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

lfryerda | High School Teacher | (Level 2) Educator

Posted on

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

jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted on

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]