How do you integrate $\int \frac{1}{\sqrt{4 x^{2} + 12 x - 5}}$ $int 1/sqrt(4x^2+12x-5)$ using trigonometric substitution?

Question

How do you integrate $\int \frac{1}{\sqrt{4 x^{2} + 12 x - 5}}$ $int 1/sqrt(4x^2+12x-5)$ using trigonometric substitution?

1 Answer

Impact of this question

4535 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-04T21:47:32

$\int \frac{d x}{\sqrt{4 x^{2} + 12 x - 5}} = \frac{1}{2} ln ∣ ∣ 2 x + 3 + \sqrt{4 x^{2} + 12 x - 5} ∣ ∣ + C$ $int dx/sqrt(4x^2+12x-5) = 1/2 ln abs (2x+3+sqrt(4x^2+12x-5))+C$

Explanation:

Complete the square at the denominator:

$\int \frac{d x}{\sqrt{4 x^{2} + 12 x - 5}} = \int \frac{d x}{\sqrt{{(2 x + 3)}^{2} - 14}} = \frac{1}{\sqrt{14}} \int \frac{d x}{\sqrt{{(\frac{2 x + 3}{\sqrt{14}})}^{2} - 1}}$ $int dx/sqrt(4x^2+12x-5) = int dx/sqrt((2x+3)^2-14) = 1/sqrt14 int dx/sqrt(((2x+3)/sqrt14)^2-1)$

Substitute for $\frac{2 x + 3}{\sqrt{14}} > 1$ $(2x+3)/sqrt 14 > 1$

$\frac{2 x + 3}{\sqrt{14}} = sec t$ $(2x+3)/sqrt14 = sect$

$d x = \frac{\sqrt{14}}{2} sec t tan t d t$ $dx = sqrt14/2 sect tantdt$

with $t \in (0, \frac{π}{2})$ $t in (0,pi/2)$

so that:

$\int \frac{d x}{\sqrt{4 x^{2} + 12 x - 5}} = \frac{1}{2} \int \frac{sec t tan t d t}{\sqrt{{sec}^{2} t - 1}}$ $int dx/sqrt(4x^2+12x-5) = 1/2 int (sect tant dt)/sqrt(sec^2t -1)$

Now:

$\sqrt{{sec}^{2} t - 1} = \sqrt{\frac{1}{{cos}^{2} t} - 1} = \sqrt{\frac{1 - {cos}^{2} t}{{cos}^{2} t}} = \sqrt{\frac{{sin}^{2} t}{{cos}^{2} t}} = \sqrt{{tan}^{2} t} = tan t$ $sqrt(sec^2t -1) = sqrt(1/cos^2t -1) = sqrt((1-cos^2t)/cos^2t) = sqrt(sin^2t/cos^2t) = sqrt(tan^2t) = tant$

as for $t \in (0, \frac{π}{2})$ $t in (0,pi/2)$ the tangent is positive.

So:

$\int \frac{d x}{\sqrt{4 x^{2} + 12 x - 5}} = \frac{1}{2} \int \frac{sec t tan t d t}{tan t}$ $int dx/sqrt(4x^2+12x-5) = 1/2 int (sect tant dt)/tant$

$\int \frac{d x}{\sqrt{4 x^{2} + 12 x - 5}} = \frac{1}{2} \int sec t d t$ $int dx/sqrt(4x^2+12x-5) = 1/2 int sect dt$

$\int \frac{d x}{\sqrt{4 x^{2} + 12 x - 5}} = \frac{1}{2} ln | sec t + tan t | + C$ $int dx/sqrt(4x^2+12x-5) = 1/2 ln abs (sect+tant)+C$

$\int \frac{d x}{\sqrt{4 x^{2} + 12 x - 5}} = \frac{1}{2} ln ∣ ∣ sec t + \sqrt{{sec}^{2} t - 1} ∣ ∣ + C$ $int dx/sqrt(4x^2+12x-5) = 1/2 ln abs (sect+sqrt(sec^2t-1))+C$

Undoing the substitution:

$\int \frac{d x}{\sqrt{4 x^{2} + 12 x - 5}} = \frac{1}{2} ln ∣ ∣ ∣ ∣ \frac{2 x + 3}{\sqrt{14}} + \sqrt{{(\frac{2 x + 3}{\sqrt{14}})}^{2} - 1} ∣ ∣ ∣ ∣ + C$ $int dx/sqrt(4x^2+12x-5) = 1/2 ln abs ((2x+3)/sqrt14+sqrt(((2x+3)/sqrt14)^2-1))+C$

$\int \frac{d x}{\sqrt{4 x^{2} + 12 x - 5}} = \frac{1}{2} ln ∣ ∣ 2 x + 3 + \sqrt{4 x^{2} + 12 x - 5} ∣ ∣ + C$ $int dx/sqrt(4x^2+12x-5) = 1/2 ln abs (2x+3+sqrt(4x^2+12x-5))+C$

and by differentiating we can verify that the solution is valid also for $\frac{2 x + 3}{\sqrt{14}} < - 1$ $(2x+3)/sqrt14 < -1$ .

Science

Math

Humanities

... and beyond

How do you integrate $\int \frac{1}{\sqrt{4 x^{2} + 12 x - 5}}$ $int 1/sqrt(4x^2+12x-5)$ using trigonometric substitution?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you integrate int 1/sqrt(4x^2+12x-5) ∫1√4x2+12x−5int 1/sqrt(4x^2+12x-5) using trigonometric substitution?

1 Answer

Explanation:

Related questions

Impact of this question

How do you integrate $\int \frac{1}{\sqrt{4 x^{2} + 12 x - 5}}$ $int 1/sqrt(4x^2+12x-5)$ using trigonometric substitution?