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Universal Substitution

April 12th, 2009 |

Author: Merrin

There is a universal transformation that can convert any rational trigonometric function into a polynomial rational function. This allows us to use of techniques to integrate rational functions such as partial fractions. Similarly there is also a universal hyperbolic transformation which can convert hyperbolic rational functions into polynomial rational functions. These transformations use the substitutions of $latex t=\tan (x/2)&s=1$ or $latex w=\tanh (x/2)&s=1$. First we will solve all the details for the trigonometric functions, and then do the same for the hyperbolic functions.

Integration by substitution requires calculating $latex dx$ in terms of the new variable. Since we will make

${t=\tan (x/2)} \qquad {x=2\arctan t} \qquad {dx=\cfrac{2dt}{1+t^{2} } }$

Now we will proceed to calculate all the trigonometric functions in terms of $latex \tan (x/2)$ so we can substitute for them also in the integral. The easiest way to proceed for $latex \tan x$ is to use the double angle formula

$\tan x=\tan \left(\cfrac{x}{2} +\cfrac{x}{2} \right)=\cfrac{2\tan (x/2)}{1-\tan ^{2} (x/2)} =\cfrac{2t}{1-t^{2} }$

Similarly we can use the double angle formula for sine to derive

$\sin x=2\sin (x/2)\cos (x/2)=\cfrac{2\sin (x/2)\cos (x/2)}{\cos ^{2} (x/2)+\sin ^{2} (x/2)}$

$\sin x=\cfrac{2\tan (x/2)}{1+\tan ^{2} (x/2)} =\cfrac{2t }{1+t^{2} }$

The last function we need cosine which can be found from

$\cos x=\cfrac{\sin x}{\tan x} =\cfrac{2t}{1+t^{2} } \cfrac{1-t^{2} }{2t} =\cfrac{1-t^{2} }{1+t^{2} }$

The other three trigonometric forms can be found from the reciprocals of these three without further calculation.

${\csc x=\cfrac{1}{\sin x} =\cfrac{1+t^{2} }{2t} } \\ \\ \\ {\sec x=\cfrac{1}{\cos x} =\cfrac{1+t^{2} }{1-t^{2} } } \\ \\ \\ {\cot x=\cfrac{1}{\tan x} =\cfrac{1-t^{2} }{2t} }$

Now we can work out the same transformation for the hyperbolic functions. The starting point is to write $latex w=\tanh (x/2)$ and to find $latex dx$.

$w=\text{tanh}\, (x/2) \qquad x=2\text{arctanh}\, w\qquad dx=\cfrac{2}{1-w^2} dw$

Now we can find the hyperbolic functions under this substitution starting with $latex \tanh x$

$\tanh x=\cfrac{2\tanh (x/2)}{1+\tanh ^{2} (x/2)} =\cfrac{2w}{1+w^{2} }$

Recall that the plus sign in the denominator is due to Osborne’s rule. Now we can find $latex \sinh x$ from the double angle formula also

$\sinh x=2\sinh (x/2)\cosh (x/2)=\cfrac{2\sinh (x/2)\cosh (x/2)}{\cosh ^{2} (x/2)-\sinh ^{2} (x/2)}$

$\sinh x=\cfrac{2w}{1-w^{2} }$

Recall that the minus sign in the denominator is due to Osborne’s rule. There is a simple formula for $latex \cosh x$ in terms of these two.

$\cosh x=\cfrac{\sinh x}{\tanh x} =\cfrac{2w}{1-w^{2} } \cfrac{1+w^{2} }{2w} =\cfrac{1+w^{2} }{1-w^{2} }$

The other three hyperbolic functions are just the inverses of these basic three. You should summarize all our calculations in a table and then try some problems. The hyperbolic counterparts to the trigonometric substitutions have some reasonably placed sign changes. All the pieces are now in place to solve some pretty hard integrals.

Example 1. Use the universal transformation to solve

$\displaystyle \int \cfrac{dx}{\cos x} =\displaystyle \int \sec xdx$

Solution 1. Substitute in the transformation from the table and we have

$\int \sec xdx=\displaystyle \int \cfrac{1+t^{2} }{1-t^{2} } \cfrac{2dt}{1+t^{2} } =\displaystyle \int \cfrac{2dt}{1-t^{2} } =2\text{arctanh}\,(t)+C } \\\\\\ {\displaystyle \int \sec xdx = 2\text{arctanh}\, (\tan (x/2)) +C$

With this transformation, alternative forms of familiar integrals can appear.

Example 2. Use the universal transformation for hyperbolic functions to integrate

$\displaystyle \int \text{sech}\, xdx$

Solution 2. According to our table we have

$\displaystyle \int \text{sech}\, xdx=\displaystyle \int \cfrac{2dw}{1-w^{2} } \cfrac{1-w^{2} }{1+w^{2} } =\displaystyle \int \cfrac{2dw}{1+w^{2} } =2\text{arctan}\, (\tanh (x/2))+C$

These first two results appear to be related by a simple repositioning of h.

Example 3. Use the universal transformation to integrate

$\displaystyle \int \cfrac{dx}{\sin x+\cos x}$

Solution 3. Using the universal transformation for trigonometric functions we can rewrite the integral as a rational function.

$\displaystyle \int \cfrac{dx}{\sin x+\cos x} \to \displaystyle \int \cfrac{2dt}{1+t^{2} } \cfrac{1}{\left(\cfrac{2t}{1+t^{2} } +\cfrac{1-t^{2} }{1+t^{2} } \right)} =\displaystyle \int \cfrac{2dt}{2t+1-t^{2} } \\ \\ \\ I=-2\displaystyle \int \cfrac{dt}{t^{2} -2t-1} =-2\displaystyle \int \cfrac{dt}{(t-1)^{2} -2} =2\displaystyle \int \cfrac{dt}{2-(t-1)^{2} }$

We recognize the rational function as an integral of $latex \text{arctanh}\, x$ up to some constants.

$\displaystyle \int \cfrac{dx}{a^{2} -(x-b)^{2} }=\cfrac{1}{a} \text{arctanh}\,\left(\cfrac{x-b}{a} \right) \\ \\ \\\\ \\ \displaystyle \int \cfrac{dx}{\sin x+\cos x} =\cfrac{2}{\sqrt{2} } \text{arctanh}\,\left(\cfrac{t-1}{\sqrt{2} } \right)+C \\ \displaystyle \int \cfrac{dx}{\sin x+\cos x} =\sqrt{2} \text{arctanh}\,\left(\cfrac{\tan (x/2)-1}{\sqrt{2} } \right)+C \\ \\ \\ \displaystyle \int \cfrac{dx}{\sin x+\cos x} =\sqrt{2} \text{arctanh}\,\left(\cfrac{\sqrt{2} \tan (x/2)-\sqrt{2} }{2} \right)+C$

Example 4. Calculate the integral of

$\displaystyle \int \cfrac{dx}{\sin x+\tan x}$

Solution 4. We don’t even have to use partial fractions for this one.

${\displaystyle \int \cfrac{2dt}{1+t^{2} } \cfrac{1}{\left(\cfrac{2t}{1+t^{2} } +\cfrac{2t}{1-t^{2} } \right)} }= {\displaystyle \int \cfrac{2dt}{2t+2t(1+t^{2} )/(1-t^{2} )} =\displaystyle \int \cfrac{2dt(1-t^{2} )}{2t(1-t^{2} +1+t^{2} )} } \\ \\ \\=\displaystyle \int\cfrac{1-t^2}{4t}dt=\cfrac{1}{4}\ln\left|\tan(x/2)\right|-\cfrac{1}{8}\tan^2(x/2)+C$

Apparently this method blazes through some difficult integrals without any trouble. Now let us try a hyperbolic integral.

Example 5. Integrate

$\displaystyle \int \cfrac{\text{csch}\, x}{\sinh x+\cosh x} dx$

Solution 5. We use the universal transformation for hyperbolic functions

$I=\displaystyle \int \cfrac{\text{csch}\, x}{\sinh x+\cosh x} dx =\displaystyle \int \cfrac{2dw}{1-w^{2} } \cfrac{1-w^{2} }{2w} \cfrac{1}{\cfrac{2w}{1-w^{2} } +\cfrac{1+w^{2} }{1-w^{2} } }$
$=\displaystyle \int \cfrac{(1-w^{2} )dw}{(1+2w+w^{2} )w} =\displaystyle \int \cfrac{1-w}{w(1+w)} dw$
$\cfrac{1-w}{w(1+w)}=\cfrac{A}{w} +\cfrac{B}{1+w}$
$(1-w)=A(1+w)+Bw \to\quad w=0\quad A=1, w=-1\quad B=-2$
$\displaystyle \int \cfrac{dt}{w} -2\displaystyle \int \cfrac{dt}{1+w} =-2\ln \left|1+w\right|+\ln \left|w\right|+C$
$\displaystyle I = \int \cfrac{\text{csch}\, x}{\sinh x+\cosh x} dx$

$I=-2\ln \left|1+\tanh (x/2)\right|+\ln \left|\tanh (x/2)\right|+C$

Example 6. Integrate the following

$\displaystyle \int \tanh x dx$

Solution 6. The quick solution is to do a u substitution.

$\displaystyle \int \cfrac{\sinh x}{\cosh x} dx=\displaystyle \int \cfrac{du}{u} =\ln \left|\cosh x\right|+C$

We can obtain the same result with the universal substitution.

$\displaystyle \int \cfrac{2dw}{1-w^{2} } \cfrac{2w}{1-w^{2} } \cfrac{1-w^{2} }{1+w^{2} } =4\displaystyle \int \cfrac{wdw}{(1-w)(1+w)(1+w^{2} )}$

By partial fractions we have

$\cfrac{4w}{(1-w)(1+w)(1+w^{2} )}=\cfrac{A}{1-w}+\cfrac{B}{1+w}+\cfrac{Cw+D}{1+w^{2} } \\A(1+w)(1+w^{2})+B(1-w)(1+w^{2} )+(Cw+D)(1-w)(1+w)=4w$

$\begin{array}{l}{w=-1\quad 4B=-4\quad B=-1} \\ {w=1\quad \quad \, 4A=4\quad \quad A=1} \\ {w=0\quad \quad \, A+B+D=0\quad \, D=0} \\ {Cw-Bw+Aw=4w\quad \, C=4-A+B=4-1+-1=2} \end{array}$

Since A = 1 and B = -1 we can combine those terms to form

$\cfrac{1}{1-w} -\cfrac{1}{1+w} =\cfrac{1+w-1+w}{1-w^{2} } =\cfrac{2w}{1-w^{2} }$

We don’t have to do this step but it might make the form of the final integral look better. Now we integrate the pieces.

$\int \cfrac{2wdw}{1-w^{2} } +\displaystyle \int \cfrac{2wdw}{1+w^{2} } =-\ln \left|1-w^{2} \right|+\ln \left|1+w^{2} \right| } \\ {\displaystyle \int \tanh xdx =\ln \left|\cfrac{1+\tanh ^{2} (x/2)}{1-\tanh ^{2} (x/2)} \right|+C$

We get a big nasty formula compared to $latex \displaystyle \int \tanh x= \ln \left|\cosh x\right|+C$. Let us get our solution into form. Multiply the top and bottom of the logarithm by $latex \cosh^{2} (x/2)$

Recall that $latex \cosh ^{2} u-\sinh ^{2} u=1$ from Osborne’s rule
for the expression in the bottom.

${\ln \left|\cfrac{1+\tanh ^{2} (x/2)}{1-\tanh ^{2} (x/2)} \right|} ={\ln \left|\cfrac{\cosh ^{2} (x/2)+\sinh ^{2} (x/2)}{\cosh ^{2} (x/2)-\sinh ^{2} (x/2)} \right|} \\ ={\ln \left|\cosh ^{2} (x/2)+\sinh ^{2} (x/2)\right|}$

Recall that $latex \cosh (A+B)=\cosh (A)\cosh (B)+\sinh (A)\sinh (B)$ with the plus sign from Osborne’s Rule.

$\ln \left|\cosh ^{2} (x/2)+\sinh ^{2} (x/2)\right|=\ln \left|\cosh x\right|$

This proves the common form a second way using the universal substitution and common identities.

Example 7. Integrate the following while considering the three cases that a > b, a $latex \displaystyle \int \cfrac{dx}{a+b\cos \theta } $

Solution 7. There will be different solutions depending on each case.

Case I (a=b): This is the simplest case so we can write immediately

$\displaystyle \int \cfrac{2dt}{1+t^2}\cfrac{1}{\left(a+b\left(\cfrac{1-t^2}{1+t^2} \right)\right)}=\displaystyle \int \cfrac{2dt}{a(1+t^2)+b(1-t^{2}) }$

$\displaystyle \int \cfrac{2dt}{(a+b)+(a-b)t^{2} } =\displaystyle \int \cfrac{2dt}{2} =t+C=\arctan (x/2)+C$

Case II ( $ a>b$): Factor out a positive quantity where $ a>b$.

$\begin{array}{c} {\displaystyle\displaystyle \int \frac{2dt}{(a+b)+(a-b)t^{2} } =\frac{1}{a-b} \displaystyle \int \frac{2dt}{\cfrac{a+b}{a-b} +t^{2} } } \\\\ {=\cfrac{2}{\sqrt{a^{2} -b^{2} } }\displaystyle \arctan \left(\frac{\tan (x/2)}{\sqrt{\frac{a+b}{a-b} } } \right)+C} \end{array}$

Case III ( a=b)

$\begin{array}{c} {\displaystyle\displaystyle \int \cfrac{2dt}{(a+b)+(a-b)t^{2} } =\cfrac{1}{b-a} \displaystyle \int \frac{2dt}{\cfrac{a+b}{b-a} -t^{2} } } \\ \\ {=\cfrac{2}{\sqrt{b^{2} -a^{2} } } \text{arctanh}\, \left(\cfrac{\tan (x/2)}{\sqrt{\frac{a+b}{b-a} } } \right)+C} \end{array}$

When $latex a=b$ the form of the integral is $latex \text{arctan}\, (x/2)$. As we can see from the integral that when $latex a>b$ the form of the integral up to some constants is $latex A \arctan (Bt)$.When $latex a<b$ the form of the integral up to some constants is $latex C\text{arctanh}\,(Dt)$. The constants in the integral definitely affect the form.

Example 8. Use the universal transformation to integrate

$\displaystyle \int \cfrac{dx}{a+b\cos x+c\sin x}$

Solution 8. This is a challenging integral, even more so since it is an indefinite integral. Use the universal transformation and see what we can do.

${I=\displaystyle \int \cfrac{dx}{a+b\cos x+c\sin x} =\displaystyle \int \cfrac{2dt}{1+t^{2} } \cfrac{1}{\left(a+b\cfrac{1-t^{2} }{1+t^{2} } +\cfrac{2ct}{1+t^{2} } \right)} } \\ {I=\displaystyle \int \cfrac{2dt}{a(1+t^{2} )+b(1-t^{2} )+2ct} =\displaystyle \int \cfrac{2dt}{(a-b)t^{2} +2ct+(a+b)} }$

To determine the form of the integral we will factor out the constant $latex a-b$.

${I=\cfrac{1}{a-b} \displaystyle \int \cfrac{2dt}{t^{2} +\cfrac{2ct}{a-b} +\cfrac{a+b}{a-b} } =\cfrac{1}{\alpha } \displaystyle \int \cfrac{2dt}{t^{2} +2\beta +\gamma ^{2} } } \\ {\alpha =a-b\quad \beta =\cfrac{c}{a-b} \quad \gamma ^{2} =\cfrac{a+b}{a-b} }$

Now we complete the square and write

$\cfrac{1}{\alpha } \displaystyle \int \cfrac{2dt}{(t+\beta )^{2} +\gamma ^{2} -\beta ^{2} }$

$\gamma ^{2} -\beta ^{2} =\cfrac{a+b}{a-b} -\left(\cfrac{c}{a-b} \right)^{2} =\cfrac{a^{2} -b^{2} -c^{2} }{(a-b)^{2} }$

So we see that when $latex a^{2} >b^{2} +c^{2} $ the integral is an $latex \text{arctan}\, (t)$ up to some constants. We are ready to write down the solution for this case.

${I} = {\cfrac{2}{\alpha \sqrt{\gamma ^{2} -\beta ^{2} } } \arctan \left(\cfrac{t+\beta }{\sqrt{\gamma ^{2} -\beta ^{2} } } \right)+C}$

${I} = {\cfrac{2}{\sqrt{a^{2} -b^{2} -c^{2} } } \arctan \left(\cfrac{(a-b)\tan (x/2)+c}{\sqrt{a^{2} -b^{2} -c^{2} } } \right)+C }$

Now we will solve the second case when …

${I=\displaystyle \int \cfrac{dt}{-k^{2} +t^{2} } =-\displaystyle \int \cfrac{dt}{k^{2} -t^{2} } =-\text{arctanh}\,(t/k)+C }$

${-\beta ^{2} +\gamma ^{2} =\cfrac{a+b}{a-b} -\left(\cfrac{c}{a-b} \right)^{2} =\cfrac{-a^{2} +b^{2} +c^{2} }{(a-b)^{2} } }$

${-\cfrac{1}{\alpha } \displaystyle \int \cfrac{2dt}{\beta ^{2} -\gamma ^{2} -(t+\beta )^{2} } =-\cfrac{1}{\alpha } \cfrac{2}{\sqrt{\beta ^{2} -\gamma ^{2} } } \text{arctanh}\,\left(\cfrac{t+\beta }{\sqrt{\beta ^{2} -\gamma ^{2} } } \right)+C}$

${=\cfrac{2}{\sqrt{c^{2} +b^{2} -a^{2} } } \text{arctanh}\,\left(\cfrac{(b-a)\text{arctan}\, (x/2)-c}{\sqrt{c^{2} +b^{2} -a^{2} } } \right)+C}$

I have used the identity that

$-\text{arctanh}\,(\tan x)=\text{arctanh}\,(-\tan x)$

The form of the integral suggests one more case that we need to consider which is when $latex a^{2} =b^{2} +c^{2}$, in which case the integral immediately reduces to

${\cfrac{1}{\alpha } \displaystyle \int \cfrac{2dt}{(t+\beta )^{2} +\gamma ^{2} -\beta ^{2} } =\cfrac{1}{\alpha } \displaystyle \int \cfrac{2dt}{(t+\beta )^{2} } =-\cfrac{1}{\alpha } \cfrac{2}{t+\beta } +C } \\ {I=\cfrac{1}{b-a} \cfrac{-2}{\cfrac{c}{b-a} -\tan (x/2)} +C=\cfrac{-2}{c+(a-b)\tan (x/2)} +C}$

Example 9. Integrate the following

$\displaystyle \int \cfrac{\sin x}{9+\sin ^{2} x} dx$

Solution 9. Is the universal substitution necessary or is there an easier way? Let us try to manipulate the denominator so the integrand is not so wound up with sine.

$\int \cfrac{\sin x}{10-\cos ^{2} x} dx =\displaystyle \int \cfrac{-du}{10-u^{2} } =\cfrac{-1}{\sqrt{10} } \text{arctanh}\,(u/\sqrt{10} )+C} \\ {\displaystyle \int \cfrac{\sin x}{10-\cos ^{2} x} dx =\cfrac{-1}{\sqrt{10} } \text{arctanh}\,(\cos (x)/\sqrt{10} )+C$

If there is an easier method try it before going to the universal substitution.

Example 10. Evaluate the following integral

$\displaystyle \int \cfrac{\sec xdx}{\sin x+\cos x}$

Solution 10. There is a trick to solving this integral, just divide the top and bottom by cosine.

$\displaystyle \int \cfrac{\sec ^{2} xdx}{\tan x+1} =\ln \left|1+\tan x\right| +C$

Let us use the universal substitution to get some practice with the method. We have

$I=\displaystyle \int \cfrac{\sec ^{2} xdx}{1+\tan x} =\displaystyle \int \cfrac{2dt}{1+t^{2} } \cdot \left(\cfrac{1+t^{2} }{1-t^{2} } \right)^{2} \cfrac{1}{1+\cfrac{2t}{1-t^{2} } }$

$I=\displaystyle \int \cfrac{2dt(1+t^{2} )}{(1-t^{2} )(2t+1-t^{2} )} \quad \quad \quad R(t)=\cfrac{-2(1+t^{2} )}{(1-t^{2} )(t^{2} -2t-1)}$

Now we have to do some work with partial fractions. The algebra of partial fractions is just busy work so I will spare us the details.

$R(t)=-\cfrac{1}{t+1} -\cfrac{1}{t-1} +\cfrac{2t-2}{t^{2} -2t-1}$

Integrating the pieces we have

${I=-\ln \left|t-1\right|-\ln \left|1+t\right|+\ln \left|t^{2} -1-2t\right|+C} \\ {\displaystyle \int \cfrac{\sec xdx}{\sin x+\cos x} =\ln \left|\cfrac{t^{2} -1-2t}{t^{2} -1} \right|+C=\ln \left|1+\cfrac{2t}{1-t^{2} } \right|+C=\ln \left|1+\tan x\right|+C}$

The universal substitutions are useful for dealing with integrals of rational functions of trigonometric or hyperbolic functions. This is particularly the case when the form of the integrand does not contain well defined forms such as $latex (1-f^{2} )^{n/2} ,(f^{2} -1)^{n/2} ,\, {\rm or} (1+f^{2} )^{n/2} $ which simplify with the normal trigonometric and hyperbolic substitutions. The universal substitutions can be used to find expressions for familiar integrals in different terms. The universal substitution might not be the first consideration for an integral but when stuck it most certainly should not be ignored as a method of solving more advanced integrals. The universal substitution is somewhat limited in practical use because partial fractions on high order polynomials don’t work so well because there is probably no way to factor them.

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