How do you integrate $\int \frac{x}{\sqrt{- x^{2} + 8 x - 32}} d x$ $int x / sqrt(-x^2+8x-32) dx$ using trigonometric substitution?

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How do you integrate $\int \frac{x}{\sqrt{- x^{2} + 8 x - 32}} d x$ $int x / sqrt(-x^2+8x-32) dx$ using trigonometric substitution?

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Answer 1 · 2016-01-07T08:55:56

$- 4 i \cdot ln (\frac{x - 4 + \sqrt{x^{2} - 8 x + 32}}{4}) - i \cdot \sqrt{x^{2} - 8 x + 32}$ $-4i* ln ((x-4+sqrt(x^2-8x+32))/4)-i*sqrt(x^2-8x+32)$ where $i = \sqrt{- 1}$ $i = sqrt(-1)$

Explanation:

The solution is really quite long but there is a solution;
start with a Right triangle.
with acute angle $θ$ $theta$
side opposite to angle $θ$ $theta$ = (x-4)
side opposite to 90 degree angle= $\sqrt{x^{2} - 8 x + 32}$ $sqrt(x^2-8x+32)$
side adjacent to $θ$ $theta$ = 4

Our trigonometric substitution goes like this:
Let $x - 4 = 4 \cdot tan θ$ $x-4 = 4*tan theta$

and $x = 4 + 4 \cdot tan θ$ $x=4+4*tan theta$

also ${(x - 4)}^{2} = 16 \cdot {tan}^{2} θ$ $(x-4)^2 = 16 *tan^2 theta$

$d x = 4 \cdot {sec}^{2} θ$ $dx = 4* sec^2 theta$ $d$ $d$ $θ$ $theta$

and $tan θ = \frac{x - 4}{4}$ $tan theta = (x-4)/4$

our $sec θ = \frac{\sqrt{x^{2} - 8 x + 32}}{4}$ $sec theta = sqrt(x^2 -8x+32)/4$

Take note that $\sqrt{x^{2} - 8 x + 32} = \sqrt{{(x - 4)}^{2} + 16}$ $sqrt(x^2-8x+32) = sqrt((x-4)^2+16)$

from the given:
$\int \frac{x}{\sqrt{- x^{2} + 8 x - 32}} d x$ $intx/sqrt(-x^2+8x-32) dx$

$\int \frac{x}{\sqrt{- (x^{2} - 8 x + 16 + 16)}} d x$ $int x/sqrt(-(x^2-8x+16+16)) dx$

$\int \frac{x}{\sqrt{- ({(x - 4)}^{2} + 16)}} d x$ $int x/sqrt(-((x-4)^2+16)) dx$

from the above it follows

$\int \frac{(4 + 4 \cdot tan θ) \cdot 4 {sec}^{2} θ}{\sqrt{- 1 (16 \cdot {tan}^{2} θ + 16)}}$ $int ((4+4*tan theta) * 4 sec^2 theta )/sqrt(-1(16*tan^2 theta+16)$ $d$ $d$ $θ$ $theta$

$\int \frac{4 (1 + tan θ) \cdot 4 {sec}^{2} θ}{\sqrt{- 1} \cdot \sqrt{16 \cdot {tan}^{2} θ + 16}}$ $int (4(1+tan theta) * 4 sec^2 theta )/(sqrt(-1) *sqrt(16*tan^2 theta+16))$ $d$ $d$ $θ$ $theta$

$\int \frac{4 (1 + tan θ) \cdot 4 {sec}^{2} θ}{\sqrt{- 1} \cdot \sqrt{16} \cdot \sqrt{{tan}^{2} θ + 1}}$ $int (4(1+tan theta) * 4 sec^2 theta )/(sqrt(-1) *sqrt(16)*sqrt(tan^2 theta+1))$ $d$ $d$ $θ$ $theta$

$\int \frac{4 (1 + tan θ) \cdot {sec}^{2} θ}{\sqrt{- 1} \cdot \sqrt{{tan}^{2} θ + 1}}$ $int (4(1+tan theta) * sec^2 theta )/(sqrt(-1) *sqrt(tan^2 theta+1))$ $d$ $d$ $θ$ $theta$

but $\sqrt{{tan}^{2} θ + 1} = sec θ$ $sqrt(tan^2 theta+1) = sec theta$

$\int \frac{4 (1 + tan θ) \cdot {sec}^{2} θ}{\sqrt{- 1} \cdot sec θ}$ $int (4(1+tan theta) * sec^2 theta )/(sqrt(-1) *sec theta)$ $d$ $d$ $θ$ $theta$

then $\int \frac{4 (1 + tan θ) \cdot sec θ}{\sqrt{- 1}}$ $int (4(1+tan theta) * sec theta )/(sqrt(-1) )$ $d$ $d$ $θ$ $theta$

it follows $\int \frac{4 (sec θ + tan θ sec θ)}{\sqrt{- 1}}$ $int (4(sec theta+tan theta sec theta) )/(sqrt(-1) )$ $d$ $d$ $θ$ $theta$

and $(\frac{4}{\sqrt{- 1}}) \int sec θ$ $(4/sqrt(-1))int sec theta$ $d$ $d$ $θ$ $theta$ + $(\frac{4}{\sqrt{- 1}}) \int tan θ sec θ$ $(4/sqrt(-1)) int tan theta sec theta$ $d$ $d$ $θ$ $theta$

but $\sqrt{- 1} = i$ $sqrt(-1) = i$ and $i^{2} = - 1$ $i^2 = -1$

$(\frac{4}{\sqrt{- 1}} \cdot \frac{\sqrt{- 1}}{\sqrt{- 1}}) \int sec θ$ $(4/sqrt(-1)*sqrt(-1)/sqrt(-1))int sec theta$ $d$ $d$ $θ$ $theta$ + $(\frac{4}{\sqrt{- 1}} \cdot \frac{\sqrt{- 1}}{\sqrt{- 1}}) \int tan θ sec θ$ $(4/sqrt(-1)*sqrt(-1)/sqrt(-1)) int tan theta sec theta$ $d$ $d$ $θ$ $theta$

$(\frac{4 i}{- 1}) \int sec θ$ $((4i)/(-1))int sec theta$ $d$ $d$ $θ$ $theta$ + $(\frac{4 i}{- 1}) \int tan θ sec θ$ $((4i)/(-1)) int tan theta sec theta$ $d$ $d$ $θ$ $theta$

after integration using the basic formulas

$(\frac{4 i}{- 1}) ln (sec θ + tan θ)$ $((4i)/-1)ln (sec theta + tan theta)$ + $(\frac{4 i}{- 1}) sec θ$ $((4i)/-1) sec theta$

returning to our original variables and simplifying, it follows,

$- 4 i \cdot ln (\frac{x - 4 + \sqrt{x^{2} - 8 x + 32}}{4}) - i \cdot \sqrt{x^{2} - 8 x + 32}$ $-4i* ln ((x-4+sqrt(x^2-8x+32))/4)-i*sqrt(x^2-8x+32)$

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How do you integrate $\int \frac{x}{\sqrt{- x^{2} + 8 x - 32}} d x$ $int x / sqrt(-x^2+8x-32) dx$ using trigonometric substitution?

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How do you integrate int x / sqrt(-x^2+8x-32) dx∫x√−x2+8x−32dxint x / sqrt(-x^2+8x-32) dx using trigonometric substitution?

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How do you integrate $\int \frac{x}{\sqrt{- x^{2} + 8 x - 32}} d x$ $int x / sqrt(-x^2+8x-32) dx$ using trigonometric substitution?