How do you find the indefinite integral of $\sqrt{25 + x^{2}}$ $sqrt(25 + x^2)$ ?

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How do you find the indefinite integral of $\sqrt{25 + x^{2}}$ $sqrt(25 + x^2)$ ?

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Answer 1 · 2018-03-26T07:04:01

$\frac{1}{2} (x \sqrt{25 + x^{2}} + 25 ln ∣ ∣ x + \sqrt{25 + x^{2}} ∣ ∣) + C$ $1/2(xsqrt(25+x^2)+25lnabs(x+sqrt(25+x^2)))+C$

Explanation:

Use trigonometric substitution.

Draw a triangle with an angle $θ$ $theta$ .

Label the opposite side as $x$ $x$ and adjacent side as $5$ $5$ .

Now examine the triangle and notice that

$tan θ = \frac{x}{5}$ $tantheta=x/5$

$\Rightarrow x = 5 tan θ$ $rArrx=5tantheta$

$\Rightarrow d x = 5 {sec}^{2} θ$ $rArrdx=5sec^2theta$ $d$ $d$ $θ$ $theta$

By the Pythagorean Theorem, the hypotenuse of the triangle is:

$\sqrt{25 + x^{2}}$ $sqrt(25+x^2)$

So we can write

$sec θ = \frac{\sqrt{25 + x^{2}}}{5}$ $sectheta=sqrt(25+x^2)/5$

$\Rightarrow \sqrt{25 + x^{2}} = 5 sec θ$ $rArrsqrt(25+x^2)=5sectheta$

Let's now rewrite the integral in terms of $θ$ $theta$

$\int \sqrt{25 + x^{2}} d x \Rightarrow \int (5 sec θ) (5 {sec}^{2} θ$ $intsqrt(25+x^2)dxrArrint(5sectheta)(5sec^2theta$ $d$ $d$ $θ)$ $theta)$

$\Rightarrow 25 \int$ $rArr25int$ ${sec}^{3} θ$ $sec^3theta$ $d$ $d$ $θ$ $theta$

Integrating ${sec}^{3} θ$ $sec^3theta$ can be tricky. There is a shortcut formula, but I'll do it the long way for teaching purposes.

Separate $({sec}^{3} θ$ $(sec^3theta$ $d$ $d$ $θ)$ $theta)$ into $(sec θ \cdot {sec}^{2} θ$ $(sectheta*sec^2theta$ $d$ $d$ $θ)$ $theta)$ and use integration by parts.

Let:

$u = sec θ$ $u=sectheta$
$d u = sec θ tan θ$ $du=secthetatantheta$ $d$ $d$ $θ$ $theta$

$d v = {sec}^{2} θ$ $dv=sec^2theta$ $d$ $d$ $θ$ $theta$
$v = tan θ$ $v=tantheta$

Then:

$\Rightarrow \int$ $rArrint$ ${sec}^{3} θ$ $sec^3theta$ $d$ $d$ $θ = sec θ tan θ - \int$ $theta=secthetatantheta-int$ ${tan}^{2} θ sec θ$ $tan^2thetasectheta$ $d$ $d$ $θ$ $theta$

$\Rightarrow \int$ $rArrint$ ${sec}^{3} θ$ $sec^3theta$ $d$ $d$ $θ = sec θ tan θ - \int$ $theta=secthetatantheta-int$ $({sec}^{2} θ - 1) sec θ$ $(sec^2theta-1)sectheta$ $d$ $d$ $θ$ $theta$

$\Rightarrow \int$ $rArrint$ ${sec}^{3} θ$ $sec^3theta$ $d$ $d$ $θ = sec θ tan θ - \int$ $theta=secthetatantheta-int$ $({sec}^{3} θ - sec θ)$ $(sec^3theta-sectheta)$ $d$ $d$ $θ$ $theta$

$\Rightarrow 2 \int$ $rArr2int$ ${sec}^{3} θ$ $sec^3theta$ $d$ $d$ $θ = sec θ tan θ + \int$ $theta=secthetatantheta+int$ $sec θ$ $sectheta$ $d$ $d$ $θ$ $theta$

$\Rightarrow 2 \int$ $rArr2int$ ${sec}^{3} θ$ $sec^3theta$ $d$ $d$ $θ = sec θ tan θ + ln | sec θ + tan θ |$ $theta=secthetatantheta+lnabs(sectheta+tantheta)$

$\Rightarrow \int$ $rArrint$ ${sec}^{3} θ$ $sec^3theta$ $d$ $d$ $θ = \frac{1}{2} (sec θ tan θ + ln | sec θ + tan θ |) + C$ $theta=1/2(secthetatantheta+lnabs(sectheta+tantheta))+C$

And let's not forget that our integral was multiplied by 25!

$\Rightarrow 25 \int$ $rArr25int$ ${sec}^{3} θ$ $sec^3theta$ $d$ $d$ $θ = \frac{25}{2} (sec θ tan θ + ln | sec θ + tan θ |) + C$ $theta=25/2(secthetatantheta+lnabs(sectheta+tantheta))+C$

Now put everything back in terms of $x$ $x$

$\Rightarrow \frac{25}{2} (\frac{\sqrt{25 + x^{2}}}{5} \frac{x}{5} + ln ∣ ∣ ∣ \frac{\sqrt{25 + x^{2}}}{5} + \frac{x}{5} ∣ ∣ ∣) + C$ $rArr25/2(sqrt(25+x^2)/5x/5+lnabs(sqrt(25+x^2)/5+x/5))+C$

$\Rightarrow \frac{1}{2} (x \sqrt{25 + x^{2}} + 25 ln ∣ ∣ ∣ \frac{\sqrt{25 + x^{2}} + x}{5} ∣ ∣ ∣) + C$ $rArr1/2(xsqrt(25+x^2)+25lnabs((sqrt(25+x^2)+x)/5))+C$

$\Rightarrow \frac{1}{2} (x \sqrt{25 + x^{2}} + 25 ln ∣ ∣ x + \sqrt{25 + x^{2}} ∣ ∣ - 25 ln 5) + C$ $rArr1/2(xsqrt(25+x^2)+25lnabs(x+sqrt(25+x^2))-25ln5)+C$

Then we can absorb the constant $- \frac{25 ln 5}{2}$ $-(25ln5)/2$ into $C$ $C$ to get our final answer:

$\Rightarrow \frac{1}{2} (x \sqrt{25 + x^{2}} + 25 ln ∣ ∣ x + \sqrt{25 + x^{2}} ∣ ∣) + C$ $rArr1/2(xsqrt(25+x^2)+25lnabs(x+sqrt(25+x^2)))+C$

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