By making a suitable substitution, evaluate the integral

`int_1^2(x^3)/sqrt(x^2-1)dx`

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Let `x=sec(theta) `

Then

`dx=sec(theta) tan(theta) d(theta)`

When `x=1` then `theta=0`

when `x=2` then `theta=(pi/3)`

`int_1^2 (x^3)/sqrt(x^2-1)dx`

`=int_0^(pi/3)(sec^3 theta xx sec theta xx tan theta)/tan(theta)d(theta)`

`=int_0^(pi/3) sec^4 theta d(theta)`

`=int_0^(pi/3) (1+tan^2 theta)sec^2theta d(theta)`

We know that;

`(d(tantheta))/(d(theta)) = sec^2theta`

`d(tantheta) = sec^2thetad(theta)`

`int_1^2 (x^3)/sqrt(x^2-1)dx`

`= int_0^(pi/3) (1+tan^2theta)d(tantheta) ` same as `int(1+x^2)dx`

`=[tantheta+(tan^3 (theta))/3]_0^(pi/3)`

`=sqrt(3)+(3sqrt(3))/3`

`=2sqrt(3)`

*So the answer is;*

`int_1^2 (x^3)/sqrt(x^2-1)dx = 2sqrt3`

**Sources:**

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