# How do you integrate int 1/sqrt(2x-12sqrtx-5)  using trigonometric substitution?

May 2, 2018

I=sqrt(2x-12sqrtx-5)+3sqrt2ln|sqrt(2x)-3sqrt2+sqrt(2x-12sqrtx- 5)|+C,
There is no any trig.function in the answer .So it is difficult to proceed with trig substitution.

#### Explanation:

Here,

$I = \int \frac{1}{\sqrt{2 x - 12 \sqrt{x} - 5}} \mathrm{dx}$

Let, $\sqrt{x} = t \implies x = {t}^{2} \implies \mathrm{dx} = 2 t \mathrm{dt}$

$I = \int \frac{2 t}{\sqrt{2 {t}^{2} - 12 t - 5}} \mathrm{dt}$

$= \frac{1}{2} \int \frac{4 t - 12 + 12}{\sqrt{2 {t}^{2} - 12 t - 5}} \mathrm{dt}$

=color(red)(1/2int(4t-12)/sqrt(2t^2-12t-5)dt)+color(blue) (1/2int12/sqrt(2t^2-12t-5)dt

$I = \textcolor{red}{{I}_{1}} + \textcolor{b l u e}{{I}_{2}} \ldots \to \left(A\right)$

Now, ${I}_{1} = \frac{1}{2} \int \frac{4 t - 12}{\sqrt{2 {t}^{2} - 12 t - 5}} \mathrm{dt}$

Take, $\sqrt{2 {t}^{2} - 12 t - 5} = u \implies 2 {t}^{2} - 12 t - 5 = {u}^{2}$

$\implies 4 t - 12 = 2 u \mathrm{du}$

$\therefore {I}_{1} = \frac{1}{2} \int \frac{2 u}{u} \mathrm{du} = \int \mathrm{du} = u + {c}_{1} , \to u = \sqrt{2 {t}^{2} - 12 t - 5}$

$= \sqrt{2 {t}^{2} - 12 t - 5} + {c}_{1} , w h e r e , \sqrt{x} = t$

$\therefore {I}_{1} = \sqrt{2 x - 12 \sqrt{x} - 5} + {c}_{1}$

$A l s o , {I}_{2} = \frac{1}{2} \int \frac{12}{\sqrt{2 {t}^{2} - 12 t - 5}} \mathrm{dt}$

${I}_{2} = \frac{6}{\sqrt{2}} \int \frac{1}{\sqrt{{t}^{2} - 6 t - \frac{5}{2}}} \mathrm{dt}$

$= \frac{6}{\sqrt{2}} \int \frac{1}{\sqrt{{t}^{2} - 6 t + 9 - \frac{23}{2}}} \mathrm{dt} \to \left[- \frac{5}{2} = \frac{18 - 23}{2}\right]$

$= 3 \sqrt{2} \int \frac{1}{\sqrt{{\left(t - 3\right)}^{2} - {\left(\sqrt{\frac{23}{2}}\right)}^{2}}} \mathrm{dt}$

$= 3 \sqrt{2} \ln | t - 3 + \sqrt{{\left(t - 3\right)}^{2} - {\left(\sqrt{\frac{23}{2}}\right)}^{2}} | + c$

$= 3 \sqrt{2} \ln | t - 3 + \textcolor{v i o \le t}{\sqrt{{t}^{2} - 6 t - \frac{5}{2}}} | + c$

=3sqrt2ln|t-3+sqrt(2t^2-12t-5)/color(orange)(sqrt2)|+color(orange)(c

I_2=3sqrt2ln|sqrt2t-3sqrt2+sqrt(2t^2-12t-5)|+color(orange)(c_2

Subst .back $t = \sqrt{x}$

${I}_{2} = 3 \sqrt{2} \ln | \sqrt{2 x} - 3 \sqrt{2} + \sqrt{2 x - 12 \sqrt{x} - 5} | + {c}_{2}$

Hence, from $\left(A\right)$,

I=sqrt(2x-12sqrtx-5)+3sqrt2ln|sqrt(2x)-3sqrt2+sqrt(2x-12sqrtx- 5)|+C,

$w h e r e , C = {c}_{1} + {c}_{2}$