How do you integrate $\int \frac{x^{2}}{\sqrt{4 x^{2} + 25}} d x$ $int x^2/sqrt(4x^2+25)dx$ using trigonometric substitution?

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How do you integrate $\int \frac{x^{2}}{\sqrt{4 x^{2} + 25}} d x$ $int x^2/sqrt(4x^2+25)dx$ using trigonometric substitution?

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Answer 1 · 2018-03-27T10:34:54

$\frac{1}{20} (\sqrt{4 x + 25}) x - \frac{5}{8} ln (\frac{\sqrt{4 x + 25}}{5} + \frac{2}{5} x)$ $1/20(sqrt(4x+25))x - 5/8ln(sqrt(4x+25)/5+2/5x)$

Explanation:

$\int \frac{x^{2}}{\sqrt{4 x + 25}} d x$ $intx^2/sqrt(4x+25)dx$

$L e t x = \frac{5}{2} tan A, d x = (\frac{5}{2} {sec}^{2} a) d A$ $Let x = 5/2 tan A, dx = (5/2 sec^2a)dA$

$= \int \frac{x^{2}}{\sqrt{4 x + 25}} d x = \int {(\frac{5}{2})}^{2} \frac{{tan}^{2} A {sec}^{2} A}{\sqrt{4 {(\frac{5}{2})}^{2} {tan}^{2} A + 25}}$ $=intx^2/sqrt(4x+25)dx = int (5/2)^2(tan^2A sec^2A) /(sqrt(4(5/2)^2tan^2A+25)$ $d A$ $dA$

$= \int \frac{x^{2}}{\sqrt{4 x + 25}} d x = \int {(\frac{5}{2})}^{2} \frac{{tan}^{2} A {sec}^{2} A}{\sqrt{25 {tan}^{2} A + 25}}$ $=intx^2/sqrt(4x+25)dx = int (5/2)^2(tan^2A sec^2A) /(sqrt(25tan^2A+25)$ $d A$ $dA$

$= \int \frac{x^{2}}{\sqrt{4 x + 25}} d x = \int {(\frac{5}{2})}^{2} \frac{{tan}^{2} A {sec}^{2} A}{5} (\sqrt{{tan}^{2} A + 1})$ $=intx^2/sqrt(4x+25)dx = int (5/2)^2(tan^2A sec^2A) /5(sqrt(tan^2A+1))$ $d A$ $dA$

${sec}^{2} A - 1 = {tan}^{2} A$ $sec^2A-1= tan^2A$

$= \int (\frac{5}{4}) \frac{{tan}^{2} A {sec}^{2} A}{\sqrt{{sec}^{2} A}}$ $= int (5/4)(tan^2A sec^2A) /(sqrt(sec^2A))$ $d A$ $dA$

$= \int (\frac{5}{4}) \frac{({sec}^{2} A - 1) {sec}^{2} A}{\sqrt{{sec}^{2} A}}$ $= int (5/4)((sec^2A-1) sec^2A) /(sqrt(sec^2A))$ $d A$ $dA$
$= \int (\frac{5}{4}) (({sec}^{3} A - sec A))$ $= int (5/4)((sec^3A-secA))$ $d A$ $dA$
$= (\frac{5}{4}) (\int ({sec}^{3} A) - \int (sec A))$ $= (5/4)(int(sec^3A)-int(secA))$ $d A$ $dA$ -------- equation (1)

We need to find

$\int (sec A)$ $int(secA)$ $d A$ $dA$ = $\int sec A \frac{sec A + tan A}{sec A + tan A}$ $int sec A(secA + tanA)/(secA+tanA)$ $d A$ $dA$

= $\int sec A \frac{{sec}^{2} A + tan A sec A}{sec A + tan A}$ $int sec A(sec^2A + tanAsecA)/(secA+tanA)$ $d A$ $dA$

Let $B = sec A + tan A$ $B= secA+tanA$
$d B$ $dB$ $= (sec A tan A + {sec}^{2} A)$ $= (secAtanA+sec^2A)$ $d A$ $dA$
$= \int \frac{1}{B}$ $=int1/B$ $d B$ $dB$
$= ln | B | + c$ $=ln |B|+c$
$= ln | sec A + tan A | + c$ $=ln |secA+tanA|+c$ -----equation 2

We now need to find

$\int ({sec}^{3} A)$ $int(sec^3A)$ $d A$ $dA$

Integrate by parts

$\int {sec}^{2} A sec A$ $int sec^2AsecA$ $d A$ $dA$
Let $C = sec A$ $C= secA$
$d C$ $dC$ $= sec A tan A$ $= secAtanA$ $d A$ $dA$

$d D$ $dD$ $= {sec}^{2} A$ $= sec^2A$ $d A$ $dA$
$D = \int {sec}^{2} A$ $D= int sec^2A$ $d A$ $dA$ $= tan A + c o n s t$ $=tanA + const$

Using by parts gives
$= \int {sec}^{2} A sec A$ $=int sec^2AsecA$ $d A$ $dA$ $= C D - \int D$ $= CD - int D$ $d C$ $dC$ / $d A$ $dA$
$= sec A tan A - \int sec A tan A tan A$ $=secAtanA - int secAtanAtanA$ $d A$ $dA$
$= sec A tan A - \int ({sec}^{2} A - 1) sec A$ $=secAtanA - int (sec^2A-1)secA$ $d A$ $dA$
$= sec A tan A - \int ({sec}^{3} A - sec A)$ $=secAtanA - int (sec^3A-secA)$ $d A$ $dA$
$\int {sec}^{3} A sec A$ $int sec^3AsecA$ $d A$ $dA$ $= sec A tan A - (\int {sec}^{3} A - \int sec A)$ $=secAtanA - (int sec^3A -int secA)$ $d A$ $dA$

substituting using equation 2
$2 \int {sec}^{3} A$ $2int sec^3A$ $d A$ $dA$ = $sec A tan A + ln ∣ sec A + tan A)$ $secAtanA+ ln|secA+tanA)$
$\frac{1}{2} (sec A tan A + ln ∣ sec A + tan A))$ $1/2(secAtanA+ ln|secA+tanA))$ ----equation 3

From equation 1 we had

$\int \frac{x^{2}}{\sqrt{4 x + 25}} d x$ $intx^2/sqrt(4x+25)dx$ $= (\frac{5}{4}) (\int ({sec}^{3} A) - \int (sec A))$ $= (5/4)(int(sec^3A)-int(secA))$ $d A$ $dA$

using the results from equations 2 and 3

$= \frac{5}{4} ((\frac{1}{2} (sec A tan A + ln | (sec A + tan A) |)) - ln | (sec A + tan A) |)$ $=5/4((1/2(secAtanA + ln|(SecA+tanA)|))-ln|(SecA+tanA)|)$
$= \frac{5}{8} (sec A tan A + - ln | (sec A + tan A) |))$ $=5/8(secAtanA + -ln|(SecA+tanA)|))$

Now convert everything back to x
$x = \frac{5}{2} tan A$ $x=5/2 tan A$ , $tan A = \frac{2}{5} x$ $tan A=2/5x$ , ${tan}^{2} A = {(\frac{2}{5})}^{2} x^{2}$ $tan^2A=(2/5)^2x^2$
${sec}^{2} A = {tan}^{2} A + 1 = \frac{4 x^{2} + 25}{25}$ $sec^2A = tan^2A +1 =(4x^2+25)/25$
$sec A = = \sqrt{\frac{4 x^{2} + 25}{25}} = \frac{\sqrt{4 x + 25}}{5}$ $secA = =sqrt((4x^2+25)/(25))= sqrt(4x+25)/5$

$\frac{1}{20} (\sqrt{4 x + 25}) x - \frac{5}{8} ln (\frac{\sqrt{(4 x + 25)}}{5} + \frac{2}{5} x)$ $1/20(sqrt(4x+25))x - 5/8ln(sqrt((4x+25))/(5)+2/5x)$

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How do you integrate $\int \frac{x^{2}}{\sqrt{4 x^{2} + 25}} d x$ $int x^2/sqrt(4x^2+25)dx$ using trigonometric substitution?

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How do you integrate ∫x2√4x2+25dxint x^2/sqrt(4x^2+25)dx using trigonometric substitution?

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How do you integrate $\int \frac{x^{2}}{\sqrt{4 x^{2} + 25}} d x$ $int x^2/sqrt(4x^2+25)dx$ using trigonometric substitution?