How do you integrate #int 1/(x^2sqrt(9-x^2))dx# using trigonometric substitution?

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Answer 1 · 2018-04-06T16:39:02

#int 1/(x^2sqrt(9-x^2))dx=-1/9(sqrt(9-x^2)/x)+c#

#I_1=int 1/(x^2sqrt(9-x^2))dx#

#x=3sinu=>dx=3cosudu#

#I_1=int1/((3sinu)^2sqrt(9-9sin^2u))3cosudu#

#I_1=int1/((9sin^2u)cancel(3sqrt(1-sin^2u)))cancel(3cosu)du#

#I_1-1/9.sin^2udu=1/9intcsc^2udu#

this si a satndard integral

#I_1=-1/9cotu+c#

#because sinu=x/3, cosu=(sqrt(9-x^2))/3#

by Pythagoras, this is left for the reader to verify

#:. -1/9cotu=-1/9(sqrt(9-x^2)/3)/(x/3)+c#

#=-1/9(sqrt(9-x^2)/x)+c#

#int 1/(x^2sqrt(9-x^2))dx=-1/9(sqrt(9-x^2)/x)+c#