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How to integrate such function:

$\int_{-\pi/3}^{\pi/3}1-\tan^4(x)$

I already found a solution using the trigonometric function secant. It looks like this:

$$u=\tan(x),\quad \frac{du}{dx}=\sec^2(x)$$

$$ \begin{align} \int 1-\tan^4(x) & = \int 1 \;dx - \int\tan^4(x) \; dx \\ & = x-\int\tan^4(x) \; dx \\ & = x-\int\tan^2(x) \tan^2(x) \; dx \\ & = x-\int\tan^2(x) (\sec^2(x) - 1) \; dx \\ & = x-\int\tan^2(x) \sec^2(x) \; dx - \int\tan^2(x) \; dx \\ & = x-\int u^2 \; du - (\tan(x)-x) \\ & = x-\frac{1}{3}u^3 - (\tan(x)-x) \\ & = 2x-\frac{1}{3}\tan^3(x)-\tan(x) \end{align} $$

The only problem now is, that I have to find a solution without using secant. Can someone say me how to solve this without secant?

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    $\begingroup$ Can we use cosine instead? What does "without using secant" really mean? $\endgroup$
    – Aryabhata
    Feb 8, 2012 at 20:36
  • $\begingroup$ cosine is ok. I'm not allowed to use secant, because it is not yet "officially" introduced in my math course. $\endgroup$
    – kiritsuku
    Feb 8, 2012 at 20:48
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    $\begingroup$ +C$\phantom{\;\;\;\;\;\;\;}$ $\endgroup$
    – wckronholm
    Feb 8, 2012 at 21:21
  • $\begingroup$ @Antauras: The indefinite integral of $1-\tan^4 x$ is $-(1/3)\tan^3 x +\tan x +C$. Just a minus sign error. $\endgroup$ Feb 8, 2012 at 21:46

2 Answers 2

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Use $$ \frac{\mathrm{d} }{\mathrm{d} x} \tan(x) = 1 + \tan^2(x) $$ Then $$ \begin{eqnarray} \int \left(1-\tan^4(x)\right)\mathrm{d} x &=& \int \left(1-\tan^2(x)\right) \left(1+\tan^2(x)\right) \mathrm{d} x = \int \left(1-\tan^2(x)\right) \mathrm{d} \tan(x)\\ &=& \tan(x) - \frac{1}{3} \tan^3(x) + \color{gray}{\text{const}} \end{eqnarray} $$

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  • $\begingroup$ How did you come to $(1-tan^2(x))(1+tan^2(x))$?. The latter is clear, but not the former. $\endgroup$
    – kiritsuku
    Feb 8, 2012 at 21:54
  • $\begingroup$ $(a^2-b^2)=(a-b)(a+b)$ $\endgroup$
    – Hidde
    Feb 8, 2012 at 22:15
  • $\begingroup$ Binomial formula. Cool! One last question: How does the last step work (from $1-\tan^2(x)$ to $-(1/3)\tan^3(x)$)? $\endgroup$
    – kiritsuku
    Feb 8, 2012 at 23:10
  • $\begingroup$ @Antonas $\int ( 1- \tan^2(x)) \mathrm{d} \tan(x) = \int \mathrm{d} \tan(x) - \int \tan^2(x) \mathrm{d} \tan(x) = \tan(x) - \frac{1}{3} \tan^3(x) + C$. $\endgroup$
    – Sasha
    Feb 8, 2012 at 23:12
  • $\begingroup$ It is the $\int\tan^2(x)\mathrm{d}\tan(x)$ that make problems. Where does the $(1/3)$ come from? $\endgroup$
    – kiritsuku
    Feb 8, 2012 at 23:49
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Sasha's solution is very elegant. However, if you can't see a trick like this, you can always mechanically integrate any rational function of sine and cosine using the Weierstraß substitution.

With

$$\tan x=\frac{2t}{1-t^2}\;,$$

$$\frac{\mathrm dx}{\mathrm dt}=\frac2{1+t^2}\;,$$

$$t=\tan\frac x2\;,$$

the integral of $\tan^4 x$ becomes

$$ \begin{eqnarray} &&\int\tan^4x\mathrm dx \\ &=& \int\left(\frac{2t}{1-t^2}\right)^4\frac2{1+t^2}\mathrm dt \\ &=& \int\frac{32t^4}{(1-t^2)^4(1+t^2)}\mathrm dt \\ &=& \int\left(\frac2{t^2+1}-\frac1{(t-1)^2}-\frac1{(t+1)^2}+\frac1{(t-1)^3}-\frac1{(t+1)^3}+\frac1{(t-1)^4}+\frac1{(t+1)^4}\right)\mathrm dt \\ &=& 2\arctan t+\frac1{t-1}+\frac1{t+1}-\frac1{2(t-1)^2}+\frac1{2(t+1)^2}-\frac1{3(t-1)^3}-\frac1{3(t+1)^3}+ \color{gray}{\text{const}} \\ &=& 2\arctan t+\frac{2t}{t^2-1}-\frac{8t^3}{3(t^2-1)^3}+ \color{gray}{\text{const}} \\ &=& x-\tan x+\frac13\tan^3 x+ \color{gray}{\text{const}}\;, \end{eqnarray} $$

where I used Wolfram|Alpha for pulling apart the fractions and putting them back together again.

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