Integrals of the Form x 2 ± a 2 \sqrt{x^2 \pm a^2} x 2 ± a 2 and a 2 − x 2 \sqrt{a^2 - x^2} a 2 − x 2
Integration by parts, with the constant function 1 1 1 taken as the second function, yields three important standard integrals. These forms appear frequently when integrating expressions with a quadratic under a square root.
Derivation of ∫ x 2 − a 2 d x \int \sqrt{x^2 - a^2}\,dx ∫ x 2 − a 2 d x
Let I = ∫ x 2 − a 2 d x I = \int \sqrt{x^2 - a^2}\,dx I = ∫ x 2 − a 2 d x . Write the integrand as x 2 − a 2 ⋅ 1 \sqrt{x^2 - a^2} \cdot 1 x 2 − a 2 ⋅ 1 , take u = x 2 − a 2 u = \sqrt{x^2 - a^2} u = x 2 − a 2 and v = 1 v = 1 v = 1 , and integrate by parts, using d d x x 2 − a 2 = x x 2 − a 2 \frac{d}{dx}\sqrt{x^2 - a^2} = \frac{x}{\sqrt{x^2 - a^2}} d x d x 2 − a 2 = x 2 − a 2 x :
I = x x 2 − a 2 − ∫ x 2 x 2 − a 2 d x I = x\sqrt{x^2 - a^2} - \int \frac{x^2}{\sqrt{x^2 - a^2}}\,dx I = x x 2 − a 2 − ∫ x 2 − a 2 x 2 d x
Rewrite the numerator as x 2 = ( x 2 − a 2 ) + a 2 x^2 = (x^2 - a^2) + a^2 x 2 = ( x 2 − a 2 ) + a 2 :
I = x x 2 − a 2 − ∫ ( x 2 − a 2 + a 2 x 2 − a 2 ) d x = x x 2 − a 2 − I − a 2 ∫ d x x 2 − a 2 I = x\sqrt{x^2 - a^2} - \int \left(\sqrt{x^2 - a^2} + \frac{a^2}{\sqrt{x^2 - a^2}}\right)dx = x\sqrt{x^2 - a^2} - I - a^2 \int \frac{dx}{\sqrt{x^2 - a^2}} I = x x 2 − a 2 − ∫ ( x 2 − a 2 + x 2 − a 2 a 2 ) d x = x x 2 − a 2 − I − a 2 ∫ x 2 − a 2 d x
The first integral on the right is again I I I . Bringing it over:
2 I = x x 2 − a 2 − a 2 ∫ d x x 2 − a 2 ⟹ I = x 2 x 2 − a 2 − a 2 2 ∫ d x x 2 − a 2 2I = x\sqrt{x^2 - a^2} - a^2 \int \frac{dx}{\sqrt{x^2 - a^2}} \quad\Longrightarrow\quad I = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\int \frac{dx}{\sqrt{x^2 - a^2}} 2 I = x x 2 − a 2 − a 2 ∫ x 2 − a 2 d x ⟹ I = 2 x x 2 − a 2 − 2 a 2 ∫ x 2 − a 2 d x
Using ∫ d x x 2 − a 2 = log ∣ x + x 2 − a 2 ∣ + C \int \frac{dx}{\sqrt{x^2 - a^2}} = \log\left|x + \sqrt{x^2 - a^2}\right| + C ∫ x 2 − a 2 d x = log x + x 2 − a 2 + C :
∫ x 2 − a 2 d x = x 2 x 2 − a 2 − a 2 2 log ∣ x + x 2 − a 2 ∣ + C \int \sqrt{x^2 - a^2}\,dx = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\log\left|x + \sqrt{x^2 - a^2}\right| + C ∫ x 2 − a 2 d x = 2 x x 2 − a 2 − 2 a 2 log x + x 2 − a 2 + C
Derivation of ∫ x 2 + a 2 d x \int \sqrt{x^2 + a^2}\,dx ∫ x 2 + a 2 d x
Let I = ∫ x 2 + a 2 d x I = \int \sqrt{x^2 + a^2}\,dx I = ∫ x 2 + a 2 d x . Again take 1 1 1 as the second function and integrate by parts, using d d x x 2 + a 2 = x x 2 + a 2 \frac{d}{dx}\sqrt{x^2 + a^2} = \frac{x}{\sqrt{x^2 + a^2}} d x d x 2 + a 2 = x 2 + a 2 x :
I = x x 2 + a 2 − ∫ x 2 x 2 + a 2 d x I = x\sqrt{x^2 + a^2} - \int \frac{x^2}{\sqrt{x^2 + a^2}}\,dx I = x x 2 + a 2 − ∫ x 2 + a 2 x 2 d x
Write x 2 = ( x 2 + a 2 ) − a 2 x^2 = (x^2 + a^2) - a^2 x 2 = ( x 2 + a 2 ) − a 2 :
I = x x 2 + a 2 − I + a 2 ∫ d x x 2 + a 2 I = x\sqrt{x^2 + a^2} - I + a^2 \int \frac{dx}{\sqrt{x^2 + a^2}} I = x x 2 + a 2 − I + a 2 ∫ x 2 + a 2 d x
2 I = x x 2 + a 2 + a 2 ∫ d x x 2 + a 2 ⟹ I = x 2 x 2 + a 2 + a 2 2 ∫ d x x 2 + a 2 2I = x\sqrt{x^2 + a^2} + a^2 \int \frac{dx}{\sqrt{x^2 + a^2}} \quad\Longrightarrow\quad I = \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2}\int \frac{dx}{\sqrt{x^2 + a^2}} 2 I = x x 2 + a 2 + a 2 ∫ x 2 + a 2 d x ⟹ I = 2 x x 2 + a 2 + 2 a 2 ∫ x 2 + a 2 d x
Using ∫ d x x 2 + a 2 = log ∣ x + x 2 + a 2 ∣ + C \int \frac{dx}{\sqrt{x^2 + a^2}} = \log\left|x + \sqrt{x^2 + a^2}\right| + C ∫ x 2 + a 2 d x = log x + x 2 + a 2 + C :
∫ x 2 + a 2 d x = x 2 x 2 + a 2 + a 2 2 log ∣ x + x 2 + a 2 ∣ + C \int \sqrt{x^2 + a^2}\,dx = \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2}\log\left|x + \sqrt{x^2 + a^2}\right| + C ∫ x 2 + a 2 d x = 2 x x 2 + a 2 + 2 a 2 log x + x 2 + a 2 + C
Derivation of ∫ a 2 − x 2 d x \int \sqrt{a^2 - x^2}\,dx ∫ a 2 − x 2 d x
Let I = ∫ a 2 − x 2 d x I = \int \sqrt{a^2 - x^2}\,dx I = ∫ a 2 − x 2 d x . With d d x a 2 − x 2 = − x a 2 − x 2 \frac{d}{dx}\sqrt{a^2 - x^2} = \frac{-x}{\sqrt{a^2 - x^2}} d x d a 2 − x 2 = a 2 − x 2 − x :
I = x a 2 − x 2 + ∫ x 2 a 2 − x 2 d x I = x\sqrt{a^2 - x^2} + \int \frac{x^2}{\sqrt{a^2 - x^2}}\,dx I = x a 2 − x 2 + ∫ a 2 − x 2 x 2 d x
Write x 2 = a 2 − ( a 2 − x 2 ) x^2 = a^2 - (a^2 - x^2) x 2 = a 2 − ( a 2 − x 2 ) :
I = x a 2 − x 2 + a 2 ∫ d x a 2 − x 2 − I I = x\sqrt{a^2 - x^2} + a^2 \int \frac{dx}{\sqrt{a^2 - x^2}} - I I = x a 2 − x 2 + a 2 ∫ a 2 − x 2 d x − I
2 I = x a 2 − x 2 + a 2 ∫ d x a 2 − x 2 ⟹ I = x 2 a 2 − x 2 + a 2 2 ∫ d x a 2 − x 2 2I = x\sqrt{a^2 - x^2} + a^2 \int \frac{dx}{\sqrt{a^2 - x^2}} \quad\Longrightarrow\quad I = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\int \frac{dx}{\sqrt{a^2 - x^2}} 2 I = x a 2 − x 2 + a 2 ∫ a 2 − x 2 d x ⟹ I = 2 x a 2 − x 2 + 2 a 2 ∫ a 2 − x 2 d x
Using ∫ d x a 2 − x 2 = sin − 1 x a + C \int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac{x}{a} + C ∫ a 2 − x 2 d x = sin − 1 a x + C :
∫ a 2 − x 2 d x = x 2 a 2 − x 2 + a 2 2 sin − 1 x a + C \int \sqrt{a^2 - x^2}\,dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C ∫ a 2 − x 2 d x = 2 x a 2 − x 2 + 2 a 2 sin − 1 a x + C
Alternative Trigonometric Substitutions
Each of the three integrals can also be evaluated by trigonometric substitution:
Integral Substitution Rationale ∫ x 2 − a 2 d x \int \sqrt{x^2 - a^2}\,dx ∫ x 2 − a 2 d x x = a sec θ x = a\sec\theta x = a sec θ x 2 − a 2 = a tan θ \sqrt{x^2 - a^2} = a\tan\theta x 2 − a 2 = a tan θ ∫ x 2 + a 2 d x \int \sqrt{x^2 + a^2}\,dx ∫ x 2 + a 2 d x x = a tan θ x = a\tan\theta x = a tan θ x 2 + a 2 = a sec θ \sqrt{x^2 + a^2} = a\sec\theta x 2 + a 2 = a sec θ ∫ a 2 − x 2 d x \int \sqrt{a^2 - x^2}\,dx ∫ a 2 − x 2 d x x = a sin θ x = a\sin\theta x = a sin θ a 2 − x 2 = a cos θ \sqrt{a^2 - x^2} = a\cos\theta a 2 − x 2 = a cos θ
Handling Quadratic Expressions: Completing the Square …