Evaluate the integral with hyperbolic or trigonometric substitution. ?

Question

736 views around the world

You can reuse this answer
Creative Commons License

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#