How do I integrate this using trig substitution?

Question

#intsqrt(x^2-81)/x^4dx#

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Answer 1 · 2017-09-11T06:28:58

# (x^2-81)^(3/2)/(243x^3).#

Let, #I=intsqrt(x^2-9^2)/x^4dx.#

Let us use the Trigo. Substn. #x=9secy rArr dx=9secytanydy.#

Hence, #I=sqrt(x^2-9^2)/x^4dx,

#=intsqrt(9^2sec^2y-9^2)/(9^4sec^4y)*9secytanydy,#

#=1/9^2int(tanycos^4y)(secytany)dy,#

#=1/9^2inttan^2ycos^3ydy,#

#=1/9^2int(sin^2y/cos^2y*cos^3y)dy,#

#=1/9^2intsin^2ycosydy,#

#=1/9^2intt^2dt......[t=siny rArr cosydy=dt],#

#=1/9^2*t^3/3,#

#=1/9^2*1/3sin^3y,#

#=1/(9^2*3)(sin^2y)^(3/2),#

#=1/(9^2*3)(1-cos^2y)^(3/2),#

#=1/(9^2*3)(1-1/sec^2y)^(3/2),#

#=1/(9^2*3){1-9^2/x^2}^(3/2),#

#rArr I=(x^2-81)^(3/2)/(243x^3).#

Enjoy Maths.!