How do you integrate #int 1/(xsqrt(3 + x^2))dx# using trigonometric substitution?

Question

Christopher Agresta · Accepted Answer

The answer is #=ln(|x/(sqrt(3+x^2)+sqrt3)|)+C#

The denominator is

#xsqrt(3+x^2)=xsqrt(3(1+(x/sqrt3)^2))#

#=sqrt3xsqrt((1+(x/sqrt3)^2))#

Perform the substitution

#x/sqrt3=tantheta#

#dx/sqrt3=sec^2theta d theta#

#sqrt((1+(x/sqrt3)^2))=sqrt(1+tan^2theta)=sqrt(sec^2theta)=sectheta#

Therefore, the integral is

#I=int(dx)/(xsqrt(3+x^2))=int(sqrt3sec^2theta d theta)/(sqrt3 tan thetasectheta)#

#=int(secthetad theta)/tan theta#

#=int(d theta)/sin theta#

#=int(csctheta d theta)#

#=int(csctheta(csctheta+cottheta)d theta)/(csctheta+cottheta)#

#=int((csc^2theta+csc thetacottheta)d theta)/(csctheta+cottheta)#

Perform the substitution

#v=csctheta+cottheta#, #=>#, #dv=(-cotthetacsctheta-csc^2theta)d theta#

Therefore,

#I=int(-dv)/(v)#

#=-ln(v)#

#=-ln(csctheta+cottheta)#

#tan theta=x/sqrt3#, #=>#, #csctheta=sqrt(3+x^2)/x#

#tan theta=x/sqrt3#, #=>#, #cottheta=sqrt(3)/x#

Finally,

#I=-ln(sqrt(3+x^2)/x+sqrt(3)/x)#

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

#=ln(|x/(sqrt(3+x^2)+sqrt3)|)+C#

Elijah Abram · Answer

undefined To integrate ∫ 1 / (x√(3 + x^2)) dx using trigonometric substitution, we can let x = √3 tan(θ), where θ is the angle in the first quadrant such that tan(θ) = x / √3. Then, dx = √3 sec^2(θ) dθ. After substitution and simplification, the integral becomes ∫ sec(θ) dθ. This integrates to ln|sec(θ) + tan(θ)| + C. Finally, substitute back for θ in terms of x to get the final result.