Miscellaneous · Q18: $\int\frac{dx}{\sqrt{\sin^3x\,\sin(x+\alpha)}}$

Basics you need

Solution

1. Take $\sin^4x$ out of the root

   On the domain $\sin x \ne 0$⁠, so we may write

   $$\sin^3x\,\sin(x+\alpha) = \sin^4x\cdot\frac{\sin(x+\alpha)}{\sin x} .$$

   Both factors on the right are positive ($\sin^4x > 0$⁠, and the quotient has the sign of $\sin x\sin(x+\alpha) > 0$⁠), so

   $$\sqrt{\sin^3x\,\sin(x+\alpha)} = \sin^2x\,\sqrt{\frac{\sin(x+\alpha)}{\sin x}} .$$
   Why $\sin^4x$ and not $\sin^3x$⁠? An even power is what leaves the square root cleanly: $\sqrt{\sin^4x} = \sin^2x$ with no sign trouble. Splitting the root of a product into a product of roots is only safe when both factors are $\ge 0$⁠, which we checked.
   Basics: trigonometric identities

2. Expand $\frac{\sin(x+\alpha)}{\sin x}$

   $$\begin{aligned}\frac{\sin(x+\alpha)}{\sin x} &= \frac{\sin x\cos\alpha + \cos x\sin\alpha}{\sin x}\\ &= \cos\alpha + \sin\alpha\cot x .\end{aligned}$$

   So the integrand is

   $$\frac{1}{\sin^2x\sqrt{\cos\alpha + \sin\alpha\cot x}} = \frac{\operatorname{cosec}^2x}{\sqrt{\cos\alpha + \sin\alpha\cot x}} .$$
   Dividing each term of the numerator by $\sin x$ turns the compound angle into an expression in $\cot x$ alone, and the leftover $\frac{1}{\sin^2x} = \operatorname{cosec}^2x$ is exactly the derivative we need for a $\cot x$ substitution.
   Compound-angle formula (Class XI). Basics: trigonometric identities

3. Substitute $t = \cos\alpha + \sin\alpha\cot x$

   $$\begin{aligned}t &= \cos\alpha + \sin\alpha\cot x ,\\ dt &= -\sin\alpha\operatorname{cosec}^2x\,dx ,\\ \operatorname{cosec}^2x\,dx &= -\frac{dt}{\sin\alpha} .\end{aligned}$$

   By Step 2, $t = \frac{\sin(x+\alpha)}{\sin x} > 0$ on the domain.

   $\alpha$ is a constant, so $\cos\alpha$ differentiates to 0 and $\sin\alpha$ stays as a factor. Dividing by $\sin\alpha$ is allowed because $\sin\alpha \ne 0$⁠. We need $t > 0$ so that $\sqrt t$ and $t^{-1/2}$ make sense.
   Section 7.3.1. Lesson: integration by substitution

4. Integrate in $t$

   $$\begin{aligned}\int\frac{\operatorname{cosec}^2x\,dx}{\sqrt{\cos\alpha + \sin\alpha\cot x}} &= -\frac{1}{\sin\alpha}\int t^{-1/2}\,dt\\ &= -\frac{1}{\sin\alpha}\cdot 2t^{1/2} + C .\end{aligned}$$
   Power formula with $n = -\frac12$⁠: $\frac{t^{1/2}}{1/2} = 2\sqrt t$⁠. The constant $-\frac{1}{\sin\alpha}$ comes outside.
   Standard formula (i), Section 7.2. Basics: powers of x

5. Return to $x$

   $$\int\frac{dx}{\sqrt{\sin^3x\,\sin(x+\alpha)}} = -\frac{2}{\sin\alpha}\sqrt{\frac{\sin(x+\alpha)}{\sin x}} + C .$$
   By Step 2, $t = \frac{\sin(x+\alpha)}{\sin x}$⁠; writing it this way matches the form in the answer key.

6. Check: differentiate the answer

   Let $u = \frac{\sin(x+\alpha)}{\sin x}$⁠. By the quotient rule and the formula $\sin(A - B) = \sin A\cos B - \cos A\sin B$⁠:

   $$\begin{aligned}u' &= \frac{\cos(x+\alpha)\sin x - \sin(x+\alpha)\cos x}{\sin^2x}\\ &= \frac{\sin\left(x - (x+\alpha)\right)}{\sin^2x} = -\frac{\sin\alpha}{\sin^2x} .\end{aligned}$$

   Then

   $$\begin{aligned}\frac{d}{dx}\left(-\frac{2}{\sin\alpha}\sqrt u\right) &= -\frac{2}{\sin\alpha}\cdot\frac{u'}{2\sqrt u}\\ &= \frac{1}{\sin^2x\,\sqrt u}\\ &= \frac{1}{\sqrt{\sin^3x\,\sin(x+\alpha)}}\ \checkmark\end{aligned}$$
   The last equality is Step 1 read backwards. The derivative check holds at every point of the domain, so the answer is valid on every interval where $\sin x\sin(x+\alpha) > 0$⁠.
   Basics: integration undoes differentiation

7. Compare with the textbook's answer

   The key gives $\frac{-2}{\sin\alpha}\sqrt{\frac{\sin(x+\alpha)}{\sin x}} + C$⁠, exactly our answer.

   The key does not state the conditions; we add them: $\sin\alpha \ne 0$ (otherwise the formula divides by 0) and $\sin x\sin(x+\alpha) > 0$⁠. When $\sin\alpha = 0$ the question changes: for $\alpha = 0$ the integrand is $\frac{1}{\sin^2x}$ and the integral is $-\cot x + C$⁠.

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