How do you integrate?
#int_sqrt3^2(sqrt(x^2-3))/xdx#
2 Answers
Explanation:
We could use trigonometric substitution to deal with this integral.
Draw a right triangle with angle
Label the hypotenuse as
Label the side adjacent to
Now we can write
Also, by the Pythagorean Theorem, the side of the triangle opposite to
Now let's write the integral in terms of
Now simplify
Use the identity
Now we can take the integral
Put it back in terms of
Now let's evaluate the endpoints:
In the given problem
#I=int\ sqrt(x^2−3)/xdx#
Substitute
Differentiating both sides with respect to respective variables we get
# du=x/sqrt(x^2−3)dx#
#=> dx=sqrt(x^2−3)/xdu#
Rewriting numerator as
#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
#I_2=int\ 1/((u/sqrt3)^2+1)du#
substitute
#dv=(du)/sqrt3#
#:.I_2=sqrt3int\ 1/(v^2+1)dv#
#=>I_2=sqrt3tan^-1v# #" (using standard integral")#
Making substitution back to
#I=u-sqrt3tan^-1(u/sqrt3)+C#
where#C# is a constant of integration.
Making substitution back to
#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#