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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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Isaiah Banta
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    $\sec^3(x)/\tan^4(x) = \cos(x)/\sin^4(x)$. Now use substitution with $u = \sin(x)$. – florence Mar 06 '17 at 08:14
  • @florence Thankyou! Could you tell me what identity states this so I know how to use it in the future? – Isaiah Banta Mar 06 '17 at 08:21
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    @IsaiahBanta $sec x = \frac{1}{cos x}, tan x = \frac{sin x}{cos x} \Rightarrow \frac{sec^{3}x}{tan^{4}x} = \frac{\frac{1}{cos^{3}x}}{\frac{sin^{4}x}{cos^{4}x}} = \frac{cos^{4}x}{sin^{4}xcos^{3}x} = \frac{cosx}{sin^{4} x}$ – mrnovice Mar 06 '17 at 08:30

3 Answers3

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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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Alternatively, note that the integral is of the form of Chebyshev's differential binomial and belongs to the case $\frac{m+1}{n}+p\in\mathbb Z$(where $m=-4$, $n=2$ and $p=\frac12$). Hence, we perform the substitution $t=\sqrt{\frac{x^2+16}{x^2}}$:

$$\require{cancel}\begin{align}\int\frac{\sqrt{x^2+16}}{x^4}\mathrm dx&=\text{sgn}x\int\cancel{x^{-4}}(t\cancel{x})\left(\frac{-\cancel{x^3}t\mathrm dt}{16}\right)\\&=\frac{-\text{sgn}x}{48}\frac{(x^2+16)^\frac32}{(x^2)^\frac32}+C\\&=\frac{-(x^2+16)^\frac32}{48x^3}+C\end{align}$$

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An alternative method is to perform the substitution $x=\frac1{t}$(this provides an intuition to the approach of Chebyshev’s differential binomial):

$$\require{cancel}\begin{align}\int\frac{\sqrt{x^2+16}}{x^4}&=-\text{sgn}(t)\int\frac{\sqrt{16t^2+1}}{\cancel{t}}\frac{t\cancel{^4}\mathrm dt}{\cancel{t^2}}\\&=-\frac{\text{sgn}(t)}{32}\int\sqrt{16t^2+1}\mathrm d(16t^2+1)\\&=-\frac{\text{sgn}(t)}{48}(16t^2+1)^\frac32+C\\&=-\frac{\text{sgn}(x)}{48}\frac{(x^2+16)^\frac32}{|x|^3}+C\\&=-\frac{(x^2+16)^\frac32}{48x^3}+C\end{align}$$

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