Integration by substitution Integration by substitution, or u u u chain rule .
Contents Use Let F ( x ) F(x) F ( x ) G ( x ) G(x) G ( x ) f ( x ) f(x) f ( x ) g ( x ) g(x) g ( x )
∫ f ( G ( x ) ) ⋅ g ( x ) d x = ∫ f ( u ) d u = F ( G ( x ) ) + C \int f(G(x)) \cdot g(x)\d x = \int f(u)\d u = F(G(x)) + C ∫ f ( G ( x )) ⋅ g ( x ) d x = ∫ f ( u ) d u = F ( G ( x )) + C where
u = G ( x ) , d u = g ( x ) d x u = G(x), \d u = g(x) \d x u = G ( x ) , d u = g ( x ) d x The definite integral version is
∫ x 1 x 2 f ( G ( x ) ) ⋅ g ( x ) d x = ∫ G ( x 1 ) G ( x 2 ) f ( u ) d u = F ( G ( x 2 ) ) − F ( G ( x 1 ) ) \int_{x_1}^{x_2} f(G(x)) \cdot g(x)\d x = \int_{G(x_1)}^{G(x_2)} f(u)\d u = F(G(x_2)) - F(G(x_1)) ∫ x 1 x 2 f ( G ( x )) ⋅ g ( x ) d x = ∫ G ( x 1 ) G ( x 2 ) f ( u ) d u = F ( G ( x 2 )) − F ( G ( x 1 )) For indefinite integrals, your answer must be in terms of the original variable (typically x x x
For definite integrals, change the limits of integration accordingly.
G ( x ) = u G(x) = u G ( x ) = u
Substitutions for integrals not in the form ∫ k ⋅ f ( G ( x ) ) ⋅ g ( x ) d x \int k\cdot f(G(x)) \cdot g(x)\d x ∫ k ⋅ f ( G ( x )) ⋅ g ( x ) d x
Integrals of the form
∫ f ( a x + b ) d x \int f(ax + b)\d x ∫ f ( a x + b ) d x can directly apply the property mentioned in indefinite integrals#Properties.
These are derived using substitution. For definite integrals, the limits of integration remain same as the original integral.
∫ f ( a x + b ) d x = 1 a F ( a x + b ) + C \int f(ax+b)\d x = \frac 1a F(ax+b) + C ∫ f ( a x + b ) d x = a 1 F ( a x + b ) + C ∫ 1 x 2 + a 2 d x = 1 a arctan x a + C \int \frac{1}{x^2 + a^2}\d x = \frac 1a \arctan \frac xa + C ∫ x 2 + a 2 1 d x = a 1 arctan a x + C ∫ 1 a 2 − x 2 d x = arcsin x a + C , ∣ x ∣ < a \int \frac{1}{\sqrt{a^2 - x^2}}\d x = \arcsin \frac xa + C,\, \lvert x\rvert < a ∫ a 2 − x 2 1 d x = arcsin a x + C , ∣ x ∣ < a ∫ f ′ ( x ) e f ( x ) d x = e f ( x ) + C \int f^\prime(x) \e^{f(x)}\d x = \e^{f(x)} + C ∫ f ′ ( x ) e f ( x ) d x = e f ( x ) + C The following can be proved using substitution, but it is not assessed.
∫ a b f ( x ) d x = ∫ a b f ( a + b − x ) d x \int_a^b f(x)\d x = \int_a^b f(a + b - x)\d x ∫ a b f ( x ) d x = ∫ a b f ( a + b − x ) d x Examples Example: Simplify
∫ sin 2 x 1 + 3 cos 2 x d x \int \frac{\sin 2x}{\sqrt{1 + 3\cos^2 x}}\d x ∫ 1 + 3 cos 2 x sin 2 x d x From double angle identity for sine sin 2 x = 2 sin x cos x \sin 2x = 2 \sin x \cos x sin 2 x = 2 sin x cos x
I = ∫ sin 2 x 1 + 3 cos 2 x d x = ∫ 2 sin x cos x 1 + 3 cos 2 x d x \begin{align*}
I &= \int \frac{\sin 2x}{\sqrt{1 + 3\cos^2 x}}\d x \\
&= \int \frac{2 \sin x \cos x}{\sqrt{1 + 3\cos^2 x}}\d x
\end{align*} I = ∫ 1 + 3 cos 2 x sin 2 x d x = ∫ 1 + 3 cos 2 x 2 sin x cos x d x Also
Let u = 1 + 3 cos 2 x , d u = − 6 sin x cos x d x u = 1 + 3\cos^2 x, \d u = -6\sin x\cos x \d x u = 1 + 3 cos 2 x , d u = − 6 sin x cos x d x
We want to make a − 6 sin x cos x -6 \sin x\cos x − 6 sin x cos x − 3 -3 − 3 − 3 -3 − 3
I = − 1 3 ∫ − 6 sin x cos x 1 + 3 cos 2 x d x = − 1 3 ∫ d u u 1 2 = − 1 3 ∫ u − 1 2 d u = − 1 3 ⋅ 2 u 1 2 + C = − 2 3 1 + 3 cos 2 x + C ■ \begin{align*}
I &= -\frac13 \int \frac{-6 \sin x \cos x}{\sqrt{1 + 3\cos^2 x}}\d x \\
&= -\frac13 \int \frac{\d u}{u^{\frac12}} \\
&= -\frac13 \int {u^{-\frac12}} \d u \\
&= -\frac13 \cdot 2u^{\frac12} + C \\
&= -\frac23 \sqrt{1 + 3\cos^2x} + C \qed
\end{align*} I = − 3 1 ∫ 1 + 3 cos 2 x − 6 sin x cos x d x = − 3 1 ∫ u 2 1 d u = − 3 1 ∫ u − 2 1 d u = − 3 1 ⋅ 2 u 2 1 + C = − 3 2 1 + 3 cos 2 x + C ■ Example: Evaluate
∫ e 2 e 8 d x x ln x d x \int_{\e^2}^{\e^8} \frac{\d x}{x\ln x}\d x ∫ e 2 e 8 x ln x d x d x Let u = ln x , d u = 1 x d x u = \ln x, \d u = \frac 1x \d x u = ln x , d u = x 1 d x
For definite integrals, we also have to change the limits of integration from x 1 , x 2 x_1, x_2 x 1 , x 2 u 1 , u 2 u_1, u_2 u 1 , u 2
u 1 = ln ( e 2 ) = 2 , u 2 = ln ( e 8 ) = 8 u_1 = \ln (\e^2) = 2, u_2 = \ln (\e^8) = 8 u 1 = ln ( e 2 ) = 2 , u 2 = ln ( e 8 ) = 8 I = ∫ e 2 e 8 d x x ln x d x = ∫ 2 8 1 u d u = ln ∣ u ∣ ∣ 2 8 = ln 8 − ln 2 = ln ( 8 2 ) = ln 4 ■ \begin{align*}I &= \int_{\e^2}^{\e^8} \frac{\d x}{x\ln x}\d x \\
&= \int_{2}^{8} \frac{1}{u}\d u \\
&= \ln \lvert u\rvert\Big\vert_{2}^{8} \\
&= \ln 8 - \ln 2 \\
&= \ln \left(\frac82\right) \\
&= \ln 4 \qed
\end{align*} I = ∫ e 2 e 8 x ln x d x d x = ∫ 2 8 u 1 d u = ln ∣ u ∣ 2 8 = ln 8 − ln 2 = ln ( 2 8 ) = ln 4 ■ HL Examples HL involves integration by substitution not in the form of ∫ k ⋅ f ( G ( x ) ) ⋅ g ( x ) d x {\int k\cdot f(G(x)) \cdot g(x)\d x} ∫ k ⋅ f ( G ( x )) ⋅ g ( x ) d x
Example: By using the substitution x = 3 sin θ x = 3 \sin \theta x = 3 sin θ
∫ 9 − x 2 x 2 d x \int \frac{\sqrt{9-x^2}}{x^2} \d x ∫ x 2 9 − x 2 d x d x = 3 cos θ d θ \d x = 3\cos \theta \d \theta d x = 3 cos θ d θ
The numerator becomes 9 − ( 3 sin θ ) 2 = 9 − 9 sin 2 θ = 9 cos 2 θ = 3 cos θ \sqrt{9-(3\sin\theta)^2} = \sqrt{9-9\sin^2\theta} = \sqrt{9\cos^2\theta} = 3\cos\theta 9 − ( 3 sin θ ) 2 = 9 − 9 sin 2 θ = 9 cos 2 θ = 3 cos θ Pythagorean identity .
The denominator becomes ( 3 sin θ ) 2 = 9 sin 2 θ (3\sin\theta)^2 = 9\sin^2\theta ( 3 sin θ ) 2 = 9 sin 2 θ
I = ∫ 3 cos θ 9 sin 2 θ ( 3 cos θ d θ ) I = ∫ cos 2 θ sin 2 θ d θ I = ∫ cot 2 θ d θ \begin{align*}
I &= \int \frac{3\cos\theta}{9\sin^2\theta} \left(3\cos \theta \d \theta\right) \\
I &= \int \frac{\cos^2\theta}{\sin^2\theta} \d\theta \\
I &= \int \cot^2\theta\d\theta
\end{align*} I I I = ∫ 9 sin 2 θ 3 cos θ ( 3 cos θ d θ ) = ∫ sin 2 θ cos 2 θ d θ = ∫ cot 2 θ d θ Checking our HL derivative tables , there is no function listed with a derivative of cot 2 x \cot^2x cot 2 x
d d x cot x = − cosec 2 x \frac\d{\d x} \cot x = -\cosec^2 x d x d cot x = − cosec 2 x and a variant of Pythagorean identity
1 + cot 2 θ = cosec 2 θ 1 + \cot^2 \theta = \cosec^2 \theta 1 + cot 2 θ = cosec 2 θ I = ∫ cot 2 θ d θ = ∫ cosec 2 θ − 1 d θ = − cot θ − θ \begin{align*}
I &= \int \cot^2\theta\d\theta \\
&= \int \cosec^2\theta - 1 \d\theta \\
&= -\cot \theta - \theta \\
\end{align*} I = ∫ cot 2 θ d θ = ∫ cosec 2 θ − 1 d θ = − cot θ − θ We need the indefinite integral in terms of original variable x x x sin θ = x 3 \sin\theta = \frac{x}{3} sin θ = 3 x triangle with opposite x x x 3 3 3 . The adjacent is 9 − x 2 ≥ 0 \sqrt{9 - x^2}\geq 0 9 − x 2 ≥ 0 sin \sin sin cot \cot cot
cot θ = adjacent opposite = 9 − x 2 x \cot \theta = \frac{\text{adjacent}}{\text{opposite}} = \frac{\sqrt{9 - x^2}}{x} cot θ = opposite adjacent = x 9 − x 2 It can be confirmed that indeed sin \sin sin cot \cot cot x x x
The integral evaluates to
∫ 9 − x 2 x 2 d x = − 9 − x 2 x − arcsin x 3 + C ■ \int \frac{\sqrt{9-x^2}}{x^2} \d x = -\frac{\sqrt{9 - x^2}}{x} - \arcsin{\frac x3} + C\qed ∫ x 2 9 − x 2 d x = − x 9 − x 2 − arcsin 3 x + C ■ which conveniently involve range of arcsin \arcsin arcsin x x x x = 3 sin θ x = 3\sin\theta x = 3 sin θ x = 3 cos θ x = 3\cos\theta x = 3 cos θ x x x
Example: By using the substitution u = sin 2 x − 3 u = \sin^2 x - 3 u = sin 2 x − 3
∫ 0 π 2 − 4 sin x cos x sin 4 x − 6 sin 2 x + 25 d x \int_0^\frac{\pi}{2} \frac{-4\sin x\cos x}{\sin^4x - 6\sin^2 x + 25}\d x ∫ 0 2 π sin 4 x − 6 sin 2 x + 25 − 4 sin x cos x d x We have d u = 2 sin x cos x d x \d u = {2\sin x\cos x \d x} d u = 2 sin x cos x d x u 1 = sin 2 ( 0 ) − 3 = − 3 u_1 = {\sin^2 (0) - 3} = -3 u 1 = sin 2 ( 0 ) − 3 = − 3 u 2 = sin 2 ( π 2 ) − 3 = − 2 u_2 = {\sin^2 \left(\frac{\pi}{2}\right) - 3} = -2 u 2 = sin 2 ( 2 π ) − 3 = − 2
Rewrite the denominator as ( sin 2 x − 3 ) 2 + ( 4 ) 2 (\sin^2 x - 3)^2 +(4)^2 ( sin 2 x − 3 ) 2 + ( 4 ) 2 completing the square , then apply
∫ 1 x 2 + a 2 d x = 1 a arctan x a + C \int \frac{1}{x^2 + a^2}\d x = \frac 1a \arctan \frac xa + C ∫ x 2 + a 2 1 d x = a 1 arctan a x + C I = ∫ 0 π 2 − 4 sin x cos x sin 2 x − 6 sin x + 25 d x = ∫ 0 π 2 − 2 ( 2 sin x cos x ) ( sin 2 x − 3 ) 2 + ( 4 ) 2 d x = ∫ − 3 − 2 − 2 d u u 2 + 4 2 = 1 4 arctan u 4 ∣ − 3 − 2 ⋅ ( − 2 ) = ( arctan ( − 2 4 ) − arctan ( − 3 4 ) ) ⋅ − 1 2 = ( − arctan ( 1 2 ) + arctan ( 3 4 ) ) ⋅ − 1 2 = 1 2 arctan ( 1 2 ) − 1 2 arctan ( 3 4 ) ■ \begin{align*}I &= \int_0^\frac{\pi}{2} \frac{-4\sin x\cos x}{\sin^2x - 6\sin x + 25}\d x \\
&= \int_0^\frac{\pi}{2} \frac{-2(2\sin x\cos x)}{(\sin^2 x - 3)^2 + (4)^2}\d x \\
&= \int_{-3}^{-2} \frac{-2\d u}{u^2 + 4^2} \\
&=\frac 14\arctan \frac u4 \Big\vert_{-3}^{-2} \cdot (-2) \\
&= \left(\arctan \left(-\frac{2}{4}\right) - \arctan \left(-\frac{3}{4}\right)\right) \cdot \frac{-1}{2} \\
&= \left(-\arctan \left(\frac{1}{2}\right) + \arctan \left(\frac{3}{4}\right)\right) \cdot \frac{-1}{2} \\
&= \frac 12 \arctan \left(\frac{1}{2}\right) - \frac12\arctan \left(\frac{3}{4}\right) \qed
\end{align*} I = ∫ 0 2 π sin 2 x − 6 sin x + 25 − 4 sin x cos x d x = ∫ 0 2 π ( sin 2 x − 3 ) 2 + ( 4 ) 2 − 2 ( 2 sin x cos x ) d x = ∫ − 3 − 2 u 2 + 4 2 − 2 d u = 4 1 arctan 4 u − 3 − 2 ⋅ ( − 2 ) = ( arctan ( − 4 2 ) − arctan ( − 4 3 ) ) ⋅ 2 − 1 = ( − arctan ( 2 1 ) + arctan ( 4 3 ) ) ⋅ 2 − 1 = 2 1 arctan ( 2 1 ) − 2 1 arctan ( 4 3 ) ■ Tip: arctan \arctan arctan odd , such that arctan ( − x ) = − arctan ( x ) \arctan (-x) = -\arctan (x) arctan ( − x ) = − arctan ( x )