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Let I = ∫ − a a a − x a + x d x I = \displaystyle\int_{-a}^{a}\sqrt{\frac{a - x}{a + x}}\,dx I = ∫ − a a a + x a − x d x a > 0 a > 0 a > 0
Multiplying the numerator and denominator inside the root by a − x a - x a − x
I = ∫ − a a a − x a + x × a − x a − x d x = ∫ − a a a − x a 2 − x 2 d x \begin{aligned} I &= \int_{-a}^{a}\sqrt{\frac{a - x}{a + x} \times \frac{a - x}{a - x}}\,dx \\ &= \int_{-a}^{a}\frac{a - x}{\sqrt{a^2 - x^2}}\,dx \end{aligned} I = ∫ − a a a + x a − x × a − x a − x d x = ∫ − a a a 2 − x 2 a − x d x Why?
A root of a fraction is hard to integrate. Multiplying top and bottom by a − x a - x a − x
( a − x ) 2 ( a + x ) ( a − x ) = ( a − x ) 2 a 2 − x 2 \begin{aligned} \frac{(a - x)^2}{(a + x)(a - x)} &= \frac{(a - x)^2}{a^2 - x^2} \end{aligned} ( a + x ) ( a − x ) ( a − x ) 2 = a 2 − x 2 ( a − x ) 2 The bottom, multiplied out:
( a + x ) ( a − x ) = a 2 − a x + a x − x 2 = a 2 − x 2 \begin{aligned} (a + x)(a - x) &= a^2 - ax + ax - x^2 \\ &= a^2 - x^2 \end{aligned} ( a + x ) ( a − x ) = a 2 − a x + a x − x 2 = a 2 − x 2 Then take the root of the top and the bottom separately:
( a − x ) 2 a 2 − x 2 = ( a − x ) 2 a 2 − x 2 \sqrt{\frac{(a - x)^2}{a^2 - x^2}} = \frac{\sqrt{(a - x)^2}}{\sqrt{a^2 - x^2}} a 2 − x 2 ( a − x ) 2 = a 2 − x 2 ( a − x ) 2 Then ( a − x ) 2 = a − x \sqrt{(a - x)^2} = a - x ( a − x ) 2 = a − x x ≤ a x \le a x ≤ a a − x ≥ 0 a - x \ge 0 a − x ≥ 0
Watch out: in general u 2 = ∣ u ∣ \sqrt{u^2} = |u| u 2 = ∣ u ∣ u u u u ≥ 0 u \ge 0 u ≥ 0
I = ∫ − a a a a 2 − x 2 d x − ∫ − a a x a 2 − x 2 d x = I 1 − I 2 \begin{aligned} I &= \int_{-a}^{a}\frac{a}{\sqrt{a^2 - x^2}}\,dx \\ &\quad - \int_{-a}^{a}\frac{x}{\sqrt{a^2 - x^2}}\,dx \\ &= I_1 - I_2 \end{aligned} I = ∫ − a a a 2 − x 2 a d x − ∫ − a a a 2 − x 2 x d x = I 1 − I 2 Why?
Split the numerator a − x a - x a − x
a − x a 2 − x 2 = a a 2 − x 2 − x a 2 − x 2 \frac{a - x}{\sqrt{a^2 - x^2}} = \frac{a}{\sqrt{a^2 - x^2}} - \frac{x}{\sqrt{a^2 - x^2}} a 2 − x 2 a − x = a 2 − x 2 a − a 2 − x 2 x 1 a 2 − ( − x ) 2 = 1 a 2 − x 2 \frac{1}{\sqrt{a^2 - (-x)^2}} = \frac{1}{\sqrt{a^2 - x^2}} a 2 − ( − x ) 2 1 = a 2 − x 2 1 1 a 2 − x 2 \frac{1}{\sqrt{a^2 - x^2}} a 2 − x 2 1 ∫ − a a f ( x ) d x = 2 ∫ 0 a f ( x ) d x \int_{-a}^{a} f(x)\,dx = 2\int_0^a f(x)\,dx ∫ − a a f ( x ) d x = 2 ∫ 0 a f ( x ) d x f ( − x ) = f ( x ) f(-x) = f(x) f ( − x ) = f ( x )
Why?
To test even or odd, replace x x x − x -x − x ( − x ) 2 = x 2 (-x)^2 = x^2 ( − x ) 2 = x 2
For an even function the graph is symmetric about the y-axis, so the area from − a -a − a a a a
I 1 = a ∫ − a a 1 a 2 − x 2 d x = 2 a ∫ 0 a 1 a 2 − x 2 d x = 2 a [ sin − 1 ( x a ) ] 0 a = 2 a ( π 2 − 0 ) = π a \begin{aligned} I_1 &= a\int_{-a}^{a}\frac{1}{\sqrt{a^2 - x^2}}\,dx \\ &= 2a\int_0^a\frac{1}{\sqrt{a^2 - x^2}}\,dx \\ &= 2a\left[\sin^{-1}\left(\frac{x}{a}\right)\right]_0^a \\ &= 2a\left(\frac{\pi}{2} - 0\right) = \pi a \end{aligned} I 1 = a ∫ − a a a 2 − x 2 1 d x = 2 a ∫ 0 a a 2 − x 2 1 d x = 2 a [ sin − 1 ( a x ) ] 0 a = 2 a ( 2 π − 0 ) = πa Why?
The constant a a a
∫ d x a 2 − x 2 = sin − 1 x a + C \begin{aligned} &\int\frac{dx}{\sqrt{a^2 - x^2}} \\ &= \sin^{-1}\frac{x}{a} + C \end{aligned} ∫ a 2 − x 2 d x = sin − 1 a x + C At the limits: sin − 1 a a = sin − 1 1 = π 2 \sin^{-1}\frac{a}{a} = \sin^{-1}1 = \frac{\pi}{2} sin − 1 a a = sin − 1 1 = 2 π sin − 1 0 = 0 \sin^{-1}0 = 0 sin − 1 0 = 0
x = a : sin − 1 a a = sin − 1 1 = π 2 x = 0 : sin − 1 0 a = sin − 1 0 = 0 2 a ( π 2 − 0 ) = 2 a × π 2 = π a \begin{aligned} x = a&: \sin^{-1}\frac{a}{a} = \sin^{-1}1 = \frac{\pi}{2} \\ x = 0&: \sin^{-1}\frac{0}{a} = \sin^{-1}0 = 0 \\[4pt] 2a\left(\frac{\pi}{2} - 0\right) &= 2a \times \frac{\pi}{2} = \pi a \end{aligned} x = a x = 0 2 a ( 2 π − 0 ) : sin − 1 a a = sin − 1 1 = 2 π : sin − 1 a 0 = sin − 1 0 = 0 = 2 a × 2 π = πa For f ( x ) = x a 2 − x 2 f(x) = \frac{x}{\sqrt{a^2 - x^2}} f ( x ) = a 2 − x 2 x f ( − x ) = − x a 2 − x 2 = − f ( x ) f(-x) = \frac{-x}{\sqrt{a^2 - x^2}} = -f(x) f ( − x ) = a 2 − x 2 − x = − f ( x ) f f f ∫ − a a f ( x ) d x = 0 \int_{-a}^{a} f(x)\,dx = 0 ∫ − a a f ( x ) d x = 0 f ( − x ) = − f ( x ) f(-x) = -f(x) f ( − x ) = − f ( x )
Why?
Replacing x x x − x -x − x
f ( − x ) = ( − x ) a 2 − ( − x ) 2 = − x a 2 − x 2 = − f ( x ) \begin{aligned} f(-x) &= \frac{(-x)}{\sqrt{a^2 - (-x)^2}} \\ &= \frac{-x}{\sqrt{a^2 - x^2}} = -f(x) \end{aligned} f ( − x ) = a 2 − ( − x ) 2 ( − x ) = a 2 − x 2 − x = − f ( x ) For an odd function the area on the left of the y-axis is the negative of the area on the right, so over [ − a , a ] [-a, a] [ − a , a ]
I 2 = ∫ − a a x a 2 − x 2 d x = 0 I_2 = \int_{-a}^{a}\frac{x}{\sqrt{a^2 - x^2}}\,dx = 0 I 2 = ∫ − a a a 2 − x 2 x d x = 0 I = π a − 0 I = \pi a - 0 I = πa − 0 Why?
Put the two results into I = I 1 − I 2 I = I_1 - I_2 I = I 1 − I 2 I 1 = π a I_1 = \pi a I 1 = πa I 2 = 0 I_2 = 0 I 2 = 0
∴ I = I 1 − I 2 = π a \therefore\ I = I_1 - I_2 = \pi a ∴ I = I 1 − I 2 = πa
Where this comes from Chapter 7, Integrals : Definite integrals: even and odd functions.
Question type: Limits from − a -a − a a a a
How to spot it: click to see Type 11. Limits from − a -a − a a a a
Spot it: the two limits are the same number with opposite signs.
Here: The limits are − a -a − a a a a a − x a - x a − x a − x a 2 − x 2 \frac{a - x}{\sqrt{a^2 - x^2}} a 2 − x 2 a − x a a 2 − x 2 \frac{a}{\sqrt{a^2 - x^2}} a 2 − x 2 a a a a x a 2 − x 2 \frac{x}{\sqrt{a^2 - x^2}} a 2 − x 2 x
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