Math III Integration by substitution Substitution replaces part of the integrand with a new variable t t t t t t
Basic, Standard: Grade 12 · Term 2 / Advanced: Grade 12 · Exam prep
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Guide Examples Practice Related
How substitution works
The steps:
Call part of the integrand t t t Differentiate to write d x dx d x d t dt d t d t d x = g ′ ( x ) \frac{dt}{dx} = g'(x) d x d t = g ′ ( x ) d t = g ′ ( x ) d x dt = g'(x)\,dx d t = g ′ ( x ) d x Write everything in terms of t t t Substitute back to get an answer in x x x Linear substitutions If the inside is linear, as in ( 2 x + 1 ) 5 (2x+1)^5 ( 2 x + 1 ) 5 t = 2 x + 1 t = 2x + 1 t = 2 x + 1 d t = 2 d x dt = 2\,dx d t = 2 d x
∫ ( 2 x + 1 ) 5 d x = 1 2 ∫ t 5 d t = 1 12 t 6 + C = 1 12 ( 2 x + 1 ) 6 + C \int (2x+1)^5\,dx = \frac{1}{2}\int t^5\,dt = \frac{1}{12}t^6 + C = \frac{1}{12}(2x+1)^6 + C ∫ ( 2 x + 1 ) 5 d x = 2 1 ∫ t 5 d t = 12 1 t 6 + C = 12 1 ( 2 x + 1 ) 6 + C In general ∫ f ( a x + b ) d x = 1 a F ( a x + b ) + C \int f(ax+b)\,dx = \frac{1}{a}F(ax+b) + C ∫ f ( a x + b ) d x = a 1 F ( a x + b ) + C
The f(g(x))g'(x) and f'(x)/f(x) patterns If the integrand contains a function g ( x ) g(x) g ( x ) g ′ ( x ) g'(x) g ′ ( x ) t = g ( x ) t = g(x) t = g ( x )
∫ x ( x 2 + 1 ) 3 d x = 1 2 ∫ t 3 d t = 1 8 ( x 2 + 1 ) 4 + C ( t = x 2 + 1 , d t = 2 x d x ) \int x(x^2+1)^3\,dx = \frac{1}{2}\int t^3\,dt = \frac{1}{8}(x^2+1)^4 + C \qquad (t = x^2 + 1,\ dt = 2x\,dx) ∫ x ( x 2 + 1 ) 3 d x = 2 1 ∫ t 3 d t = 8 1 ( x 2 + 1 ) 4 + C ( t = x 2 + 1 , d t = 2 x d x ) ∫ sin 2 x cos x d x = ∫ t 2 d t = 1 3 sin 3 x + C ( t = sin x , d t = cos x d x ) \int \sin^2 x\cos x\,dx = \int t^2\,dt = \frac{1}{3}\sin^3 x + C \qquad (t = \sin x,\ dt = \cos x\,dx) ∫ sin 2 x cos x d x = ∫ t 2 d t = 3 1 sin 3 x + C ( t = sin x , d t = cos x d x ) In particular, when the numerator is the derivative of the denominator, use this formula directly.
Example: ∫ tan x d x = ∫ sin x cos x d x = − ∫ ( cos x ) ′ cos x d x = − log ∣ cos x ∣ + C \displaystyle\int \tan x\,dx = \int \frac{\sin x}{\cos x}\,dx = -\int \frac{(\cos x)'}{\cos x}\,dx = -\log|\cos x| + C ∫ tan x d x = ∫ cos x sin x d x = − ∫ cos x ( cos x ) ′ d x = − log ∣ cos x ∣ + C
Trigonometric substitutions For definite integrals containing a 2 − x 2 \sqrt{a^2 - x^2} a 2 − x 2 1 x 2 + a 2 \dfrac{1}{x^2 + a^2} x 2 + a 2 1
Form Substitution Identity used
a 2 − x 2 \sqrt{a^2 - x^2} a 2 − x 2 x = a sin θ x = a\sin\theta x = a sin θ 1 − sin 2 θ = cos 2 θ 1 - \sin^2\theta = \cos^2\theta 1 − sin 2 θ = cos 2 θ 1 x 2 + a 2 \dfrac{1}{x^2 + a^2} x 2 + a 2 1 x = a tan θ x = a\tan\theta x = a tan θ 1 + tan 2 θ = 1 cos 2 θ 1 + \tan^2\theta = \dfrac{1}{\cos^2\theta} 1 + tan 2 θ = cos 2 θ 1
When you substitute in a definite integral, change the limits to the new variable too. For x = a sin θ x = a\sin\theta x = a sin θ x : 0 → a x: 0 \to a x : 0 → a θ : 0 → π 2 \theta: 0 \to \frac{\pi}{2} θ : 0 → 2 π
Common mistakes Not converting d x dx d x : with t = 2 x + 1 t = 2x + 1 t = 2 x + 1 d x dx d x d t dt d t 1 2 \frac{1}{2} 2 1 Not substituting back : an indefinite integral must be written in terms of x x x Limits of a definite integral : change them when you substitute.Absolute value in log \log log : ∫ d x x = log ∣ x ∣ + C \int \frac{dx}{x} = \log|x| + C ∫ x d x = log ∣ x ∣ + C x 2 + 1 x^2 + 1 x 2 + 1 Worked examples
Find the indefinite integral.
∫ 1 4 x − 1 d x \int \frac{1}{\sqrt{4x - 1}}\,dx ∫ 4 x − 1 1 d x Check answer Hint Answer Solution + Worksheet ✓ Added
Hint Let t t t t = t 1 2 \sqrt{t} = t^{\frac{1}{2}} t = t 2 1
Answer ∫ 1 4 x − 1 d x = 1 2 4 x − 1 + C \int \frac{1}{\sqrt{4x - 1}}\,dx = \frac{1}{2}\sqrt{4x - 1} + C ∫ 4 x − 1 1 d x = 2 1 4 x − 1 + C (C C C
Solution Let t = 4 x − 1 t = 4x - 1 t = 4 x − 1 d x = 1 4 d t dx = \frac{1}{4}\,dt d x = 4 1 d t t = t 1 2 \sqrt{t} = t^{\frac{1}{2}} t = t 2 1 t 3 = t 1 3 \sqrt[3]{t} = t^{\frac{1}{3}} 3 t = t 3 1
∫ 1 4 x − 1 d x = 1 4 ∫ 1 t d t = 1 4 ⋅ 2 t + C = 1 2 t + C = 1 2 4 x − 1 + C \begin{aligned}\int \frac{1}{\sqrt{4x - 1}}\,dx &= \frac{1}{4}\int \frac{1}{\sqrt{t}}\,dt \\ &= \frac{1}{4} \cdot 2\sqrt{t} + C \\ &= \frac{1}{2}\sqrt{t} + C \\ &= \frac{1}{2}\sqrt{4x - 1} + C\end{aligned} ∫ 4 x − 1 1 d x = 4 1 ∫ t 1 d t = 4 1 ⋅ 2 t + C = 2 1 t + C = 2 1 4 x − 1 + C
Find the indefinite integral.
∫ x ( x 2 + 2 ) 2 d x \int x(x^{2} + 2)^{2}\,dx ∫ x ( x 2 + 2 ) 2 d x Check answer Hint Answer Solution + Worksheet ✓ Added
Hint Let t t t d t = 2 x d x dt = 2x\,dx d t = 2 x d x x x x
Answer ∫ x ( x 2 + 2 ) 2 d x = 1 6 ( x 2 + 2 ) 3 + C \int x(x^{2} + 2)^{2}\,dx = \frac{1}{6}(x^{2} + 2)^{3} + C ∫ x ( x 2 + 2 ) 2 d x = 6 1 ( x 2 + 2 ) 3 + C (C C C
Solution Let t = x 2 + 2 t = x^{2} + 2 t = x 2 + 2 d t = 2 x d x dt = 2x\,dx d t = 2 x d x x d x = 1 2 d t x\,dx = \dfrac{1}{2}\,dt x d x = 2 1 d t
∫ x ( x 2 + 2 ) 2 d x = 1 2 ∫ t 2 d t = 1 6 t 3 + C = 1 6 ( x 2 + 2 ) 3 + C \begin{aligned}\int x(x^{2} + 2)^{2}\,dx &= \frac{1}{2}\int t^{2}\,dt \\ &= \frac{1}{6}t^{3} + C \\ &= \frac{1}{6}(x^{2} + 2)^{3} + C\end{aligned} ∫ x ( x 2 + 2 ) 2 d x = 2 1 ∫ t 2 d t = 6 1 t 3 + C = 6 1 ( x 2 + 2 ) 3 + C
Find the indefinite integral.
∫ e 2 x e x + 2 d x \int \frac{e^{2x}}{e^{x} + 2}\,dx ∫ e x + 2 e 2 x d x Check answer Hint Answer Solution + Worksheet ✓ Added
Hint Let t = e x t = e^x t = e x e 2 x d x = t d t e^{2x}\,dx = t\,dt e 2 x d x = t d t
Answer ∫ e 2 x e x + 2 d x = e x − 2 log ( e x + 2 ) + C \int \frac{e^{2x}}{e^{x} + 2}\,dx = e^{x} - 2\log(e^{x} + 2) + C ∫ e x + 2 e 2 x d x = e x − 2 log ( e x + 2 ) + C (C C C
Solution Let t = e x t = e^x t = e x d t = e x d x dt = e^x\,dx d t = e x d x e 2 x d x = e x ⋅ e x d x = t d t e^{2x}\,dx = e^x \cdot e^x\,dx = t\,dt e 2 x d x = e x ⋅ e x d x = t d t
∫ e 2 x e x + 2 d x = ∫ t t + 2 d t = ∫ ( 1 − 2 t + 2 ) d t = t − 2 log ( t + 2 ) + C = e x − 2 log ( e x + 2 ) + C \begin{aligned}\int \frac{e^{2x}}{e^{x} + 2}\,dx &= \int \frac{t}{t + 2}\,dt \\ &= \int \left(1 - \frac{2}{t + 2}\right)\,dt \\ &= t - 2\log(t + 2) + C \\ &= e^{x} - 2\log(e^{x} + 2) + C\end{aligned} ∫ e x + 2 e 2 x d x = ∫ t + 2 t d t = ∫ ( 1 − t + 2 2 ) d t = t − 2 log ( t + 2 ) + C = e x − 2 log ( e x + 2 ) + C Practice problems
Find the indefinite integral.
∫ 1 2 x + 3 d x \int \frac{1}{\sqrt{2x + 3}}\,dx ∫ 2 x + 3 1 d x Check answer Hint Answer Solution + Worksheet ✓ Added
Hint Let t t t t = t 1 2 \sqrt{t} = t^{\frac{1}{2}} t = t 2 1
Answer ∫ 1 2 x + 3 d x = 2 x + 3 + C \int \frac{1}{\sqrt{2x + 3}}\,dx = \sqrt{2x + 3} + C ∫ 2 x + 3 1 d x = 2 x + 3 + C (C C C
Solution Let t = 2 x + 3 t = 2x + 3 t = 2 x + 3 d x = 1 2 d t dx = \frac{1}{2}\,dt d x = 2 1 d t t = t 1 2 \sqrt{t} = t^{\frac{1}{2}} t = t 2 1 t 3 = t 1 3 \sqrt[3]{t} = t^{\frac{1}{3}} 3 t = t 3 1
∫ 1 2 x + 3 d x = 1 2 ∫ 1 t d t = 1 2 ⋅ 2 t + C = t + C = 2 x + 3 + C \begin{aligned}\int \frac{1}{\sqrt{2x + 3}}\,dx &= \frac{1}{2}\int \frac{1}{\sqrt{t}}\,dt \\ &= \frac{1}{2} \cdot 2\sqrt{t} + C \\ &= \sqrt{t} + C \\ &= \sqrt{2x + 3} + C\end{aligned} ∫ 2 x + 3 1 d x = 2 1 ∫ t 1 d t = 2 1 ⋅ 2 t + C = t + C = 2 x + 3 + C
Find the indefinite integral.
∫ cos 5 x sin x d x \int \cos^{5} x\sin x\,dx ∫ cos 5 x sin x d x Check answer Hint Answer Solution + Worksheet ✓ Added
Hint Let t = cos x t = \cos x t = cos x sin x d x \sin x\,dx sin x d x − d t -dt − d t
Answer ∫ cos 5 x sin x d x = − 1 6 cos 6 x + C \int \cos^{5} x\sin x\,dx = -\frac{1}{6}\cos^{6} x + C ∫ cos 5 x sin x d x = − 6 1 cos 6 x + C (C C C
Solution Let t = cos x t = \cos x t = cos x d t = − sin x d x dt = -\sin x\,dx d t = − sin x d x sin x d x = − d t \sin x\,dx = -dt sin x d x = − d t
∫ cos 5 x sin x d x = − ∫ t 5 d t = − 1 6 t 6 + C = − 1 6 cos 6 x + C \begin{aligned}\int \cos^{5} x\sin x\,dx &= -\int t^{5}\,dt \\ &= -\frac{1}{6}t^{6} + C \\ &= -\frac{1}{6}\cos^{6} x + C\end{aligned} ∫ cos 5 x sin x d x = − ∫ t 5 d t = − 6 1 t 6 + C = − 6 1 cos 6 x + C
Find the indefinite integral.
∫ 1 e x + 2 d x \int \frac{1}{e^{x} + 2}\,dx ∫ e x + 2 1 d x Check answer Hint Answer Solution + Worksheet ✓ Added
Hint Let t = e x t = e^x t = e x d x = d t t dx = \frac{dt}{t} d x = t d t
Answer ∫ 1 e x + 2 d x = 1 2 ( x − log ( e x + 2 ) ) + C \int \frac{1}{e^{x} + 2}\,dx = \frac{1}{2}(x - \log(e^{x} + 2)) + C ∫ e x + 2 1 d x = 2 1 ( x − log ( e x + 2 )) + C (C C C
Solution Let t = e x t = e^x t = e x d t = e x d x = t d x dt = e^x\,dx = t\,dx d t = e x d x = t d x d x = d t t dx = \dfrac{dt}{t} d x = t d t
∫ 1 e x + 2 d x = ∫ 1 t ( t + 2 ) d t = 1 2 ∫ ( 1 t − 1 t + 2 ) d t = 1 2 ( log t − log ( t + 2 ) ) + C = 1 2 ( x − log ( e x + 2 ) ) + C \begin{aligned}\int \frac{1}{e^{x} + 2}\,dx &= \int \frac{1}{t(t + 2)}\,dt \\ &= \frac{1}{2}\int \left(\frac{1}{t} - \frac{1}{t + 2}\right)\,dt \\ &= \frac{1}{2}(\log t - \log(t + 2)) + C \\ &= \frac{1}{2}(x - \log(e^{x} + 2)) + C\end{aligned} ∫ e x + 2 1 d x = ∫ t ( t + 2 ) 1 d t = 2 1 ∫ ( t 1 − t + 2 1 ) d t = 2 1 ( log t − log ( t + 2 )) + C = 2 1 ( x − log ( e x + 2 )) + C Since t = e x > 0 t = e^x > 0 t = e x > 0 log \log log