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I have a problem requiring trigonometric substitution to evaluate $\int \frac{\sqrt {x^2+16}}{x^4}dx$.

This is how far I have gotten:

Let $ x = 4\tan \theta $

Then $dx = 4\sec^2\theta$

$$\int \frac{\sqrt {16\tan^2\theta+16}}{(4\tan\theta)^4}4\sec^2\theta d\theta$$

Then doing some cancellations and the trig identity of $ 1 + \tan^2x = \sec^2x$ we get,

$$=\int \frac {4\sec\theta}{(4\tan\theta)^4}4\sec^2\theta d\theta$$

$$=\int \frac{\sec^3\theta}{16\tan^4\theta}d\theta$$

I'm not sure what to do after this point, do I split my $sec^3\theta $ into $sec^2\theta * sec\theta$ in order to split the integral after using the above identity a second time? Thanks for the help in advance.

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    $\begingroup$ $\sec^3(x)/\tan^4(x) = \cos(x)/\sin^4(x)$. Now use substitution with $u = \sin(x)$. $\endgroup$ – florence Mar 6 '17 at 8:14
  • $\begingroup$ @florence Thankyou! Could you tell me what identity states this so I know how to use it in the future? $\endgroup$ – Isaiah Banta Mar 6 '17 at 8:21
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    $\begingroup$ @IsaiahBanta $sec x = \frac{1}{cos x}, tan x = \frac{sin x}{cos x} \Rightarrow \frac{sec^{3}x}{tan^{4}x} = \frac{\frac{1}{cos^{3}x}}{\frac{sin^{4}x}{cos^{4}x}} = \frac{cos^{4}x}{sin^{4}xcos^{3}x} = \frac{cosx}{sin^{4} x}$ $\endgroup$ – mrnovice Mar 6 '17 at 8:30
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$$=\int \frac{\sec^3\theta}{16\tan^4\theta}d\theta$$

I'm not sure what to do after this point, do I split my $sec^3\theta $ into $sec^2\theta * sec\theta$ in order to split the integral after using the above identity a second time?

You're almost there. Remember that $\sec$ and $\tan$ are just other names for (simple) functions of $\sin$ and $\cos$ so you can always switch to those to try and simplify as much as possible. Since: $$\frac{\color{blue}{\sec^3x}}{\color{red}{\tan^4x}}=\color{blue}{\frac{1}{\cos^3x}}\color{red}{\frac{\cos^4x}{\sin^4x}}=\frac{\cos x}{\sin^4 x}$$ you can now proceed with the substitution $u=\sin x$ and thus $\mbox{d}u=\cos x \,\mbox{d}x$: $$\int\frac{\sec^3x}{16\tan^4x}\,\mbox{d}x=\int\frac{\cos x}{16\sin^4 x}\,\mbox{d}x\to\int\frac{1}{16u^4}\,\mbox{d}u=-\frac{1}{48u^3} +C= \ldots$$

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  • $\begingroup$ Could you explain why the $cosx$ disappears in the numerator? $\endgroup$ – Isaiah Banta Mar 6 '17 at 9:08
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    $\begingroup$ That's a consequence of substitution rule; letting $u=\sin x$, you get $\mbox{d}u=\cos x \,\mbox{d}x$ (I'll add this). $\endgroup$ – StackTD Mar 6 '17 at 9:09
  • $\begingroup$ Oh I see now, ok thank you! $\endgroup$ – Isaiah Banta Mar 6 '17 at 9:11
  • $\begingroup$ Alright, you're welcome! $\endgroup$ – StackTD Mar 6 '17 at 9:13

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