Let I= integral 1 to 0 9÷(3+x^2)^2 dx
(i) Using the substitution x=(root 3) tan theta, show that I= (root 3)
Integral 1÷6pie to 0 cos^2 theta dtheta.
(ii) Hence find the exact value of I
Can someone please solve this question?

Question

Let I= integral 1 to 0 9÷(3+x^2)^2 dx
(i) Using the substitution x=(root 3) tan theta, show that I= (root 3) 
Integral 1÷6pie to 0 cos^2 theta dtheta.
(ii) Hence find the exact value of I
Can someone please solve this question?

Charles Arocha · Accepted Answer

Kindly see the Explanation.

We have, #I=int_1^0 9/(3+x^2)^2dx#. Part (i) : We use the subst. #x=sqrt3tantheta," so that, "dx=sqrt3sec^2thetad theta#. Also, #x=1rArrsqrt3tantheta=1rArrtheta=pi/6, and, # #x=0rArrtantheta=0/sqrt3rArrtheta=0#. #:. I=int_(pi/6)^0 9/(3+3tan^2theta)^2sqrt3sec^2thetad theta#, #=int_(pi/6)^0 (9sqrt3)/(3sec^2theta)^2sec^2thetad theta#, #=int_(pi/6)^0 (9sqrt3)/(9sec^4theta)sec^2thetad theta#, #rArr I=sqrt3int_(pi/6)^o cos^2thetad theta#. Part (ii) : Recall that, #cos^2theta=(1+cos2theta)/2#. #:. I=1/2*sqrt3int_(pi/6)^o (1+cos2theta)d theta#, #=sqrt3/2[theta+(sin2theta)/2]_(pi/6)^0#, #=sqrt3/2[theta+sinthetacostheta]_(pi/6)^0#, #=sqrt3/2[0+0-(pi/6+sin(pi/6)cos(pi/6)]#, #=sqrt3/2(pi/6+1/2*sqrt3/2)#. # rArr I=sqrt3/24(2pi+3sqrt3)#.

Let I= integral 1 to 0 9÷(3+x^2)^2 dx (i) Using the substitution x=(root 3) tan theta, show that I= (root 3) Integral 1÷6pie to 0 cos^2 theta dtheta. (ii) Hence find the exact value of I Can someone please solve this question?