Example 15: $\int\frac{(3\sin\phi-2)\cos\phi}{5-\cos^2\phi-4\sin\phi}\,d\phi$

Basics you need

Solution

1. Write the bottom in terms of $\sin\phi$

   Using $\cos^2\phi = 1 - \sin^2\phi$⁠:

   $$\begin{aligned} 5 - \cos^2\phi - 4\sin\phi &= 5 - (1 - \sin^2\phi) - 4\sin\phi \\ &= \sin^2\phi - 4\sin\phi + 4 \\ &= (\sin\phi - 2)^2. \end{aligned}$$

   Since $\sin\phi \le 1$⁠, we have $\sin\phi - 2 \le -1$⁠, so the bottom is at least $1$⁠. It is never zero, and the integrand is defined for every real $\phi$⁠.

   We want the integrand in terms of $\sin\phi$ only (plus the single $\cos\phi$ that will become part of $dy$⁠). Recognising the perfect square $s^2 - 4s + 4 = (s-2)^2$ also tells us at once where the integrand is defined.
   $\sin^2\phi + \cos^2\phi = 1$⁠; and $-1 \le \sin\phi \le 1$ for every $\phi$⁠. Basics: trigonometric identities

2. Substitute $y = \sin\phi$

   Let $y = \sin\phi$⁠. Then $dy = \cos\phi\,d\phi$⁠, and

   $$\begin{aligned} &\int \frac{(3\sin\phi - 2)\cos\phi}{5 - \cos^2\phi - 4\sin\phi}\,d\phi \\ &= \int \frac{(3y-2)\,dy}{5 - (1 - y^2) - 4y} \\ &= \int \frac{3y-2}{y^2 - 4y + 4}\,dy = \int \frac{3y-2}{(y-2)^2}\,dy. \end{aligned}$$

   Call this integral $I$⁠. The new variable $y$ takes values in $[-1, 1]$⁠.

   The factor $\cos\phi\,d\phi$ in the question is replaced by $dy$⁠; nothing involving $\phi$ is left. That is the test of a good substitution. The range $-1 \le y \le 1$ will matter in Step 5, when we decide the sign of $y - 2$⁠.
   Substitution rule (Section 7.3.1); $\frac{d}{d\phi}\sin\phi = \cos\phi$⁠. Lesson: substitution

3. Split $\frac{3y-2}{(y-2)^2}$ into partial fractions

   The fraction is proper (top degree 1, bottom degree 2) and the bottom is a repeated linear factor. By Table 7.2, row 2,

   $$\frac{3y-2}{(y-2)^2} = \frac{A}{y-2} + \frac{B}{(y-2)^2},$$ $$3y - 2 = A(y-2) + B.$$

   Comparing coefficients: of $y$⁠, $A = 3$⁠; constant terms, $B - 2A = -2$⁠, so $B = 4$⁠. (Check: $y = 2$ gives $4 = B$⁠.) So

   $$\frac{3y-2}{(y-2)^2} = \frac{3}{y-2} + \frac{4}{(y-2)^2}.$$
   Another way to see it: $3y - 2 = 3(y - 2) + 4$⁠, and dividing by $(y-2)^2$ gives the same two pieces. Either way, both powers of the repeated factor are needed.
   Equal polynomials have equal coefficients. Lesson: partial fractions

4. Integrate in $y$

   $$\begin{aligned} I &= 3\int\frac{dy}{y-2} + 4\int\frac{dy}{(y-2)^2} \\ &= 3\log|y-2| + 4\left(-\frac{1}{y-2}\right) + C. \end{aligned}$$
   The first piece is a shifted $\frac1y$⁠, so it gives a logarithm. The second is the power $(y-2)^{-2}$⁠, which by the power rule integrates to $\frac{(y-2)^{-1}}{-1} = -\frac{1}{y-2}$⁠.
   Formulae (xii) and (i) with the substitution $t = y - 2$⁠, $dt = dy$⁠. Basics: powers of x

5. Return to $\phi$ and remove the absolute value

   Put $y = \sin\phi$ back:

   $$I = 3\log|\sin\phi - 2| + \frac{4}{2 - \sin\phi} + C,$$

   using $-\frac{4}{\sin\phi - 2} = \frac{4}{2 - \sin\phi}$⁠. Since $\sin\phi \le 1$⁠, the number $\sin\phi - 2$ is always negative, so $|\sin\phi - 2| = 2 - \sin\phi$⁠, which is always positive. Therefore

   $$I = 3\log(2 - \sin\phi) + \frac{4}{2 - \sin\phi} + C.$$
   The answer to an integral in $\phi$ must be written in $\phi$⁠. The absolute value can be removed only because we know the sign of what is inside: $|u| = -u$ when $u < 0$⁠. This is the book's remark "since $2 - \sin\phi$ is always positive".
   $|u| = -u$ for $u < 0$⁠; $-1 \le \sin\phi \le 1$⁠. Basics: logarithms and 1/x

6. Check: differentiate the answer

   By the chain rule, $\frac{d}{d\phi}\log(2 - \sin\phi) = \frac{-\cos\phi}{2 - \sin\phi}$ and $\frac{d}{d\phi}(2 - \sin\phi)^{-1} = \frac{\cos\phi}{(2 - \sin\phi)^2}$⁠. So

   $$\begin{aligned} &\frac{d}{d\phi}\left[3\log(2 - \sin\phi) + \frac{4}{2 - \sin\phi}\right] \\ &= \frac{-3\cos\phi}{2 - \sin\phi} + \frac{4\cos\phi}{(2 - \sin\phi)^2} \\ &= \frac{\cos\phi\,[-3(2 - \sin\phi) + 4]}{(2 - \sin\phi)^2} \\ &= \frac{(3\sin\phi - 2)\cos\phi}{(2 - \sin\phi)^2}. \end{aligned}$$

   By Step 1, $(2 - \sin\phi)^2 = (\sin\phi - 2)^2$⁠, and this equals $5 - \cos^2\phi - 4\sin\phi$⁠. So the derivative is exactly the integrand.

   Differentiating returns the original integrand for every $\phi$⁠, which confirms the substitution, the partial fractions and the sign handling all at once.
   Chain rule; $\frac{d}{d\phi}\sin\phi = \cos\phi$⁠. Basics: integration undoes differentiation

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