

#x=sqrt5sintheta#

#dx=sqrt5costheta*d(theta)#

Substitute

#int1/(sqrt(5-x^2))^3dx#=#int1/(sqrt(5-5sin^2theta))^3sqrt5costhetad(theta)#

#color(green) (5-5sin^2theta=5cos^2theta#

=#int1/((sqrt(5cos^2theta))^3)sqrt5costhetad(theta)#

Simplify

=#1/5int1/cos^2thetad(theta)#=#1/5intsec^2theta*d(theta)#

=#1/5tantheta+C#

#=1/5sintheta/costheta+C#

#=1/5sintheta/sqrt((1-sin^2theta))+C...to where,sintheta=x/sqrt5#

Reverse the substitution

#=1/5(x/sqrt5)/sqrt(1-(x/sqrt5)^2)+C#

#=1/5(x/sqrt5)/sqrt(1-x^2/5)+C#

#=1/5*x/sqrt(5-x^2)+C#