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January 14th, 2016, 03:20 AM  #1 
Newbie Joined: Oct 2012 Posts: 29 Thanks: 0  Please explain. (Trigonometric substitution)
Attached I tried $\displaystyle x^2 + 16 = 16sec^2\theta\\ 2x dx = d/dx (16sec^2\theta)$ and got stuck here 
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 $\displaystyle v= sin(\theta)$.

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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explain, substitution, trigonometric 
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