# Evaluate integral 2,1 (1/x^2 squareroot 4x^2+9) using trig substitution

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To solve the integral `\int_2^1 1/{x^2\sqrt{4x^2+9}} dx`, we need to use the trigonometric substitution `x=3/2 tan u`. This can be differentiated to find `dx=3/2 sec^2 u du` and also from the trig identity `tan^u+1=sec^2u`, we see that `\sqrt{4x^2+9}=3 sec u`.

This means that the integral becomes

`\int_{x=2}^{x=1} 1/3 \cdot 3/2 \cdot 4/9 \cdot {sec^2 u du}/{tan^u sec u}`

`=2/9\int_{x=2}^{x=1} {sec u du}/{tan^2 u}` now simplify the trig components

`=2/9\int_{x=2}^{x=1} {cos u du}/{sin^2 u}` now have the new subs `z=sin u`, so `dz=cos u du`.

`=2/9\int_{x=2}^{x=1} {dz}/{z^2}`

`=2/9(- 1/z|_{x=2}^{x=1}`

`=2/9(- 1/{sin u}|_{x=2}^{x=1}`

To evaluate the limits, note that when `x=1`, then `tan u = 2/3`. This is the same as `sin u =2/\sqrt 13` by the Pythagorean Theorem. Also, when `x=2`, then `tan u = 4/3`, which is the same as `sin u =4/5`.

Putting these values into the limits of the integral gives a final value of

`{5-2\sqrt 13}/18`

∫

Let x=(3/2)tanU

1 = 3/2*sec^2U * dU/dx

dx = (3/2)sec^2(U)*dU

x^2 = 9/4 tan^2(U) = 9/4(sec^2(U)-1)

4x^2+9 = 4*((3/2)tanU)^2+9

= 9(1+tan^2(U))

= 9*sec^2(U)

sqrt(4x^2+9) = 3secU

1/x^2 squareroot 4x^2+9 = 1/[x^2*squareroot (4x^2+9)]

= 1/[9/4(sec^2(U)-1)*3secU

∫1/x^2 squareroot 4x^2+9 dx

=∫1/[9/4(sec^2(U)-1)*3secU] *(3/2)sec^2(U)*dU

= ∫1/[(3/2)(sec^2(U)-1)] *sec(U)*dU

secU/(sec^2(U)-1) = (1/cosU)/[(1-cos^2(U))/cos^2(U)]

= cosU/sin^2(U)

= cosec U*cotU

∫1/x^2 squareroot 4x^2+9 dx

=∫(2/3)*cosec U*cotU dU

= -(2/3) [cosecU](2,1)

When x=2 then tanU=4/3

When x=1 then tanU = 2/3

cosecU = sqrt(cot^2(U)+1) = sqrt[(1+tan^2(U))/tan^2(U)]

∫1/x^2 squareroot 4x^2+9 dx

= -(2/3) [cosecU] limits (arctan(4/3),arctan(2/3))

=-(2/3)sqrt[(1+tan^2(U))/tan^2(U)] limits (arctan(4/3),arctan(2/3))

= -(2/3) {[sqrt(1+16/9)/(16/9)]-sqrt[(1+4/9)/(4/9)]}

**= -(2/3)[5/4-sqrt13/2]**