Let #I=int_0^1 9/(3+x^2)^2\ dx#. Using the substitution #x=sqrt(3) tan(theta)#, show that #I= sqrt(3) int_0^(pi/6) cos^2(theta)\ d theta#. What is the exact value of #I#?

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Jan 28, 2018

Answer:

#int_0^1 9/(3+x^2)^2\ dx=((2pi+3)sqrt(3))/24#

Explanation:

#\ \ \ \ \ \ int_0^1 9/(3+x^2)^2\ dx#

Substitute #x=sqrt(3)tan(theta)# and #dx=sqrt(3)sec^2(theta)d\theta#:
#=int_arctan(0)^arctan(1/sqrt(3))9/(3+(sqrt(3)tan(theta))^2)^2sqrt(3)sec^2(theta)\ d\theta#

#=sqrt(3)int_0^(pi/6) (9sec^2(theta))/(3+3tan^2(theta))^2\ d\theta#

#=sqrt(3)int_0^(pi/6) sec^2(theta)/(1+tan^2(theta))^2\ d\theta#

Using the fact that #tan^2(theta)+1=sec^2(theta)#:

#=sqrt(3)int_0^(pi/6) sec^2(theta)/sec^4(theta)\ d\theta#

#=sqrt(3)int_0^(pi/6) 1/sec^2(theta)\ d\theta#

Since #sec(theta)=1/cos(theta)#, we have

#=sqrt(3)int_0^(pi/6) cos^2(theta)\ d\theta#

Using the fact that #cos^2(theta)=(1+cos(2theta))/2#:

#=sqrt(3)/2int_0^(pi/6) 1+cos(2theta)\ d\theta#

#=sqrt(3)/2[theta+sin(2theta)/2]_0^(pi/6)#

#=sqrt(3)/2(pi/6+1/4)#

#=((2pi+3)sqrt(3))/24#

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