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[parent] integration of rational function of sine and cosine (Topic)

The integration task

$\displaystyle \int\!R(\sin{x},\,\cos{x})\,dx,$ (1)

where the integrand is a rational function of $ \sin{x}$ and $ \cos{x}$, changes via the Weierstrass substitution
$\displaystyle \tan\frac{x}{2} = t$ (2)

to a form having an integrand that is a rational function of $ t$. Namely, since $ x = 2\arctan{t}$, we have
$\displaystyle dx = 2\cdot\frac{1}{1+t^2}\,dt,$ (3)

and we can substitute
$\displaystyle \sin{x} = \frac{2t}{1+t^2}, \;\; \cos{x} = \frac{1-t^2}{1+t^2},$ (4)

getting
$\displaystyle \int\!R(\sin{x},\,\cos{x})\,dx\; =\; 2\int\!R\!\left(\frac{2t}{1\!+\!t^2},\,\frac{1\!-\!t^2}{1\!+\!t^2}\right)\frac{dt}{1\!+\!t^2}.$

Proof of the formulae (4): Using the double angle formulas of sine and cosine and then dividing the numerators and the denominators by $ \cos^2\frac{x}{2}$ we obtain

$\displaystyle \sin{x} = \frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{\sin^2\frac{x}{2... ...\frac{x}{2}} = \frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}} = \frac{2t}{1+t^2},$
$\displaystyle \cos{x} = \frac{\cos^2\frac{x}{2}-\sin^2\frac{x}{2}}{\sin^2\frac{... ...x}{2}} = \frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}} = \frac{1-t^2}{1+t^2}.$

Example. The above formulae give from $ \displaystyle \int\frac{dx}{\sin{x}}$ the result

$\displaystyle \int\frac{dx}{\sin{x}} = \int\frac{1\!+\!t^2}{2t}\cdot2\cdot\frac... ...nt\frac{dt}{t} = \ln\vert t\vert+C = \ln\left\vert\tan\frac{x}{2}\right\vert+C.$

Note. The substitution (2) is sometimes called the “universal trigonometric substitution”. In practice, it often gives rational functions that are too complicated. In many cases, it is more profitable to use other substitutions:

  • In the case $ \int\!R(\sin{x})\cos{x}\,dx$ the substitution $ \sin{x} = t$ is simpler.
  • Similarly, in the case $ \int\!R(\cos{x})\sin{x}\,dx$ the substitution $ \cos{x} = t$ is simpler.
  • If the integrand depends only on $ \tan{x}$, the substitution $ \tan{x} = t$ is simpler.
  • If the integrand is of the form $ R(\sin^2{x},\, \cos^2{x})$, one can use the substitution $ \tan{x} = t$; then
    $ \displaystyle \cos^2{x} = \frac{1}{1+\tan^2{x}} = \frac{1}{1+t^2}$, $ \displaystyle \sin^2{x} = 1-\cos^2{x} = \frac{t^2}{1+t^2}$, $ \displaystyle dx = \frac{dt}{1+t^2}.$

Example. The integration of $ \displaystyle \int\!\frac{dx}{\cos^4{x}}\,dx$ is of the last case:

$\displaystyle \int\!\frac{dx}{\cos^4{x}}\,dx = \int\!\frac{1}{(\cos^2{x})^2}\,d... ...+t^2} = \int\!(1+t^2)\,dt = \frac{t^3}{3}+t+C = \frac{1}{3}\tan^3{x}+\tan{x}+C.$

Example. The integral $ \displaystyle I = \int\!\frac{dx}{\cos^3{x}}\,dx = \int\! \sec^3{x}\,dx $ is a peculiar case in which one does not have to use the substitutions mentioned above, as integration by parts is a simpler method for evaluating this integral. Thus,

$\displaystyle u = \sec{x}\; \Rightarrow\; du = \sec{x}\;\tan{x}\,dx; \qquad dv = \sec^2{x}\,dx \; \Rightarrow \; v = \tan{x}.$
Therefore,
\begin{displaymath}\begin{array}{rl} I & \displaystyle = \int\! \sec^3{x}\,dx \\... ...ystyle = \sec{x}\;\tan{x} - I + \int\! \sec{x}\,dx. \end{array}\end{displaymath}

Hence,

$\displaystyle \int\!\frac{dx}{\cos^3{x}}\,dx = \frac{1}{2} \big( \sec{x}\;\tan{x}\; + \ln\;\vert\sec{x}\; + \; \tan{x}\vert \big)+C.$



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"integration of rational function of sine and cosine" is owned by pahio. [ full author list (3) ]
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See Also: goniometric formulas, substitution for integration, Weierstrass substitution formulas, Euler's substitutions for integration

Other names:  universal trigonometric substitution

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Cross-references: integration by parts, integral, denominators, numerators, cosine, sine, double angle formulas, proof, Weierstrass substitution, rational function
There are 3 references to this entry.

This is version 18 of integration of rational function of sine and cosine, born on 2007-05-14, modified 2007-05-23.
Object id is 9380, canonical name is IntegrationOfRationalFunctionOfSineAndCosine.
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AMS MSC: 26A36 (Real functions :: Functions of one variable :: Antidifferentiation)
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Open question in the entry 9380 by pahio on 2007-05-14 09:21:08
Stevecheng's correction question (if I understood it):
R(sin{x}, cos{x})
is supposed continuous in odd multiples of pi. If we integrate through such a point using t = tan(x/2), can we obtain right results?
If anybody knows the thing, please supplement the entry.
Jussi


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