Math III Definite integrals (advanced) Besides substitution and integration by parts, definite integrals reward using symmetry. These techniques show up often in exams.
Basic, Standard: Grade 12 · Term 2 / Advanced: Grade 12 · Exam prep
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Guide Examples Practice Related
Basic evaluation As before, find an antiderivative F ( x ) F(x) F ( x ) F ( b ) − F ( a ) F(b) - F(a) F ( b ) − F ( a ) sin π 6 = 1 2 \sin\frac{\pi}{6} = \frac{1}{2} sin 6 π = 2 1 e 0 = 1 e^0 = 1 e 0 = 1 log 1 = 0 \log 1 = 0 log 1 = 0 log e = 1 \log e = 1 log e = 1
∫ 0 π 3 cos x d x = [ sin x ] 0 π 3 = 3 2 , ∫ 1 e d x x = [ log x ] 1 e = 1 \int_0^{\frac{\pi}{3}} \cos x\,dx = \Bigl[\sin x\Bigr]_0^{\frac{\pi}{3}} = \frac{\sqrt{3}}{2},\qquad \int_1^{e} \frac{dx}{x} = \Bigl[\log x\Bigr]_1^{e} = 1 ∫ 0 3 π cos x d x = [ sin x ] 0 3 π = 2 3 , ∫ 1 e x d x = [ log x ] 1 e = 1 Using even and odd functions sin x \sin x sin x x cos x x\cos x x cos x x x 2 + 1 \dfrac{x}{x^2+1} x 2 + 1 x cos x \cos x cos x x sin x x\sin x x sin x x 2 x^2 x 2 − a -a − a a a a
∫ − π 2 π 2 ( x cos x + cos x ) d x = 2 ∫ 0 π 2 cos x d x = 2 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (x\cos x + \cos x)\,dx = 2\int_0^{\frac{\pi}{2}} \cos x\,dx = 2 ∫ − 2 π 2 π ( x cos x + cos x ) d x = 2 ∫ 0 2 π cos x d x = 2 Absolute values Split ∣ cos x ∣ |\cos x| ∣ cos x ∣ ∣ log x ∣ |\log x| ∣ log x ∣
∫ 0 π ∣ cos x ∣ d x = ∫ 0 π 2 cos x d x − ∫ π 2 π cos x d x = 1 + 1 = 2 \int_0^{\pi} |\cos x|\,dx = \int_0^{\frac{\pi}{2}} \cos x\,dx - \int_{\frac{\pi}{2}}^{\pi} \cos x\,dx = 1 + 1 = 2 ∫ 0 π ∣ cos x ∣ d x = ∫ 0 2 π cos x d x − ∫ 2 π π cos x d x = 1 + 1 = 2 The Wallis formula Let I n = ∫ 0 π 2 sin n x d x I_n = \displaystyle\int_0^{\frac{\pi}{2}} \sin^n x\,dx I n = ∫ 0 2 π sin n x d x I n = n − 1 n I n − 2 I_n = \dfrac{n-1}{n}I_{n-2} I n = n n − 1 I n − 2 I 0 = π 2 I_0 = \dfrac{\pi}{2} I 0 = 2 π I 1 = 1 I_1 = 1 I 1 = 1
I n = { n − 1 n ⋅ n − 3 n − 2 ⋯ 1 2 ⋅ π 2 ( n even ) n − 1 n ⋅ n − 3 n − 2 ⋯ 2 3 ⋅ 1 ( n odd ) I_n = \begin{cases}\dfrac{n-1}{n}\cdot\dfrac{n-3}{n-2}\cdots\dfrac{1}{2}\cdot\dfrac{\pi}{2} & (n\ \text{even})\\[2ex] \dfrac{n-1}{n}\cdot\dfrac{n-3}{n-2}\cdots\dfrac{2}{3}\cdot 1 & (n\ \text{odd})\end{cases} I n = ⎩ ⎨ ⎧ n n − 1 ⋅ n − 2 n − 3 ⋯ 2 1 ⋅ 2 π n n − 1 ⋅ n − 2 n − 3 ⋯ 3 2 ⋅ 1 ( n even ) ( n odd ) cos n x \cos^n x cos n x
The substitution x = π/2 − t For integrals like ∫ 0 π 2 sin x sin x + cos x d x \displaystyle\int_0^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x}\,dx ∫ 0 2 π sin x + cos x sin x d x x = π 2 − t x = \frac{\pi}{2} - t x = 2 π − t sin \sin sin cos \cos cos 1 1 1 2 I = π 2 2I = \frac{\pi}{2} 2 I = 2 π I = π 4 I = \frac{\pi}{4} I = 4 π
This uses the general fact ∫ a b f ( x ) d x = ∫ a b f ( a + b − x ) d x \displaystyle\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx ∫ a b f ( x ) d x = ∫ a b f ( a + b − x ) d x
Worked examples
Evaluate the definite integral.
∫ 0 π 6 1 cos 2 x d x \int_{0}^{\frac{\pi}{6}} \frac{1}{\cos^{2} x}\,dx ∫ 0 6 π cos 2 x 1 d x Check answer Hint Answer Solution + Worksheet ✓ Added
Hint Find an antiderivative and compute (value at the upper limit) − (value at the lower limit).
Answer ∫ 0 π 6 1 cos 2 x d x = 3 3 \int_{0}^{\frac{\pi}{6}} \frac{1}{\cos^{2} x}\,dx = \frac{\sqrt{3}}{3} ∫ 0 6 π cos 2 x 1 d x = 3 3 Solution Find an antiderivative and subtract its value at the lower limit from its value at the upper limit.
∫ 0 π 6 1 cos 2 x d x = [ tan x ] 0 π 6 = tan π 6 − tan 0 = 3 3 \begin{aligned}\int_{0}^{\frac{\pi}{6}} \frac{1}{\cos^{2} x}\,dx &= \left[\tan x\right]_{0}^{\frac{\pi}{6}} \\ &= \tan\frac{\pi}{6} - \tan 0 \\ &= \frac{\sqrt{3}}{3}\end{aligned} ∫ 0 6 π cos 2 x 1 d x = [ tan x ] 0 6 π = tan 6 π − tan 0 = 3 3
Evaluate the definite integral.
∫ 0 π 4 tan x d x \int_{0}^{\frac{\pi}{4}} \tan x\,dx ∫ 0 4 π tan x d x Check answer Hint Answer Solution + Worksheet ✓ Added
Hint Check whether the numerator is the derivative of the denominator (the f ′ ( x ) f ( x ) \frac{f'(x)}{f(x)} f ( x ) f ′ ( x )
Answer ∫ 0 π 4 tan x d x = 1 2 log 2 \int_{0}^{\frac{\pi}{4}} \tan x\,dx = \frac{1}{2}\log 2 ∫ 0 4 π tan x d x = 2 1 log 2 Solution ∫ tan x d x = − log ∣ cos x ∣ + C \displaystyle\int \tan x\,dx = -\log|\cos x| + C ∫ tan x d x = − log ∣ cos x ∣ + C cos x > 0 \cos x > 0 cos x > 0
∫ 0 π 4 tan x d x = [ − log ( cos x ) ] 0 π 4 = 1 2 log 2 \begin{aligned}\int_{0}^{\frac{\pi}{4}} \tan x\,dx &= \left[-\log(\cos x)\right]_{0}^{\frac{\pi}{4}} \\ &= \frac{1}{2}\log 2\end{aligned} ∫ 0 4 π tan x d x = [ − log ( cos x ) ] 0 4 π = 2 1 log 2
Evaluate the definite integral.
∫ 1 e ( log x ) 3 x d x \int_{1}^{e} \frac{(\log x)^{3}}{x}\,dx ∫ 1 e x ( log x ) 3 d x Check answer Hint Answer Solution + Worksheet ✓ Added
Hint When you substitute, change the limits of integration to the range of the new variable as well.
Answer ∫ 1 e ( log x ) 3 x d x = 1 4 \int_{1}^{e} \frac{(\log x)^{3}}{x}\,dx = \frac{1}{4} ∫ 1 e x ( log x ) 3 d x = 4 1 Solution Let t = log x t = \log x t = log x d t = 1 x d x dt = \dfrac{1}{x}\,dx d t = x 1 d x x : 1 → e x: 1 \to e x : 1 → e t : 0 → 1 t: 0 \to 1 t : 0 → 1
∫ 1 e ( log x ) 3 x d x = ∫ 0 1 t 3 d t = [ 1 4 t 4 ] 0 1 = 1 4 \begin{aligned}\int_{1}^{e} \frac{(\log x)^{3}}{x}\,dx &= \int_{0}^{1} t^{3}\,dt \\ &= \left[\frac{1}{4}t^{4}\right]_{0}^{1} \\ &= \frac{1}{4}\end{aligned} ∫ 1 e x ( log x ) 3 d x = ∫ 0 1 t 3 d t = [ 4 1 t 4 ] 0 1 = 4 1 Practice problems
Evaluate the definite integral.
∫ 0 1 1 2 x + 1 d x \int_{0}^{1} \frac{1}{2x + 1}\,dx ∫ 0 1 2 x + 1 1 d x Check answer Hint Answer Solution + Worksheet ✓ Added
Hint Recall ∫ d x a x + b = 1 a log ∣ a x + b ∣ \int \frac{dx}{ax+b} = \frac{1}{a}\log|ax+b| ∫ a x + b d x = a 1 log ∣ a x + b ∣
Answer ∫ 0 1 1 2 x + 1 d x = 1 2 log 3 \int_{0}^{1} \frac{1}{2x + 1}\,dx = \frac{1}{2}\log 3 ∫ 0 1 2 x + 1 1 d x = 2 1 log 3 Solution Find an antiderivative with the formula ∫ f ( a x + b ) d x = 1 a F ( a x + b ) + C \displaystyle\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
∫ 0 1 1 2 x + 1 d x = [ 1 2 log ( 2 x + 1 ) ] 0 1 = 1 2 log 3 \begin{aligned}\int_{0}^{1} \frac{1}{2x + 1}\,dx &= \left[\frac{1}{2}\log(2x + 1)\right]_{0}^{1} \\ &= \frac{1}{2}\log 3\end{aligned} ∫ 0 1 2 x + 1 1 d x = [ 2 1 log ( 2 x + 1 ) ] 0 1 = 2 1 log 3
Evaluate the definite integral.
∫ 0 π 3 sin 2 x d x \int_{0}^{\frac{\pi}{3}} \sin^{2} x\,dx ∫ 0 3 π sin 2 x d x Check answer Hint Answer Solution + Worksheet ✓ Added
Hint Lower the degree with the half-angle formula, then integrate.
Answer ∫ 0 π 3 sin 2 x d x = π 6 − 3 8 \int_{0}^{\frac{\pi}{3}} \sin^{2} x\,dx = \frac{\pi}{6} - \frac{\sqrt{3}}{8} ∫ 0 3 π sin 2 x d x = 6 π − 8 3 Solution Using the half-angle formula sin 2 x = 1 − cos 2 x 2 \sin^2 x = \dfrac{1 - \cos 2x}{2} sin 2 x = 2 1 − cos 2 x
∫ 0 π 3 sin 2 x d x = ∫ 0 π 3 1 2 ( 1 − cos 2 x ) d x = [ 1 2 x − 1 4 sin 2 x ] 0 π 3 = π 6 − 3 8 \begin{aligned}\int_{0}^{\frac{\pi}{3}} \sin^{2} x\,dx &= \int_{0}^{\frac{\pi}{3}} \frac{1}{2}(1 - \cos 2x)\,dx \\ &= \left[\frac{1}{2}x - \frac{1}{4}\sin 2x\right]_{0}^{\frac{\pi}{3}} \\ &= \frac{\pi}{6} - \frac{\sqrt{3}}{8}\end{aligned} ∫ 0 3 π sin 2 x d x = ∫ 0 3 π 2 1 ( 1 − cos 2 x ) d x = [ 2 1 x − 4 1 sin 2 x ] 0 3 π = 6 π − 8 3
Evaluate the definite integral.
∫ 0 π 2 sin x sin x + cos x d x \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x}\,dx ∫ 0 2 π sin x + cos x sin x d x Check answer Hint Answer Solution + Worksheet ✓ Added
Hint Substituting x = π 2 − t x = \frac{\pi}{2} - t x = 2 π − t sin \sin sin cos \cos cos
Answer ∫ 0 π 2 sin x sin x + cos x d x = π 4 \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x}\,dx = \frac{\pi}{4} ∫ 0 2 π sin x + cos x sin x d x = 4 π Solution Let I = ∫ 0 π 2 sin x sin x + cos x d x I = \displaystyle\int_0^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x}\,dx I = ∫ 0 2 π sin x + cos x sin x d x x = π 2 − t x = \dfrac{\pi}{2} - t x = 2 π − t d x = − d t dx = -dt d x = − d t sin x = cos t \sin x = \cos t sin x = cos t cos x = sin t \cos x = \sin t cos x = sin t x : 0 → π 2 x: 0 \to \dfrac{\pi}{2} x : 0 → 2 π t : π 2 → 0 t: \dfrac{\pi}{2} \to 0 t : 2 π → 0
Hence
I = ∫ π 2 0 cos t cos t + sin t ( − d t ) = ∫ 0 π 2 cos x cos x + sin x d x I = \int_{\frac{\pi}{2}}^{0} \frac{\cos t}{\cos t + \sin t}\,(-dt) = \int_0^{\frac{\pi}{2}} \frac{\cos x}{\cos x + \sin x}\,dx I = ∫ 2 π 0 cos t + sin t cos t ( − d t ) = ∫ 0 2 π cos x + sin x cos x d x Adding this to the original integral, the integrands add up to 1, so
2 I = ∫ 0 π 2 sin x + cos x sin x + cos x d x = ∫ 0 π 2 d x = π 2 2I = \int_0^{\frac{\pi}{2}} \frac{\sin x + \cos x}{\sin x + \cos x}\,dx = \int_0^{\frac{\pi}{2}} dx = \frac{\pi}{2} 2 I = ∫ 0 2 π sin x + cos x sin x + cos x d x = ∫ 0 2 π d x = 2 π Therefore
I = π 4 I = \frac{\pi}{4} I = 4 π