Example 5: Substitution: $\sin mx$⁠, $2x\sin(x^2+1)$⁠, $\frac{\tan^4\sqrt x\,\sec^2\sqrt x}{\sqrt x}$⁠, $\frac{\sin(\tan^{-1}x)}{1+x^2}$

Basics you need

Part (i): $\int\sin mx\,dx$

1. Choose the substitution

   The inner expression is $mx$⁠, with derivative $m$⁠, a constant. Assume $m \ne 0$ and put

   $$t = mx, \qquad dt = m\,dx, \qquad dx = \frac{dt}{m}.$$
   The missing factor $m$ is a constant, and a constant factor can always be supplied by dividing by it. This needs $m \ne 0$⁠; if $m = 0$ the integrand is $\sin 0 = 0$ and the integral is just a constant, so that case is trivial.
   Substitution, Section 7.3.1. Basics: substitution

2. Integrate in $t$ and return to $x$

   $$\begin{aligned} \int\sin mx\,dx &= \frac1m\int\sin t\,dt \\ &= -\frac1m\cos t + C \\ &= -\frac1m\cos mx + C. \end{aligned}$$
   $\int\sin t\,dt = -\cos t$ is formula (iii). The answer must be in the original variable, so $t$ is replaced by $mx$ at the end.
   Standard formula (iii), Section 7.2. Lesson: integration by substitution

3. Check: differentiate the answer

   $$\begin{aligned} \frac{d}{dx}\left(-\frac1m\cos mx\right) &= -\frac1m\cdot(-\sin mx)\cdot m \\ &= \sin mx. \ \checkmark \end{aligned}$$

   Valid for every real $x$⁠.

   The chain rule produces the factor $m$ (derivative of the inner $mx$⁠), which cancels the $\frac1m$⁠: exactly the factor we divided by in Part (i), Step 1.

Part (ii): $\int 2x\sin(x^2 + 1)\,dx$

1. Spot the inner function and its derivative

   The sine is applied to $x^2 + 1$⁠, whose derivative $2x$ is the other factor. Put

   $$t = x^2 + 1, \qquad dt = 2x\,dx.$$
   This is the ideal situation for substitution: the whole of $2x\,dx$ is replaced by $dt$⁠, and nothing is left over in $x$⁠.
   Basics: substitution, the chain rule backwards

2. Integrate and return to $x$

   $$\begin{aligned} \int 2x\sin(x^2 + 1)\,dx &= \int\sin t\,dt \\ &= -\cos t + C \\ &= -\cos(x^2 + 1) + C. \end{aligned}$$
   Formula (iii) again, then $t = x^2 + 1$ is put back.
   Standard formula (iii), Section 7.2. Lesson: integration by substitution

3. Check: differentiate the answer

   $$\frac{d}{dx}\left(-\cos(x^2 + 1)\right) = \sin(x^2 + 1)\cdot 2x. \ \checkmark$$

   Valid for every real $x$⁠.

   Outer: $-\cos$ differentiates to $\sin$⁠; inner: $x^2 + 1$ differentiates to $2x$⁠. The product is the integrand.

Part (iii): $\int\frac{\tan^4\sqrt x\,\sec^2\sqrt x}{\sqrt x}\,dx$

1. Where the integrand exists

   $\sqrt x$ is in a denominator, so $x > 0$⁠; and $\sec\sqrt x$⁠, $\tan\sqrt x$ need $\cos\sqrt x \ne 0$⁠, that is $\sqrt x \ne \frac{\pi}{2}, \frac{3\pi}{2}, \dots$

   These are the conditions under which every symbol in the integrand has a value; the answer will be valid on each interval where they hold.

2. First substitution: $t = \sqrt x$

   $\frac{d}{dx}\sqrt x = \frac12 x^{-\frac12} = \frac{1}{2\sqrt x}$⁠. Put

   $$\begin{gathered} t = \sqrt x \ (t > 0), \qquad dt = \frac{dx}{2\sqrt x}, \\ \frac{dx}{\sqrt x} = 2\,dt. \end{gathered}$$

   Then

   $$\int\frac{\tan^4\sqrt x\,\sec^2\sqrt x}{\sqrt x}\,dx = 2\int\tan^4 t\,\sec^2 t\,dt.$$
   The $\frac{1}{\sqrt x}$ in the integrand is, up to the factor 2, the derivative of the inner expression $\sqrt x$⁠. So $\frac{dx}{\sqrt x}$ becomes $2\,dt$⁠, and every $\sqrt x$ becomes $t$⁠. The book writes this as $dx = 2t\,dt$ and then cancels the $t$ against the $\sqrt x = t$ in the denominator; the result is the same.
   Power rule with $n = \frac12$⁠. Basics: the power rule

3. Second substitution: $u = \tan t$

   Now $\sec^2 t$ is the derivative of $\tan t$⁠. Put

   $$u = \tan t, \qquad du = \sec^2 t\,dt.$$ $$\begin{aligned} 2\int\tan^4 t\,\sec^2 t\,dt &= 2\int u^4\,du \\ &= 2\cdot\frac{u^5}{5} + C. \end{aligned}$$
   Again an inner function ($\tan t$⁠) appears together with its derivative ($\sec^2 t$⁠), so the integral collapses to a power of $u$⁠. Formula (i) with $n = 4$ finishes it.
   Standard formulae (iv) and (i), Section 7.2. Lesson: integration by substitution

4. Return to $x$

   $$\frac25 u^5 + C = \frac25\tan^5 t + C = \frac25\tan^5\sqrt x + C.$$
   Undo the substitutions in reverse order: $u = \tan t$⁠, then $t = \sqrt x$⁠. The book also suggests ("Alternatively") the single substitution $t = \tan\sqrt x$⁠, without working it out. It does both jobs at once: then $dt = \sec^2\sqrt x\cdot\frac{1}{2\sqrt x}\,dx$⁠, and the integral is $\int 2t^4\,dt$ directly.

5. Check: differentiate the answer

   $$\begin{aligned} &\frac{d}{dx}\left(\frac25\tan^5\sqrt x\right) \\ &= \frac25\cdot 5\tan^4\sqrt x\cdot\sec^2\sqrt x\cdot\frac{1}{2\sqrt x} \\ &= \frac{\tan^4\sqrt x\,\sec^2\sqrt x}{\sqrt x}. \ \checkmark \end{aligned}$$
   Three layers, three factors from the chain rule: the fifth power gives $5\tan^4$⁠, the tangent gives $\sec^2$⁠, and the square root gives $\frac{1}{2\sqrt x}$⁠. The constants $\frac25\cdot 5\cdot\frac12 = 1$⁠.

Part (iv): $\int\frac{\sin(\tan^{-1}x)}{1 + x^2}\,dx$

1. Choose the substitution

   $\frac{d}{dx}\tan^{-1}x = \frac{1}{1 + x^2}$⁠, and the integrand contains exactly this factor. Put

   $$t = \tan^{-1}x, \qquad dt = \frac{dx}{1 + x^2},$$

   with $t$ in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$⁠, the principal range.

   The sine is applied to $\tan^{-1}x$⁠, and the derivative of $\tan^{-1}x$ is the rest of the integrand. The integrand is defined for every real $x$⁠, since $1 + x^2 \ge 1$⁠.
   Standard formula (x), Section 7.2. Basics: substitution

2. Integrate and return to $x$

   $$\begin{aligned} \int\frac{\sin(\tan^{-1}x)}{1 + x^2}\,dx &= \int\sin t\,dt \\ &= -\cos t + C \\ &= -\cos(\tan^{-1}x) + C. \end{aligned}$$
   Formula (iii), then $t = \tan^{-1}x$ is put back. This is the book's final form.
   Lesson: integration by substitution

3. A simpler form (optional)

   If $t = \tan^{-1}x$ then $\tan t = x$ and $\cos t > 0$ (because $-\frac{\pi}{2} < t < \frac{\pi}{2}$⁠). From $\sec^2 t = 1 + \tan^2 t = 1 + x^2$⁠:

   $$\cos(\tan^{-1}x) = \frac{1}{\sqrt{1 + x^2}},$$

   so the answer can also be written $-\dfrac{1}{\sqrt{1 + x^2}} + C$⁠.

   The positive square root is correct only because the principal range keeps $\cos t$ positive. This form is identical to the book's, not merely equal up to a constant.
   Principal value branch of $\tan^{-1}$⁠, Chapter 2. Basics: trigonometric identities

4. Check: differentiate the answer

   $$\begin{aligned} &\frac{d}{dx}\left(-\cos(\tan^{-1}x)\right) \\ &= \sin(\tan^{-1}x)\cdot\frac{1}{1 + x^2}. \ \checkmark \end{aligned}$$

   Valid for every real $x$⁠.

   Outer $-\cos$ gives $\sin$⁠; inner $\tan^{-1}x$ gives $\frac{1}{1 + x^2}$⁠.

Every fact used, and where it comes from