# How to solve the integration question int1/(4+9x^2)^(1/2) using trigonometric substitution?

## $\int \frac{1}{4 + 9 {x}^{2}} ^ \left(\frac{1}{2}\right)$

Apr 11, 2017

$\int \frac{\mathrm{dx}}{4 + 9 {x}^{2}} ^ \left(\frac{1}{2}\right) = \frac{1}{3} \ln \left\mid \left(\sqrt{4 + 9 {x}^{2}} + 3 x\right) \right\mid + C$

#### Explanation:

Substitute:

$x = \frac{2}{3} \tan t$

$\mathrm{dx} = \frac{2 \mathrm{dt}}{3 {\cos}^{2} t} \mathrm{dt}$

As then $t = \arctan \left(\frac{3 x}{2}\right)$ we have that $t \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$

we have:

$\int \frac{\mathrm{dx}}{4 + 9 {x}^{2}} ^ \left(\frac{1}{2}\right) = \frac{2}{3} \int \frac{\mathrm{dt}}{\cos} ^ 2 t \frac{1}{4 + 9 \cdot \frac{4}{9} {\tan}^{2} t} ^ \left(\frac{1}{2}\right) = \frac{1}{3} \int \frac{\mathrm{dt}}{\cos} ^ 2 t \frac{1}{1 + {\tan}^{2} t} ^ \left(\frac{1}{2}\right)$

Use now the trigonometric identity:

$1 + {\tan}^{2} t = 1 + {\sin}^{2} \frac{t}{\cos} ^ 2 t = \frac{{\cos}^{2} t + {\sin}^{2} t}{\cos} ^ 2 t = \frac{1}{\cos} ^ 2 t$

$\int \frac{\mathrm{dx}}{4 + 9 {x}^{2}} ^ \left(\frac{1}{2}\right) = \frac{2}{3} \int \frac{\mathrm{dt}}{\cos} ^ 2 t \frac{1}{\frac{1}{\cos} ^ 2 t} ^ \left(\frac{1}{2}\right)$

For $t \in \left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$, $\cos t > 0$ so:

${\left(\frac{1}{\cos} ^ 2 t\right)}^{\frac{1}{2}} = \frac{1}{\cos} t$

and:

$\int \frac{\mathrm{dx}}{4 + 9 {x}^{2}} ^ \left(\frac{1}{2}\right) = \frac{2}{3} \int \frac{\mathrm{dt}}{\cos} ^ 2 t \frac{1}{\frac{1}{\cos} t} = \frac{2}{3} \int \frac{\cos t \mathrm{dt}}{\cos} ^ 2 t$

we do not simplify but substitute again:

$u = \sin t$

$\mathrm{du} = \cos t \mathrm{dt}$

${\cos}^{2} t = 1 - {\sin}^{2} t = 1 - {u}^{2}$

$\int \frac{\mathrm{dx}}{4 + 9 {x}^{2}} ^ \left(\frac{1}{2}\right) = \frac{2}{3} \int \frac{\mathrm{du}}{1 - {u}^{2}}$

which can be solved by partial fractions:

$\frac{1}{1 - {u}^{2}} = \frac{1}{\left(1 + u\right) \left(1 - u\right)} = \frac{1}{2} \frac{1}{1 + u} + \frac{1}{2} \frac{1}{1 - u}$

so:

$\int \frac{\mathrm{du}}{1 - {u}^{2}} = \frac{1}{2} \int \frac{\mathrm{du}}{1 - u} + \frac{1}{2} \int \frac{\mathrm{du}}{1 + u} = - \frac{1}{2} \ln \left\mid 1 - u \right\mid + \frac{1}{2} \ln \left\mid 1 + u \right\mid + C$

and using the properties of logarithms:

$\int \frac{\mathrm{du}}{1 - {u}^{2}} = \ln {\left\mid \frac{1 + u}{1 - u} \right\mid}^{\frac{1}{2}} + C$

Reversing the substitution we have:

$u = \sin t = \tan t \cos t = \tan \frac{t}{\sqrt{1 + {\tan}^{2} t}} = \frac{\frac{3}{2} x}{\sqrt{1 + {\left(\frac{3}{2} x\right)}^{2}}} = \frac{3 x}{\sqrt{4 + 9 {x}^{2}}}$

So:

$\int \frac{\mathrm{dx}}{4 + 9 {x}^{2}} ^ \left(\frac{1}{2}\right) = \frac{2}{3} \ln \sqrt{\frac{1 + \left(\frac{3 x}{\sqrt{4 + 9 {x}^{2}}}\right)}{1 - \left(\frac{3 x}{\sqrt{4 + 9 {x}^{2}}}\right)}} + C$

$\int \frac{\mathrm{dx}}{4 + 9 {x}^{2}} ^ \left(\frac{1}{2}\right) = \frac{2}{3} \ln \sqrt{\frac{\sqrt{4 + 9 {x}^{2}} + 3 x}{\sqrt{4 + 9 {x}^{2}} - 3 x}} + C$

Rationalize the denominator of the argument:

$\int \frac{\mathrm{dx}}{4 + 9 {x}^{2}} ^ \left(\frac{1}{2}\right) = \frac{2}{3} \ln \sqrt{\frac{\left(\sqrt{4 + 9 {x}^{2}} + 3 x\right) \left(\sqrt{4 + 9 {x}^{2}} + 3 x\right)}{\left(\sqrt{4 + 9 {x}^{2}} - 3 x\right) \left(\sqrt{4 + 9 {x}^{2}} + 3 x\right)}} + C$

int dx/(4+9x^2)^(1/2) = 2/3 ln sqrt ( ((sqrt(4+9x^2)+3x)( sqrt(4+9x^2) + 3x))/ ( (4+9x^2-9x^2))+ C

$\int \frac{\mathrm{dx}}{4 + 9 {x}^{2}} ^ \left(\frac{1}{2}\right) = \frac{2}{3} \ln \frac{\sqrt{{\left(\sqrt{4 + 9 {x}^{2}} + 3 x\right)}^{2}}}{\sqrt{\left(4 + 9 {x}^{2} - 9 {x}^{2}\right)}} + C$

$\int \frac{\mathrm{dx}}{4 + 9 {x}^{2}} ^ \left(\frac{1}{2}\right) = \frac{2}{3} \ln \frac{\sqrt{{\left(\sqrt{4 + 9 {x}^{2}} + 3 x\right)}^{2}}}{\sqrt{4}} + C$

$\int \frac{\mathrm{dx}}{4 + 9 {x}^{2}} ^ \left(\frac{1}{2}\right) = \frac{1}{3} \ln \left\mid \left(\sqrt{4 + 9 {x}^{2}} + 3 x\right) \right\mid + C$