My Math Forum Please explain. (Trigonometric substitution)

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January 14th, 2016, 03:20 AM   #1
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I tried
$\displaystyle x^2 + 16 = 16sec^2\theta\\ 2x dx = d/dx (16sec^2\theta)$
and got stuck here
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 January 14th, 2016, 05:09 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 What, exactly is the question? To integrate $\displaystyle \sqrt{x^2+ 16}$? Then you want to use the trig identity $\displaystyle tan^2(\theta)+ 1= sec^2(\theta)$ to "get rid of" that square root: if $\displaystyle x= 4 tan(\theta)$ then $\displaystyle \sqrt{x^2+ 16}= \sqrt{16tan^2(\theta)+ 16}= 4\sqrt{tan^2(\theta)+ 1}= 4\sqrt{sec^2(\theta)}= 4 sec(\theta)$. Differentiating both sides of $\displaystyle x= 4tan(\theta)$ (not $\displaystyle x^2+ 16= 4 sec^2(\theta))$ we have $\displaystyle dx= 4 sec^2(\theta) d\theta$ so $\displaystyle \int \sqrt{x^2+ 16} dx= 16 \int sec^3(\theta) d\theta$. To integrate that, I think I would write it as $\displaystyle 16\int \frac{d\theta}{cos^3(\theta)}$, then multiply both numerator and denominator by $\displaystyle cos(\theta)$ to get $\displaystyle 16\int \frac{cos(\theta)d\theta}{cos^4(\theta)}= 16\int \frac{cos(\theta)}{(1- sin^2(\theta))^2}$ and let $\displaystyle v= sin(\theta)$. Thanks from Volle
 January 14th, 2016, 05:14 AM #3 Global Moderator   Joined: Dec 2006 Posts: 20,835 Thanks: 2162 Let $x = 4\tan(θ)$, where $|θ| < \pi/2$, so that $dx = 4\sec^2(θ)dθ$. $\displaystyle \int\!\sqrt{x^2+16}\,dx = 16\!\int\!\sec^3(θ)dθ$ Integrating by parts, $\displaystyle \int\!\sec^3\!(θ) dθ = \sec(θ)\tan(θ) - \!\int\!\sec(θ)\tan^2(θ)dθ = \sec(θ)\tan(θ) - \!\int\!\sec(θ)(\sec^2(θ) - 1)dθ$, etc. It's a bit easier to use the substitution $x = \sinh(u)$ instead of the above.

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