............. I
When x=3,t=103Squaring both side, we get
⇒x2+1x2=t2−2.............. II
Differentiating both sides of equation I, we get
⇒(1−1x2)dx=dt
⇒(x−1x)dxx=dt................. III
Now, (x−1x)2=x2+1x2−2
=t2−2−2
=t2−4
⇒(x−1x)=√t2−4................. IV
From equation III and IV, we get
⇒dxx=dt√t2−4.............. V
Substituting equations I, II and V in the equation given in question and changing the limits accordingly, we get
I=∫10/310/3tan(t2−2)dt√t2−4sint
=0 [∵ upper and lower limits of integration become equal]
Hence, the answer is 0