# How to integrate $1/(16+x^2)^2$ using trig substitution

My answer: $1/128 (\tan^{-1}(x/4) + 4x/(16+x^2)^4)$

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Have you checked your answer? If you have then what are you asking? And if you haven't, the why not? – DonAntonio Sep 30 '12 at 18:17

Consider the substitution $x=4\sin y$. Then $dx=4\cos y\ dy$, and $$\frac 1{(16-x^2)^2} = \frac 1{(16\cos^2 y)^2} = \frac 1{(4\cos y)^4}$$ So, $$\int\frac 1{(16-x^2)^2}dx = \int\frac{4\cos y}{(4\cos y)^4} dy$$ So, finally you need $\displaystyle\int\frac1{\cos^3}$.

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 $\displaystyle\int\frac{dx}{\cos^3 x}=\int\sec^3x\,dx$ is the topic of a Wikipedia article: en.wikipedia.org/wiki/Integral_of_secant_cubed – Michael Hardy Sep 30 '12 at 18:28 thanks,.. I just couldn't so easily continue.. – Berci Sep 30 '12 at 18:34 Wouldn't I use x=tan y? My teacher said that we use sin if its a a^2-x^2 pattern. – kitana97 Sep 30 '12 at 18:34 Yes, try it, $x=4\tan y$, why not? – Berci Sep 30 '12 at 18:35 @DonAntonio - Yes, I have checked my answer and it is wrong. I don't know what I did wrong.. – kitana97 Sep 30 '12 at 18:36

$$\int\frac{dx}{(16+x^2)^2} = \int\frac{4\sec^2\theta\,d\theta}{(16\sec^2\theta)^2} = \frac{1}{64} \int\frac{d\theta}{\sec^2\theta} = \frac{1}{64}\int \cos^2\theta\,d\theta = \cdots\cdots$$ \begin{align} x & = 4\tan\theta \\[6pt] dx & = 4\sec^2\theta\,d\theta \\[6pt] 16+x^2 & = 16 + 16\tan^2\theta = 16\sec^2\theta \end{align}

After getting a function of $\theta$, it may be useful to know that $$\text{If }x=4\tan\theta\text{ then }\sin\theta = \frac{x}{\sqrt{x^2+16}}\text{ and }\cos\theta = \frac{4}{\sqrt{x^2+16}}.$$ I.e., remember trigonometry.

Appendix on trigonometry in response to comments:

If $\tan\theta=\dfrac x4$, then you can construct a triangle in which $\text{opposite}=x$ and $\text{adjacent}=4$. Consequently $\text{hypotenuse}=\sqrt{x^2+4^2}$. So $\sin\theta=\dfrac{\text{opposite}}{\text{hypotenus}}=\dfrac{x}{\sqrt{x^2+4^2}}$ and $\cos\theta=\dfrac{\text{adjacent}}{\text{hypotenus}}=\dfrac{4}{\sqrt{x^2+4^2}}$.

I'm not sure whether this actually addresses the concerns raised in the comments.

Later edit in order to compare answers: \begin{align} \frac{1}{64}\int\cos^2\theta\,d\theta & = \frac{1}{64}\int \frac12 + \frac12 \cos(2\theta)\,d\theta = \frac{1}{64} \left(\frac\theta2 + \frac14\sin(2\theta)\right)+C \\[12pt] & = \frac{1}{64}\left(\frac12\arctan\left(\frac x4\right)+\frac12\sin\theta\cos\theta\right)+C \\[12pt] & = \frac{1}{128}\arctan\left(\frac x4\right)+\frac{1}{128}\cdot\frac{4x}{x^2+16} + C \\[12pt] & = \frac{1}{128}\arctan\left(\frac x4\right)+\frac{1}{32}\cdot\frac{x}{x^2+16} + C. \end{align}

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 So I finally got it right. Its: 1/128(tan^-1(x/4) + 4x/16+x^2) However, I'm confused how it's 16+x^2. For sin theta I got x/(16+x^2)^2 and for cos theta I got 4/(16+x^2)^2. Wouldnt it be (16+x^2)^4? – kitana97 Sep 30 '12 at 20:02 Typos: I wrote $\sqrt{x^2+4}$ where I needed $\sqrt{x^2+16}$. If $\tan\theta=x/4$, you can construct a right triangle in which $x=\text{opposite}$ and $4=\text{adjacent}$, so $\sqrt{x^2+4^2}=\text{hypotenuse}$. – Michael Hardy Sep 30 '12 at 20:48

The main question is how to write $16-x^2$ as a square. Note that $\cosh^2 \theta-\sinh^2 \theta = 1$ hence $1-\tanh^2\theta = \text{sech}^2 \theta$ thus $16-16\tanh^2\theta = 16\text{sech}^2 \theta$. This suggests a $x=4\tanh \theta$ substitution. Observe $dx = 4\text{sech}^2(\theta) d\theta$. Putting this all together, $$\int \frac{dx}{(16-x^2)^2} = \int \frac{4\text{sech}^2(\theta) d\theta}{(16)^2\text{sech}^4 \theta} = \frac{1}{64}\int \cosh^2 \theta d\theta$$ Now, $\cosh^2 \theta = \frac{1}{2}(\cosh 2 \theta+1)$ $$\frac{1}{64}\int \cosh^2 \theta d\theta = \frac{1}{128}\int (\cosh 2 \theta +1)d\theta = \frac{1}{256}(\sinh 2 \theta + 2\theta)+c$$ Finally, $\theta = \tanh^{-1}(x/4)$ thus, $$\int \frac{dx}{(16-x^2)^2} = \frac{1}{256}\biggl(\sinh \bigl[2 \tanh^{-1}(x/4)\bigr] + 2\tanh^{-1}(x/4)\biggr)+c$$ Of course you can write the first term as an algebraic function by studying the identities for hyperbolic functions and the given hyperbolic subsitution.

( I like the sine or cosine substitution for this problem, but I include this answer for breadth)

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Substition

$x=4\sin y$ $\Rightarrow$ $dx=4\cos y dy$

$\sin y=\frac{x}{4}$$\Rightarrow y=\arcsin\frac{x}{4} \int\frac{1}{(16-x^2)^2}dx=\int\frac{1}{(4^2-x^2)^2}dx=\int\frac{4\cos y dy}{(16-(4\sin y)^2)^2}=\int\frac{4\cos y dy}{(16-16\sin^2y)^2} = \int\frac{4\cos y dy}{16^2(1-\sin^2 y)^2} =\frac{1}{64} \int\frac{\cos y dy}{(\cos^2 y)^2}=\frac{1}{64}\int\frac{dy}{\cos^3 y}=\frac{1}{64}\int{\sec^3 y dy}=\frac{1}{64}\cdot\frac{\sin y}{2\cos^2 y} +\frac{1}{2}\int\frac{dy}{\cos y}=\frac{1}{32}\cdot\frac{\sin y}{\cos^2 y} +\frac{1}{2}\ln|\tan(\frac{y}{2}+\frac{\pi}{4})|=\frac{1}{32}\cdot\frac{\sin y}{1-\sin^2 y} +\frac{1}{2}\ln|\tan(\frac{y}{2}+\frac{\pi}{4})|=\frac{1}{32}\cdot\frac{\frac{x}{4}}{1-(\frac{x}{4})^2 } +\frac{1}{2}\ln|\tan(\frac{\arcsin\frac{x}{4}}{2}+\frac{\pi}{4})|=\frac{x}{8(16-x^2)} +\frac{1}{2}\ln|\tan(\frac{\arcsin\frac{x}{4}}{2}+\frac{\pi}{4})|+C We have implement the formula: 1) \int\frac{dx}{\cos^n x}=\frac{\sin x}{(n-1)\cos^{n-1} x} +\frac{n-2}{n-1}\int\frac{dx}{\cos^{n-2}x} and 2) \int\frac{dx}{\cos x}=\ln|\tan(\frac{x}{2}+\frac{\pi}{4})| - Substition$$x=4\tan y \Rightarrow dx=\frac{4}{\cos^2 y} dy\tan y=\frac{x}{4} \Rightarrow y=\arctan\frac{x}{4}$\int\frac{1}{(16+x^2)^2}dx=\int\frac{1}{(4^2+x^2)^2}dx$=$\int\frac{\frac{4}{\cos^2 y}}{(16+(4\tan y)^2)^2}dy$=$\int\frac{\frac{4}{\cos^2 y}}{(16+16\tan^2 y)^2}dy= \int\frac{\frac{4}{\cos^2 y}}{16^2(1+\tan^2 y)^2}dy =\int\frac{\frac{4}{\cos^2 y}}{\frac{256}{(\cos^2 y)^2}}=\int\frac{\frac{4}{\cos^2 y}}{\frac{256}{\cos^4 y}}=\frac{1}{64} \int\frac{\cos^4 y dy}{\cos^2 }=\frac{1}{64} \int{\cos^2 y dy}=\frac{1}{64}(\frac{1}{2}\sin y\cos y+\frac{1}{2}y)=\frac{1}{128}(\frac{1}{\tan y+\cot y}+y)=\frac{1}{128}(\frac{1}{\tan y+\frac{1}{\tan y}}+y)=\frac{1}{128}(\frac{\tan y}{1+\tan^2 y}+y)$=$\frac{1}{128}(\frac{4x}{16+x^2}+\arctan\frac{x}{4})+C

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