Evaluate the integral with hyperbolic or trigonometric substitution. ?

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Answer 1 · 2018-03-29T03:21:43

$int1/sqrt(x^6-x^4)dx=cos(sec^-1x)+c$

$I=int1/sqrt(x^6-x^4)dx$

Let

$x=sect$

$dx/dt=sect tant$

$dx=sect tant dt$

$sqrt(x^6-x^4)=sqrt(sec^6t-sec^4t)$

$=sqrt(sec^4t(sec^2t-1))$

$sqrt(x^6-x^4)=sec^2t tant$

$I=int1/sqrt(x^6-x^4)dx$

$I=int(sect tant dt)/(sec^2t tant)$

$I=intcostdt$

$I=sint$

$x=sect$

$t=sec^-1x$

$I=cos(sec^-1x)+c$

$int1/sqrt(x^6-x^4)dx=cos(sec^-1x)+c$