EVALUATE THE DEFINITE INTEGRAL 2 ∫dx/(x^2(√4x^2 + 9)) 1 INTEGRATION, SINGLE VARIABLE CALCULUS,TRIGONOMETRIC SUBSTITUTION

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thilina-g | College Teacher | (Level 1) Educator

Posted on

`y = int_1^2 1/(x^2sqrt(4x^2+9)) dx`

First we will integrate the indefinite integral as below.

`y = int1/(x^2sqrt(4x^2+9)) dx`

Let `x = 3/2tan(u)`

Then,

`dx = 3/2 sec^2(u) du`

Now we can change the integral as below.

`y = int(1/((3/2tan(u))^2sqrt(4(3/2tan(u))^2+9))) 3/2sec^2(u)du`

`y = 2/9int(sec^2(u))/(tan^2(u)sqrt(tan^2(u)+1)) du`

`y = 2/9int(sec^2(u))/(tan^2(u)sec(u)) du`

 `y = 2/9int(1/(cos^2(u)))/((sin^2(u))/(cos^2(u))xx 1/cos(u)) du`

`y = 2/9 intcos(u)/(sin^2(u)) du`

We know,

`d(sin(u)) = cos(u) du`

Therefore,

`y = 2/9 int1/(sin^2(u)) d(sin(u))`

`y = 2/9[(-1)/sin(u)] + c`

Where c is an arbitrary constant.

`y = -2/(9sin(u)) +c`

`x = 3/2 tan(u)`

`tan(u) = (2x)/3`

`u = tan^(-1)(2x)/3`

 

Therefore,

`y = -2/(9sin(tan^(-1)(2x)/3))+c`

Therefore,

`int_1^2 1/(x^2sqrt(4x^2+9)) dx = y(2) -y(1)`

`y(2) = -2/(9sin(tan^(-1)(4)/3))+c`

`y(2) = -2/(9xx0.8)+c`

`y(2) = -5/18 +c`

`y(1) = -2/(9sin(tan^(-1)(2)/3))+c`

`y(1) = -2/(9 xx 0.5547) + c`

`y(1) = -0.4 + c`

 

Therefore,

`int_1^2 1/(x^2sqrt(4x^2+9)) dx = -5/18 +c - (-0.4+c)`

`int_1^2 1/(x^2sqrt(4x^2+9)) dx = 0.4 -5/18 = 0.12222`

 

Therefore the value of the integral is 0.12222

 

The value I got from calculator is 0.12284. Therefore this is correct.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You should factor out 9 under the square root such that:

`int_1^2 (dx)/(x^2sqrt((4/9)x^2+1))`

You should come up with the substitution `x = (3/2)(cot x) =gt x^2 = (9/4)cot^2 x.`

Differentiating`x = (3/2)(cot x)`  with respect to x yields:

`dx = -(3/2)(1/(sin^2 x))`

`int_1^2 (dx)/(x^2sqrt((4/9)x^2+1)) = int_1^2 (-(3/2)(1/sin^2 x)dx)/(((9/4)cot^2 x)sqrt(cot^2 x+1))`

You ned to substitute `1/(sin^2 x)`  for `(cot^2 x+1)`  such that:

`int_1^2 (-(3/2)(1/sin^2 x)dx)/(((9/4)cot^2 x)sqrt(cot^2 x+1)) =int_1^2 (-(3/2)(1/sin^2 x))/(((9/4)cot^2 x)sqrt(1/(sin^2 x)))`

`int_1^2 (-(1/sin^2 x))/(((3/2)cot^2 x)(1/(sin x)))`

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

You need to remember that `cot x = cos x/sin x`

`int_1^2 (-(1/sin x))/((3/2)(cos^2 x)/(sin^2 x))`

`int_1^2 (-dx)/((3/2)(cos^2 x)/(sin x))`

`(2/3)int_1^2 (-sin x dx)/(cos^2 x)`

You need to substitute cos x by y such that:

`cos x = y =gt - sin x dx = dy`

You need to change the limits of integration such that:

`x = 1 =gt cos 1 = y`

`x = 2 =gt cos 2 = y`

`(2/3)int_(cos 1)^(cos 2) (dy)/(y^2) = (-2/3)(1/y)|_(cos 1)^(cos 2)`

`int_1^2 (dx)/(x^2sqrt((4/9)x^2+1)) = (-2/3)(1/cos 2 - 1/cos 1)`

Hence, evaluating the given definite integral yields `int_1^2 (dx)/(x^2sqrt((4/9)x^2+1)) = (2/3)(cos 2 - cos 1)/(cos 1*cos 2).`

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