Example 6: $\int\sin^3x\cos^2x\,dx$⁠, $\int\frac{\sin x}{\sin(x+a)}\,dx$⁠, $\int\frac{dx}{1+\tan x}$

Basics you need

Part (i): $\int\sin^3x\cos^2x\,dx$

1. Save one $\sin x$ and convert the rest to cosines

   $$\begin{aligned} \sin^3x\cos^2x &= \sin^2x\cos^2x\,(\sin x) \\ &= (1 - \cos^2x)\cos^2x\,(\sin x). \end{aligned}$$
   The odd power of $\sin x$ is the key. One factor $\sin x$ is kept aside because it is (minus) the derivative of $\cos x$⁠; the remaining even power $\sin^2x$ is converted into cosines with $\sin^2x = 1 - \cos^2x$⁠. Now everything except $\sin x\,dx$ is a function of $\cos x$⁠.
   $\sin^2x + \cos^2x = 1$⁠. Basics: trigonometric identities

2. Substitute $t = \cos x$

   $$\begin{gathered} t = \cos x, \qquad dt = -\sin x\,dx, \\ \sin x\,dx = -dt. \end{gathered}$$ $$\begin{aligned} &\int(1 - \cos^2x)\cos^2x\,(\sin x)\,dx \\ &= -\int(1 - t^2)\,t^2\,dt \\ &= -\int(t^2 - t^4)\,dt. \end{aligned}$$
   $\frac{d}{dx}\cos x = -\sin x$⁠, so $\sin x\,dx$ is exactly $-dt$⁠. The minus sign must travel with it.
   Substitution, Section 7.3.1. Lesson: integration by substitution

3. Integrate the polynomial and return to $x$

   $$\begin{aligned} -\int(t^2 - t^4)\,dt &= -\left(\frac{t^3}{3} - \frac{t^5}{5}\right) + C \\ &= -\frac13\cos^3x + \frac15\cos^5x + C. \end{aligned}$$
   Power formula (i) for $t^2$ and $t^4$⁠; then $t = \cos x$⁠. Multiplying the bracket by $-1$ flips both signs.
   Standard formula (i), Section 7.2. Basics: the power rule

4. Check: differentiate the answer

   $$\begin{aligned} &\frac{d}{dx}\left(-\frac13\cos^3x + \frac15\cos^5x\right) \\ &= \cos^2x\sin x - \cos^4x\sin x \\ &= \cos^2x\sin x\,(1 - \cos^2x) \\ &= \sin^3x\cos^2x. \ \checkmark \end{aligned}$$

   Valid for every real $x$⁠.

   $\frac{d}{dx}\cos^n x = n\cos^{n-1}x\cdot(-\sin x)$ by the chain rule. Then $1 - \cos^2x = \sin^2x$ rebuilds $\sin^3x$⁠.

Part (ii): $\int\frac{\sin x}{\sin(x + a)}\,dx$

1. Shift the variable: $t = x + a$

   Put $t = x + a$⁠. Then $dt = dx$ and $x = t - a$⁠, so

   $$\int\frac{\sin x}{\sin(x + a)}\,dx = \int\frac{\sin(t - a)}{\sin t}\,dt.$$

   The integrand needs $\sin(x + a) \ne 0$⁠, that is $\sin t \ne 0$⁠.

   The awkward part is the denominator $\sin(x + a)$⁠. Naming $x + a$ as $t$ makes the denominator the simple $\sin t$⁠, and since $a$ is a constant, $dt = dx$ with no extra factor.
   Substitution, Section 7.3.1. Basics: substitution

2. Expand $\sin(t - a)$ and split

   $$\begin{aligned} \frac{\sin(t - a)}{\sin t} &= \frac{\sin t\cos a - \cos t\sin a}{\sin t} \\ &= \cos a - \sin a\,\cot t. \end{aligned}$$
   The subtraction formula separates $t$ from $a$⁠. Dividing each term by $\sin t$ leaves a constant, $\cos a$⁠, and a constant multiple of $\cot t$⁠, both of which we can integrate. Remember $\cos a$ and $\sin a$ are constants, because $a$ is.
   $\sin(A - B)$ $= \sin A\cos B - \cos A\sin B$⁠. Basics: trigonometric identities

3. Integrate

   $$\begin{aligned} &\int(\cos a - \sin a\cot t)\,dt \\ &= (\cos a)\,t - (\sin a)\log|\sin t| + C_1. \end{aligned}$$
   $\int\cos a\,dt = (\cos a)t$⁠, since $\cos a$ is a constant. $\int\cot t\,dt = \log|\sin t|$ is result (ii) proved in Section 7.3.1 (put $\sin t = s$⁠, $ds = \cos t\,dt$⁠).
   $\int\cot x\,dx = \log|\sin x| + C$⁠, Section 7.3.1. Lesson: integration by substitution

4. Return to $x$ and absorb the constant

   $$\begin{aligned} &(\cos a)(x + a) \\ &\qquad - (\sin a)\log|\sin(x + a)| + C_1 \\ &= x\cos a - \sin a\log|\sin(x + a)| \\ &\qquad + (a\cos a + C_1). \end{aligned}$$

   $a\cos a + C_1$ is a constant; call it $C$⁠. So

   $$\begin{aligned} &\int\frac{\sin x}{\sin(x + a)}\,dx \\ &= x\cos a - \sin a\log|\sin(x + a)| + C. \end{aligned}$$
   The term $a\cos a$ contains no $x$⁠, so it is part of the arbitrary constant. The book places its $C_1$ inside the bracket multiplied by $\sin a$⁠, so its constant is $C = -C_1\sin a + a\cos a$⁠; both bookkeepings describe the same family of answers.
   Basics: integration undoes differentiation

5. Check: differentiate the answer

   $$\begin{aligned} &\frac{d}{dx}\left(x\cos a - \sin a\log|\sin(x + a)|\right) \\ &= \cos a - \sin a\,\frac{\cos(x + a)}{\sin(x + a)} \\ &= \frac{\sin(x + a)\cos a - \cos(x + a)\sin a}{\sin(x + a)} \\ &= \frac{\sin\big((x + a) - a\big)}{\sin(x + a)} = \frac{\sin x}{\sin(x + a)}. \ \checkmark \end{aligned}$$
   $\frac{d}{dx}\log|u| = \frac{u'}{u}$⁠. The numerator is the subtraction formula read backwards. Valid wherever $\sin(x + a) \ne 0$⁠.

Part (iii): $\int\frac{dx}{1 + \tan x}$

1. Clear the fraction inside the fraction

   Where $\cos x \ne 0$⁠, multiply top and bottom by $\cos x$⁠:

   $$\frac{1}{1 + \tan x} = \frac{\cos x}{\cos x + \sin x}.$$

   The integrand also needs $1 + \tan x \ne 0$⁠, which is the same as $\cos x + \sin x \ne 0$⁠.

   $\tan x = \frac{\sin x}{\cos x}$⁠, so multiplying by $\cos x$ removes the inner fraction. This is allowed only where $\cos x \ne 0$⁠, which is required anyway for $\tan x$ to exist.
   Basics: trigonometric identities

2. Write the numerator as denominator plus derivative of denominator

   The denominator is $D = \cos x + \sin x$⁠; its derivative is $D' = \cos x - \sin x$⁠. Look for constants $A$⁠, $B$ with $\cos x = A\,D + B\,D'$⁠:

   $$\begin{aligned} \cos x &= A(\cos x + \sin x) \\ &\quad + B(\cos x - \sin x). \end{aligned}$$

   Comparing coefficients: of $\cos x$⁠, $A + B = 1$⁠; of $\sin x$⁠, $A - B = 0$⁠. So $A = B = \frac12$⁠:

   $$\begin{aligned} \cos x &= \tfrac12(\cos x + \sin x) \\ &\quad + \tfrac12(\cos x - \sin x). \end{aligned}$$
   $\frac{D}{D} = 1$ is trivial to integrate, and $\frac{D'}{D}$ integrates to $\log|D|$⁠. So writing the numerator as a mix of $D$ and $D'$ splits the integral into two easy pieces. The book writes this split directly; the comparison of coefficients shows where the $\frac12$s come from.
   Comparing coefficients. Basics: splitting fractions

3. Split the integral

   $$\begin{aligned} &\int\frac{\cos x\,dx}{\cos x + \sin x} \\ &= \frac12\int dx + \frac12\int\frac{\cos x - \sin x}{\cos x + \sin x}\,dx \\ &= \frac x2 + \frac{C_1}{2} + \frac12\,I, \end{aligned}$$

   where $I = \displaystyle\int\frac{\cos x - \sin x}{\cos x + \sin x}\,dx$⁠.

   The first fraction is $\frac12\cdot\frac{D}{D} = \frac12$⁠. The second is $\frac12\cdot\frac{D'}{D}$⁠, which is set aside as $I$⁠.
   Property V, Section 7.2.1. Lesson: integration as anti-differentiation

4. Find $I$ with $t = \cos x + \sin x$

   $$\begin{gathered} t = \cos x + \sin x, \\ dt = (\cos x - \sin x)\,dx. \end{gathered}$$ $$\begin{aligned} I = \int\frac{dt}{t} &= \log|t| + C_2 \\ &= \log|\cos x + \sin x| + C_2. \end{aligned}$$
   The numerator is exactly the derivative of the denominator, the ideal case for substitution. $t \ne 0$ because $\cos x + \sin x \ne 0$ on the domain (Part (iii), Step 1).
   Standard formula (xii), Section 7.2. Basics: logarithms and 1/x

5. Assemble the answer

   $$\begin{aligned} \int\frac{dx}{1 + \tan x} &= \frac x2 + \frac12\log|\cos x + \sin x| \\ &\qquad + \frac{C_1}{2} + \frac{C_2}{2} \\ &= \frac x2 + \frac12\log|\cos x + \sin x| + C, \end{aligned}$$

   with $C = \frac{C_1}{2} + \frac{C_2}{2}$⁠.

   The two half-constants merge into one arbitrary constant, as the book does.

6. Check: differentiate the answer

   $$\begin{aligned} &\frac{d}{dx}\left(\frac x2 + \frac12\log|\cos x + \sin x|\right) \\ &= \frac12 + \frac12\cdot\frac{\cos x - \sin x}{\cos x + \sin x} \\ &= \frac{2\cos x}{2(\cos x + \sin x)} = \frac{\cos x}{\cos x + \sin x}. \end{aligned}$$

   Dividing top and bottom by $\cos x$ gives $\frac{1}{1 + \tan x}$⁠. $\checkmark$

   Over the common denominator, the sum $(\cos x + \sin x)$ $+ (\cos x - \sin x)$ is $2\cos x$⁠. The last division is legitimate because $\cos x \ne 0$ on the domain.

Every fact used, and where it comes from