# Integration by Partial Fractions $\int \sec^3xdx$

I have looked at multiple ways to do partial fractions to integrate $$\sec^3x$$, but there is a part where I keep getting stuck when I see the partial fractions get split up.

$$\int \frac{\cos x}{\cos^4 x} = \int \frac{1}{(1-y^2)^2}$$

The next step I have seen is shown like this on Wikipedia:

$$\frac{1}{(1-u^2)^2}=\frac{1/4}{1-u}+\frac{1/4}{(1-u)^2}+\frac{1/4}{1+u}+\frac{1/4}{(1+u)^2}.$$

Where does the $$\frac{1}{4}$$ come from in the numerator?

And for the denominator, why are there two additional positive fraction decompositions $$1+u$$ and $$(1+u)^2$$?

• See this page for the general theorem in use. – Bernard May 2 at 12:35
• – cgiovanardi May 2 at 14:54

And for the denominator, why are there two additional positive fraction decompositions

Because $$1-u^2 = (1-u)(1+u).$$ It's the "difference of squares" factorization. And so, squaring both sides, we also have $$(1-u^2)^2 = (1-u)^2(1+u)^2$$.

In partial fractions decomposition, all irreducible factors must appear as denominators, to the highest power that appears, as well as to all lower powers. So that means we need a denominator for $$(1-u)$$ and $$(1-u)^2,$$ but also for $$(1+u)$$ and $$(1+u)^2.$$

Where does the $$\frac{1}{4}$$ come from in the numerator?

For partial fraction decomposition, the numerators must be chosen so that if you were to give every term common denominator and combine the numerators, you would get the starting fraction back. It takes a bit of algebra to find what numerators do the job, but that is where the $$\dfrac{1}{4}$$ comes from.

But let's do the algebra. The normal way to find these numerators is to introduce variables for them and solve. Like so:

$$\dfrac{1}{(1-u^2)^2} = \dfrac{1}{(1+u)^2(1-u)^2} = \dfrac{A}{1+u} + \dfrac{B}{(1+u)^2} + \dfrac{C}{1-u} + \dfrac{D}{(1-u)^2}$$

here the "variable" numerators are called $$A, B, C,$$ and $$D.$$

We solve for them by clearing denominators:

$$1 = A(1-u)^2(1+u) + B(1-u)^2 + C(1-u)(1+u)^2 + D(1+u)^2$$

Then in general you may have to expand all those binomials and match like terms of the resulting polynomials in $$u$$. But a shortcut can be to sub in the values of the roots in $$u$$. For example, at $$u=1,$$ the above equation becomes

$$1 = A(1-1)^2(1+1) + B(1-1)^2 + C(1-1)(1+1)^2+ D(1+1)^2 = D(1+1)^2.$$

So we have $$D=1/4.$$

Similarly, by next subbing $$u=-1,$$

$$1 = B(1-(-1))^2$$

So we have $$B=1/4.$$

To figure out that $$A$$ and $$C$$ are also $$1/4$$, we can either sub in some other, non-root values of $$u$$, or bite the bullet and do the algebra. (But if this were a partial fractions problem with no powers of irreducibles, we'd be done).

I'll do the algebra, I guess. Expand, and gather like terms in $$u$$, treating the coefficients $$A, B, C,$$ and $$D$$ as variables.

$$1 = A(1-u-u^2+u^3) + B(1-2u+u^2) + C(1+u-u^2-u^3) + D(1+2u+u^2) \\ = (A+B+C+D) + (-A-2B+C+2D)u + (-A+B-C+D)u^2 + (A-C)u^3.$$

Now for two polynomials in $$u$$ to be equal for all values of $$u$$, every coefficient must be equal. On the left-hand side we have the polynomial $$1$$, which has a constant coefficient of $$1$$ and all higher coefficients are $$0$$.

So this gives us the following equations:

$$1 = A+B+C+D\\ 0 = -A-2B+C+2D\\ 0 = -A+B-C+D\\ 0 = A-C.$$

We could solve this system of four linear equations in four unknowns, using Gaussian elimination or substitution or matrix methods. But first let's remember that we already know that $$B=D=1/4.$$ So we don't have four unknowns any more, our only unknowns left are $$A$$ and $$C$$. By the fourth equation, $$A=C$$, so our only unknown is really $$A$$. The first equation becomes

$$1 = A+B+C+D = A + \dfrac{1}{4} + A + \dfrac{1}{4}.$$

So $$2A=1-\dfrac{2}{4} = \dfrac{1}{2}.$$ And so $$A=B=C=D=\dfrac{1}{4}.$$

• Oh, seeing the way you wrote that makes sense (going from $1-u^2$ to $(1-u^2)^2)$, just adding ^2 to the factors – Evan Kim May 2 at 12:38

This is actually a problem of partial fractions.

When dividing one fraction to partial fractions, all possible cases of denominator should be considered. In this case, the denominator can be factorised to $$(1-u)^2(1+u)^2$$, so there are actually in total 8 possible fractions, with denominators being $$1-u$$, $$(1-u)^2$$, etc. Setting all the numerators to A, B, C, D, etc. for different denominator, layout the expression and compare coefficients, the result should then be that 1 fraction equal to 4 partial fractions.

Note that $$(1-y^2)^2=(y-1)^2(y+1)^2$$. On the other hand, every rational function of the form$$\frac{p(x)}{(x-r_1)^{m_1}\times\cdots\times(x-r_k)^{m_k}},$$where $$\deg p(x), can be written as a linear combination of rational functions of the type $$\frac1{(x-r_j)^N}$$, with $$N\in\{1,2,\ldots,m_j\}$$. The rest is a matter of computations.