How do you integrate #int sqrt(-x^2-10x)/xdx# using trigonometric substitution?

2 Answers
Eric S.
Mar 20, 2018

Use the substitution #x+5=5sintheta#.

Explanation:

Let

#I=intsqrt(-x^2-10x)/xdx#

Complete the square in the square root:

#I=intsqrt(25-(x+5)^2)/xdx#

Apply the substitution #x+5=5sintheta#:

#I=int(5costheta)/(5sintheta-5)(5costhetad theta)#

Simplify:

#I=5intcos^2theta/(sintheta-1)d theta#

Apply the identity #sin^2theta+cos^2theta=1#:

#I=5int(1-sin^2theta)/(sintheta-1)d theta#

Apply the difference of squares #a^2-b^2=(a-b)(a+b)#:

#I=-5int(1+sintheta)d theta#

Integrate directly:

#I=-5(theta-costheta)+C#

Reverse the substitution:

#I=sqrt(25-(x+5)^2)-5sin^(-1)((x+5)/5)+C#

Narad T.
Mar 20, 2018

The answer is #=-5arcsin(1/5(x+5))+5sqrt(1-((x+5)/5)^2)+C#

Explanation:

Complete the square :

#-x^2-10x=25-(x+5)^2#

Therefore, the integral is

#I=int(sqrt(-x^2-10x)dx)/(x)=int(sqrt(25-(x+5)^2)dx)/x#

Let #u=x+5#, #=>#, #du=dx#

#I=int(sqrt(25-u^2)du)/(u-5)#

Let #u=5sinv#, #=>#, #du=5cosvdv#

Therefore,

#I=int((5cosv)*5cosvdv)/(5(sinv-1))#

#=5int(cos^2vdv)/(-(1-sinv))#

#=-5int(1+sinv)dv#

#=-5v+5cosv#

#=-5arcsin(u/5)+5sqrt(1-(u/5)^2)#

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

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