How do you integrate $\int \frac{1}{\sqrt{3 x - 12 \sqrt{x} + 29}}$ $int 1/sqrt(3x-12sqrtx+29)$ using trigonometric substitution?

Question

How do you integrate $\int \frac{1}{\sqrt{3 x - 12 \sqrt{x} + 29}}$ $int 1/sqrt(3x-12sqrtx+29)$ using trigonometric substitution?

1 Answer

Impact of this question

2035 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-16T21:31:32

$= \frac{2}{3} (\sqrt{3 x - 12 \sqrt{x} + 29} + 2 \sqrt{3} {sinh}^{- 1} (\sqrt{\frac{3}{17}} (\sqrt{x} - 2))) + C$ $=2/3(sqrt(3x-12sqrt(x)+29)+2sqrt(3) sinh^-1(sqrt(3/17)(sqrt(x)-2)))+C$

Explanation:

For the integrand $\frac{1}{\sqrt{3 x - 12 \sqrt{x} + 29}}$ $1/(sqrt(3x-12sqrt(x)+29))$ , substitute

$u = \sqrt{x}$ $u=sqrt(x)$ and $d u = \frac{1}{2 \sqrt{x}} d x$ $du=1/(2sqrt(x))dx$

$= 2 \int (\frac{u}{\sqrt{3 u^{2} - 12 u + 29}}) d u$ $=2int(u/(sqrt(3u^2-12u+29)))du$

Completing the square, we get

$= 2 \int ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{u}{\sqrt{{(\sqrt{3} u - 2 \sqrt{3})}^{2} + 17}} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠ d u$ $=2int(u/(sqrt((sqrt(3)u-2sqrt(3))^2+17)))du$

Let $s = \sqrt{3} u - 2 \sqrt{3}$ $s=sqrt(3)u-2sqrt(3)$ and $d s = \sqrt{3} d u$ $ds=sqrt(3)du$

$= \frac{2}{\sqrt{3}} \int (\frac{s + 2 \sqrt{3}}{\sqrt{3} \sqrt{s^{2} + 17}} d s)$ $=2/sqrt(3) int ((s+2sqrt(3))/(sqrt(3)sqrt(s^2+17))ds)$

Factor out the constant $\frac{1}{\sqrt{3}}$ $1/sqrt(3)$

$= \frac{2}{3} \int \frac{s + 2 \sqrt{3}}{\sqrt{s^{2} + 17}} d s$ $=2/3 int(s+2sqrt(3))/(sqrt(s^2+17)) ds$

Next, substitute $s = \sqrt{17} tan (p)$ $s=sqrt(17) tan(p)$ and $d s = \sqrt{17} {sec}^{2} (p) d p$ $ds=sqrt(17)sec^2(p)dp$

Then, $\sqrt{s^{2} + 17} = \sqrt{17 {tan}^{2} (p) + 17} = \sqrt{17} sec (p)$ $sqrt(s^2+17)=sqrt(17tan^2(p)+17)=sqrt(17)sec(p)$ and $p = {tan}^{- 1} (\frac{s}{\sqrt{17}})$ $p=tan^-1(s/sqrt(17))$ . This gives us

$= \frac{2 \sqrt{17}}{3} \int \frac{\sqrt{17} tan (p) + 2 \sqrt{3} sec (p)}{\sqrt{17}} d p$ $=(2sqrt(17))/3 int (sqrt(17)tan(p)+2sqrt(3)sec(p))/sqrt(17) dp$

Factoring out $\frac{1}{\sqrt{17}}$ $1/sqrt(17)$ gives

$= \frac{2}{3} \int (\sqrt{17} tan (p) + 2 \sqrt{3}) sec (p) d p$ $=2/3 int (sqrt(17)tan(p) + 2sqrt(3))sec(p) dp$

$= \frac{2}{3} \int (\sqrt{17} tan (p) sec (p) + 2 \sqrt{3} sec (p)) d p$ $=2/3 int (sqrt(17)tan(p)sec(p) + 2sqrt(3)sec(p)) dp$

$= \frac{2 \sqrt{17}}{3} \int (\sqrt{17} tan (p) sec (p)) d p + \frac{4}{\sqrt{3}} \int sec (p)) d p$ $=(2sqrt(17))/3 int (sqrt(17)tan(p)sec(p))dp + 4/sqrt(3) int sec(p)) dp$

Substitute $w = sec (p)$ $w=sec(p)$ and $d w = tan (p) sec (p) d p$ $dw=tan(p)sec(p)dp$

$= \frac{2 \sqrt{17}}{3} \int 1 d w + \frac{4}{\sqrt{3}} \int sec (p) d p$ $=(2sqrt(17))/3 int 1dw + 4/sqrt(3) intsec(p)dp$

$= \frac{2 \sqrt{17} w}{3} + \frac{4}{\sqrt{3}} \int sec (p) d p$ $=(2sqrt(17)w)/3+4/sqrt(3) int sec(p)dp$

$= \frac{4 ln (tan (p) + sec (p))}{\sqrt{3}} + \frac{2 \sqrt{17} 3}{3} + C$ $=(4 ln(tan(p)+sec(p)))/sqrt(3) + (2sqrt(17)3)/3+C$

Backwards substituting all the variables eventually gives

$= \frac{2}{3} (\sqrt{3 x - 12 \sqrt{x} + 29} + 2 \sqrt{3} {sinh}^{- 1} (\sqrt{\frac{3}{17}} (\sqrt{x} - 2))) + C$ $=2/3(sqrt(3x-12sqrt(x)+29)+2sqrt(3) sinh^-1(sqrt(3/17)(sqrt(x)-2)))+C$

Science

Math

Humanities

... and beyond

How do you integrate $\int \frac{1}{\sqrt{3 x - 12 \sqrt{x} + 29}}$ $int 1/sqrt(3x-12sqrtx+29)$ using trigonometric substitution?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you integrate int 1/sqrt(3x-12sqrtx+29) ∫1√3x−12√x+29int 1/sqrt(3x-12sqrtx+29) using trigonometric substitution?

1 Answer

Explanation:

Related questions

Impact of this question

How do you integrate $\int \frac{1}{\sqrt{3 x - 12 \sqrt{x} + 29}}$ $int 1/sqrt(3x-12sqrtx+29)$ using trigonometric substitution?