How do you integrate?

#int_sqrt3^2(sqrt(x^2-3))/xdx#

2 Answers
Mar 26, 2018

#int_sqrt3^2(sqrt(x^2-3))/xdx=1-(pisqrt3)/6#

Explanation:

We could use trigonometric substitution to deal with this integral.

Draw a right triangle with angle #theta#.

Label the hypotenuse as #x#.

Label the side adjacent to #theta# as #sqrt3#

Now we can write

#sectheta=x/sqrt3#

#rArrcolor(green)(x=sqrt3sectheta)#

#rArrcolor(red)(dx=sqrt3secthetatantheta# #color(red)d##color(red)theta#

Also, by the Pythagorean Theorem, the side of the triangle opposite to #theta# is #sqrt(x^2-3)#. So we can write

#tantheta=sqrt(x^2-3)/sqrt3#

#rArrcolor(blue)(sqrt(x^2-3)=sqrt3tantheta)#

Now let's write the integral in terms of #theta#

#int_sqrt3^2color(blue)sqrt(x^2-3)/color(green)xcolor(red)dxrArrint_sqrt3^2color(blue)(sqrt3tantheta)/color(green)(sqrt3sectheta)color(red)(sqrt3secthetatantheta# #color(red)d##color(red)theta)#

Now simplify

#rArrint_sqrt3^2(sqrt3tantheta)/cancel(sqrt3sectheta)cancel(sqrt3sectheta)tantheta# #d##theta#

#rArrint_sqrt3^2sqrt3tan^2theta# #d##theta#

Use the identity #sec^2theta=1+tan^2theta#

#rArrsqrt3int_sqrt3^2(sec^2theta-1)##d##theta#

Now we can take the integral

#rArrsqrt3[tantheta-theta]_sqrt3^2#

Put it back in terms of #x#

#rArrsqrt3[(sqrt(x^2-3))/sqrt3-arctan((sqrt(x^2-3))/sqrt3)]_sqrt3^2#

Now let's evaluate the endpoints:

#rArrsqrt3[(sqrt(4-3)/sqrt3-arctan(sqrt(4-3)/sqrt3))-(sqrt(3-3)/sqrt3-arctan(sqrt(3-3)/sqrt3))]#

#rArrsqrt3[(1/sqrt3-arctan(1/sqrt3))-(0-arctan(0))]#

#rArrsqrt3[1/sqrt3-pi/6]#

#rArr1-(pisqrt3)/6#

Mar 26, 2018

In the given problem

#I=int\ sqrt(x^2−3)/xdx#

Substitute #u=sqrt(x^2−3)#
Differentiating both sides with respect to respective variables we get

# du=x/sqrt(x^2−3)dx#
#=> dx=sqrt(x^2−3)/xdu#

#:.I=int\ (sqrt(x^2−3)/x)(sqrt(x^2−3)/xdu)#
#=>int\ u^2/(u^2+3)du#

Rewriting numerator as #(u^2+3)-3# and simplifying we get

#I=int\ (1-3/(u^2+3))du#
#I=I_1-I_2=int\ du-int3/(u^2+3)du#

Now

#I_1=u#

and by dividing both numerator and denominator with #3,##I_2# can be rewritten as

#I_2=int\ 1/((u/sqrt3)^2+1)du#

substitute #v=u/sqrt3# to make it a standard integral. We get

#dv=(du)/sqrt3#
#:.I_2=sqrt3int\ 1/(v^2+1)dv#
#=>I_2=sqrt3tan^-1v# #" (using standard integral")#

Making substitution back to #u# we get

#I=u-sqrt3tan^-1(u/sqrt3)+C#
where #C# is a constant of integration.

Making substitution back to #x# and applying limits we get

#I=[sqrt(x^2−3)-sqrt3tan^-1((sqrt(x^2−3))/sqrt3)+C]_sqrt3^2#
#I=[sqrt(2^2−3)-sqrt3tan^-1((sqrt(2^2−3))/sqrt3)+C]-[sqrt((sqrt3)^2−3)-sqrt3tan^-1((sqrt((sqrt3)^2−3))/sqrt3)+C]#
#=>I=1-sqrt3tan^-1(1/sqrt3)#
#=>I=1-sqrt3pi/6#