Exercise 7.9 · Q2: $\int_0^{\pi/2}\sqrt{\sin\phi}\,\cos^5\phi\,d\phi$

Basics you need

Solution

1. Check the integrand is defined and continuous

   For $0 \le \phi \le \frac{\pi}{2}$ we have $\sin\phi \ge 0$⁠, so $\sqrt{\sin\phi}$ is a real number, and it is continuous there (a square root of a continuous non-negative function). $\cos^5\phi$ is continuous everywhere. Their product is continuous on $\left[0, \frac{\pi}{2}\right]$⁠.

   The square root is the only possible trouble: $\sqrt{\sin\phi}$ is not a real number when $\sin\phi < 0$⁠. On this interval sine is never negative, so the integrand makes sense at every point, including the end points where $\sin 0 = 0$⁠.
   Sine is $\ge 0$ in the first quadrant: $\sin\phi$ rises from 0 to 1 as $\phi$ goes from 0 to $\frac{\pi}{2}$⁠. Basics: trigonometry

2. Spare one $\cos\phi$ and rewrite the rest in sines

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

   So the integrand is $\sqrt{\sin\phi}\,(1 - \sin^2\phi)^2\cos\phi$⁠.

   $\cos\phi\,d\phi$ is exactly the differential of $\sin\phi$⁠. Keeping one factor $\cos\phi$ aside prepares $dt$⁠; everything else must then be expressed through $\sin\phi$⁠, and the even power $\cos^4\phi$ can be, by the identity.
   $\cos^2\phi = 1 - \sin^2\phi$ from $\sin^2\phi + \cos^2\phi = 1$⁠. Basics: trigonometric identities for integrals

3. Substitute $t = \sin\phi$ and change the limits

   Then $dt = \cos\phi\,d\phi$⁠, $\sqrt{\sin\phi} = \sqrt{t} = t^{1/2}$⁠, and $1 - \sin^2\phi = 1 - t^2$⁠.

   New limits. When $\phi = 0$⁠, $t = \sin 0 = 0$⁠. When $\phi = \frac{\pi}{2}$⁠, $t = \sin\frac{\pi}{2} = 1$⁠.

   $$\int_0^{\frac{\pi}{2}}\sqrt{\sin\phi}\,\cos^5\phi\,d\phi = \int_0^1 t^{1/2}(1 - t^2)^2\,dt.$$

   Why it is valid on the whole interval. $t = \sin\phi$ has the continuous derivative $\cos\phi$⁠, and $\cos\phi \ge 0$ on $\left[0, \frac{\pi}{2}\right]$⁠, so $t$ increases steadily from 0 to 1. The new integrand $t^{1/2}(1 - t^2)^2$ is continuous on $[0, 1]$⁠.

   Every piece of the integrand has been rewritten in $t$⁠, and the leftover $\cos\phi\,d\phi$ became $dt$⁠. The limits are $t$-values because we now integrate with respect to $t$⁠.
   $\frac{d}{d\phi}\sin\phi = \cos\phi$⁠. Basics: substitution (the chain rule backwards)

4. Expand into powers of $t$

   $(1 - t^2)^2 = 1 - 2t^2 + t^4$⁠, so

   $$\begin{aligned}t^{1/2}(1 - t^2)^2 &= t^{1/2} - 2t^{1/2}t^2 + t^{1/2}t^4 \\ &= t^{1/2} - 2t^{5/2} + t^{9/2}.\end{aligned}$$
   The power rule integrates single powers of $t$⁠, not products, so we expand first. Adding exponents: $\frac{1}{2} + 2 = \frac{5}{2}$ and $\frac{1}{2} + 4 = \frac{9}{2}$⁠.
   $(a - b)^2 = a^2 - 2ab + b^2$ and $t^m\cdot t^n = t^{m+n}$⁠. Basics: powers of x

5. Integrate each power and evaluate

   Using $\int t^n dt = \frac{t^{n+1}}{n+1}$⁠: $\int t^{1/2}dt = \frac{2}{3}t^{3/2}$⁠, $\int t^{5/2}dt = \frac{2}{7}t^{7/2}$⁠, $\int t^{9/2}dt = \frac{2}{11}t^{11/2}$⁠. So

   $$\begin{aligned}\int_0^1 &\left(t^{1/2} - 2t^{5/2} + t^{9/2}\right)dt \\ &= \left[\tfrac{2}{3}t^{3/2} - \tfrac{4}{7}t^{7/2} + \tfrac{2}{11}t^{11/2}\right]_0^1 \\ &= \frac{2}{3} - \frac{4}{7} + \frac{2}{11}.\end{aligned}$$

   Every power of $t$ is 1 at $t = 1$ and 0 at $t = 0$⁠, so the lower limit contributes nothing.

   Dividing by $\frac{3}{2}$ is multiplying by $\frac{2}{3}$⁠, and so on; the middle term carries the factor 2 from the expansion: $2\cdot\frac{2}{7} = \frac{4}{7}$⁠.
   Standard formula (i), Section 7.2, then Theorem 2. Basics: the power rule

6. Add the fractions

   The least common denominator of 3, 7 and 11 is $3\times 7\times 11 = 231$⁠:

   $$\begin{aligned}\frac{2}{3} - \frac{4}{7} + \frac{2}{11} &= \frac{154}{231} - \frac{132}{231} + \frac{42}{231} \\ &= \frac{154 - 132 + 42}{231} = \frac{64}{231}.\end{aligned}$$

   In decimals, $\frac{64}{231} \approx 0.2770563$⁠.

   $\frac{2}{3} = \frac{2\cdot 77}{231}$⁠, $\frac{4}{7} = \frac{4\cdot 33}{231}$⁠, $\frac{2}{11} = \frac{2\cdot 21}{231}$⁠. The fraction is in lowest terms: $64 = 2^6$ and 231 is odd.
   Fractions are added over a common denominator. Basics: splitting fractions

7. Same result by staying in $\phi$

   Putting $t = \sin\phi$ back into the antiderivative of Step 5:

   $$\begin{aligned}F(\phi) = \tfrac{2}{3}\sin^{3/2}\phi &- \tfrac{4}{7}\sin^{7/2}\phi \\ &+ \tfrac{2}{11}\sin^{11/2}\phi.\end{aligned}$$

   Check: differentiate. By the chain rule each term gives (power) $\times$ $\sin^{\text{power} - 1}\phi\cdot\cos\phi$⁠:

   $$\begin{aligned}F'(\phi) &= \cos\phi\left(\sin^{1/2}\phi - 2\sin^{5/2}\phi + \sin^{9/2}\phi\right) \\ &= \cos\phi\,\sqrt{\sin\phi}\,(1 - \sin^2\phi)^2 \\ &= \sqrt{\sin\phi}\,\cos\phi\cdot\cos^4\phi \\ &= \sqrt{\sin\phi}\,\cos^5\phi,\end{aligned}$$

   the integrand. And $F\left(\frac{\pi}{2}\right) - F(0) = \frac{2}{3} - \frac{4}{7} + \frac{2}{11} - 0 = \frac{64}{231}$⁠.

   Differentiating back confirms both the antiderivative and the algebra of Steps 2 and 4 in one go.
   Chain rule: outer $u^{n}$⁠, inner $u = \sin\phi$⁠. Basics: the chain rule backwards

8. The region

   The integrand is $\ge 0$ on the interval, so the value is an area. The bump is at most about 0.43 high on an interval of length $\frac{\pi}{2} \approx 1.57$⁠, but it is narrow ($\cos^5\phi$ is tiny near $\frac{\pi}{2}$⁠), so an area of about 0.28 is reasonable.
   For a function that is never negative, the definite integral is the area under its graph. Basics: area as a sum of thin strips

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