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January 14th, 2016, 03:20 AM   #1
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Please explain. (Trigonometric substitution)

I tried
$\displaystyle x^2 + 16 = 16sec^2\theta\\
2x dx = d/dx (16sec^2\theta)$
and got stuck here
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Volle is offline  
January 14th, 2016, 05:09 AM   #2
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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 $\displaystyle v= sin(\theta)$.
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January 14th, 2016, 05:14 AM   #3
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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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