Integrating with Trigonometric Identities (AQA A-Level Mathematics): Revision Notes

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8.2.7 Integrating with Trigonometric Identities

Step-by-step method:

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To integrate using trigonometric identities:

  1. Simplify the expression with identities like half-angle, double-angle, or product-to-sum.
  2. Apply standard integrals (e.g., ∫sin⁡(x) dx=−cos⁡(x)+C\int \sin(x) \, dx = -\cos(x) + C ).
  3. Simplify the result and add the constant CC . This process reduces complex trigonometric integrals to manageable forms.

Trigonometric Integration

Integrating trig functions requires in-depth knowledge and quick recall of trigonometric identities.

Key Trigonometric Identities:

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sin⁡2θ+cos⁡2θ≡1\sin^2 \theta + \cos^2 \theta \equiv 1tan⁡2θ+1≡sec⁡2θ\tan^2 \theta + 1 \equiv \sec^2 \theta1+cot⁡2θ≡csc⁡2θ1 + \cot^2 \theta \equiv \csc^2 \thetacos⁡2θ≡cos⁡2θ−sin⁡2θ=2cos⁡2θ−1=1−2sin⁡2θ\cos 2\theta \equiv \cos^2 \theta - \sin^2 \theta = 2 \cos^2 \theta - 1 = 1 - 2 \sin^2 \thetasin⁡2θ≡2sin⁡θcos⁡θ\sin 2\theta \equiv 2 \sin \theta \cos \thetatan⁡θ≡sin⁡θcos⁡θ\tan \theta \equiv \frac{\sin \theta}{\cos \theta}

Examples of Trigonometric Integration:

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  1. Example: ∫tan⁡2θ dθ\int \tan^2 \theta \, d\theta
(tan⁡2θ+1≡sec⁡2θ)(\tan^2 \theta +1 \equiv \sec^2 \theta)=∫sec⁡2θ−1 dθ⇒tan⁡θ−θ+c=\int \sec^2\theta-1 \,d\theta \Rightarrow \tan \theta - \theta+ c
  1. Example: ∫tan⁡θ dθ=∫sin⁡θcos⁡θ dθ(tan⁡θ≡sin⁡θcos⁡θ)\int \tan \theta \, d\theta = \int \frac{\sin \theta}{\cos \theta} \, d\theta \quad (\tan \theta \equiv \frac {\sin \theta}{\cos \theta}) Let u=cos⁡θu = \cos \theta
⇒dudθ=−sin⁡θ⇒dθ=−1sin⁡θ du\Rightarrow \frac{du}{d\theta} = -\sin \theta \Rightarrow d\theta = \frac{-1}{\sin \theta} \, du⇒I=∫sin⁡θu×−1sin⁡θ du=−∫1u du=−ln⁡∣u∣\Rightarrow I = \int \frac{\cancel\sin \theta}{u} \times \frac{-1}{\cancel\sin \theta} \, du = -\int \frac{1}{u} \, du = -\ln |u|=−ln⁡∣cos⁡θ∣+c(or −1ln⁡∣cos⁡θ∣=ln⁡∣(cos⁡θ)−1∣=ln⁡∣sec⁡θ∣)= -\ln |\cos \theta| + c \quad \text{(or } -1 \ln |\cos \theta | =\ln |(\cos \theta)^{-1} | = \ln |\sec \theta| \text{)}
  1. Example: ∫sin⁡θcos⁡θ dθ=∫12sin⁡2θ dθ \int \sin \theta \cos \theta \, d\theta = \int \frac{1}{2} \sin 2\theta \, d\theta
=−14cos⁡2θ+c= \frac{-1}{4} \cos 2\theta + c
  1. Example: ∫sin⁡θcos⁡θ dθ\int \sin \theta \cos \theta \, d\theta
⇒Let u=sin⁡θ⇒dudθ=cos⁡θ⇒dθ=1cos⁡θ du\Rightarrow \text{Let } u = \sin \theta \Rightarrow \frac{du}{d\theta} = \cos \theta \Rightarrow d\theta = \frac{1}{\cos \theta} \, du=∫u cos⁡θ×1cos⁡θ du=\int u \,\cancel\cos \theta \times \frac{1}{\cancel\cos \theta} \, du=∫u du=12u2+c=12sin⁡2θ+c=\int u \, du = \frac {1}{2}u^2 + c = \frac {1}{2}\sin^2 \theta +c

Verification of Identity:

12sin⁡2θ+c14cos⁡2θ+c2=14cos⁡2θ+c2\frac{1}{2} \sin^2 \theta + \frac{c_1}{4} \cos 2\theta + c_2 = \frac{1}{4} \cos 2\theta + c_2⇒14(1−2sin⁡2θ)+c2=14+12sin⁡2θ+c2\Rightarrow \frac{1}{4} (1 - 2\sin^2 \theta) + c_2 = \frac{1}{4} + \frac{1}{2} \sin^2 \theta + c_2=12sin⁡2θ+14+c2✓= \frac{1}{2} \sin^2 \theta + \frac{1}{4}+c_2 \quad \checkmark
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Example

∫tan⁡3x dx=∫tan⁡2xtan⁡2x dx=∫tan⁡2x(sec⁡2x−1) dx\int \tan^3 x \, dx = \int \tan^2 x \tan^2 x \, dx = \int \tan^2 x (\sec^2 x - 1) \, dx=∫tan⁡2xsec⁡2x(A◯)−∫tan⁡2x dx(B◯)= \int \tan^2 x \sec^2 x \quad(\text{\textcircled A}) - \int \tan^2 x \, dx \quad(\text{\textcircled B})

A◯\text{\textcircled A}: Let u=tan⁡xu = \tan x ⇒dudx=sec⁡2x⇒dx=1sec⁡2x du\Rightarrow \frac{du}{dx} = \sec^2 x \Rightarrow dx = \frac{1}{\sec^2 x} \, du

A◯⇒∫u sec⁡2x×1sec⁡2x du=13u3+c=13tan⁡3x+c\text{\textcircled A}\Rightarrow \int u\, \cancel\sec^2 x \times \frac {1}{\cancel\sec^2 x}\, du = \frac {1}{3}u^3+c = \frac {1}{3}\tan^3x+c

B◯\text{\textcircled B}:

∫tan⁡2x dx=∫sec⁡2x−1 dx=tan⁡x−x+c2\int \tan^2 x \, dx = \int \sec^2x-1 \, dx= \tan x - x +c_2∴I=13tan⁡3x−tan⁡x+x+d←integration constant\therefore \quad I = \frac{1}{3}\tan^3 x - \tan x +x+d \leftarrow \text{integration constant}

Integrating sin⁡2x and cos⁡2x\sin^2 x \ and\ \cos^2 x

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  1. Expand cos⁡2x\cos 2x and write only in terms of what we are asked to integrate.
  2. Make our integral the subject.
  3. Replace our integral with the new integrable expression.
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Example:

∫sin⁡2x dx\int \sin^2 x \, dxcos⁡2x=1−2sin⁡2x2sin⁡2x=1−cos⁡2xsin⁡2x=12−12cos⁡2x\cos 2x = 1 - 2 \sin^2 x \\ 2\sin^2x = 1-\cos 2x\\ \sin^2 x = \frac{1}{2}-\frac {1}{2}\cos2xI=∫12−12cos⁡(2x) dxI = \int \frac{1}{2} - \frac{1}{2} \cos (2x) \, dx=12x−14sin⁡(2x)+c= \frac{1}{2}x - \frac{1}{4} \sin(2x) + c
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Example:

∫cos⁡2x dx\int \cos^2 x \, dxcos⁡2x=2cos⁡2x−1⇒2cos⁡2x=cos⁡2x+1⇒cos⁡2x=12cos⁡2x+12\cos 2x = 2 \cos^2 x - 1 \\ \Rightarrow 2\cos^2x=\cos2x +1\\ \Rightarrow \cos^2x=\frac {1}{2}\cos2x+\frac {1}{2}=∫12cos⁡2x+12 dx= \int \frac {1}{2}\cos2x+\frac {1}{2}\, dx=12x+14sin⁡(2x)+c= \frac{1}{2}x + \frac{1}{4} \sin(2x) + c
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Example:

∫cos⁡2(10x) dx\int \cos^2 (10x) \, dx

cos⁡2θ=2cos⁡2θ−1\cos 2\theta = 2\cos^2 \theta -1

Multiply by 2 and Divided by 2

cos⁡20x=2cos⁡2(10x)−1⇒2cos⁡2(10x)=cos⁡(20x)+1\cos 20x = 2 \cos^2(10x) - 1\\ \Rightarrow 2\cos^2(10x) = \cos(20x) +1⇒cos⁡2(10x)=12cos⁡(20x)+12\Rightarrow \cos^2(10x) = \frac{1 }{2}\cos (20x)+\frac {1}{2}=∫12cos⁡(20x)+12 dx=140sin⁡(20x)+12x+c=\int \frac{1 }{2}\cos (20x)+\frac {1}{2} \, dx \\ = \frac {1}{40}\sin (20x)+\frac {1}{2}x+c∫(x−sin⁡2x)2 dx=∫(x−sin⁡2x)(x−sin⁡2x) dx\int (x - \sin 2x)^2 \, dx = \int (x -\sin 2x)(x- \sin 2x) \, dx=∫x2−xsin⁡2x−xsin⁡2x+sin⁡22x dx= \int x^2-x\sin2x-x\sin2x+\sin^22x\, dx=∫x2−2xsin⁡2x+sin⁡22x dx=\int x^2 - 2x\sin2x + \sin^22x \, dxA◯=x2∣B◯=2xsin⁡2x∣C◯=sin⁡22x dx\boxed {\text{\textcircled A} = x^2 \bigg | \text{\textcircled B} = 2x\sin2x \bigg | \text{\textcircled C}= \sin^22x \, dx}

A◯\text{\textcircled A} ∫x2 dx\int x^2 \, dx

=13x3+c= \frac{1}{3}x^3 + c

B◯\text{\textcircled B} ∫2xsin⁡2x dx\int 2x \sin 2x \, dx By parts: u=2xu = 2x

⇒dudx=2\Rightarrow \frac{du}{dx} = 2v=−12cos⁡2x⇒dvdx=sin⁡2xv = -\frac{1}{2} \cos 2x \quad \Rightarrow \quad \frac{dv}{dx} = \sin 2x⇒I=−xcos⁡2x−∫−cos⁡2x dx\Rightarrow I = -x \cos 2x - \int -\cos 2x \, dx=−xcos⁡2x+∫cos⁡2x dx= -x \cos 2x + \int \cos 2x \, dx=−xcos⁡2x+12sin⁡2x+c= -x \cos 2x + \frac{1}{2} \sin 2x + c

C◯\text{\textcircled C} ∫sin⁡22x dx\int \sin^2 2x \, dx

cos⁡(4x)=1−2sin⁡2(2x)⇒2sin⁡2(2x)=1−cos⁡(4x)⇒sin⁡2(2x)=12−12cos⁡(4x)\cos(4x) = 1 - 2\sin^2(2x) \\ \Rightarrow 2\sin^2(2x) = 1 - \cos (4x)\\ \Rightarrow \sin^2(2x) = \frac {1}{2}-\frac {1}{2}\cos(4x)=∫12−12cos⁡(4x) dx= \int \frac {1}{2}-\frac {1}{2}\cos(4x)\, dx\\=12x−18sin⁡(4x)= \frac{1}{2}x - \frac{1}{8} \sin (4x)∴I=13x3−xcos⁡2x−12sin⁡2x+12x−18sin⁡4x+c\therefore \quad I = \frac{1}{3}x^3 - x \cos 2x - \frac{1}{2} \sin 2x + \frac{1}{2}x - \frac{1}{8}\sin 4x + c

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