Porisma

Use trigonometric identities

Rewrite products and powers before integrating.

Trigonometric standard forms and identities

Check the inner derivative before choosing a formula. If no direct match appears, simplify with an identity: lower even sine/cosine powers, or replace tangent/cotangent squares using the corresponding secant/cosecant identity. Recheck signs and chain-rule factors after integrating.

Reverse ddxsin⁡[f(x)]=f′(x)cos⁡[f(x)]\frac{d}{dx}\sin[f(x)]=f^{\prime}(x)\cos[f(x)] to get the first composite rule below. The same reasoning gives all six source pairs (each with an arbitrary CC):

Basic integral: ∫cos⁡x dx=sin⁡x+C\int\cos x\,dx=\sin x+C

Composite integral: ∫f′(x)cos⁡[f(x)] dx=sin⁡[f(x)]+C\int f^{\prime}(x)\cos[f(x)]\,dx=\sin[f(x)]+C

Basic integral: ∫sin⁡x dx=−cos⁡x+C\int\sin x\,dx=-\cos x+C

Composite integral: ∫f′(x)sin⁡[f(x)] dx=−cos⁡[f(x)]+C\int f^{\prime}(x)\sin[f(x)]\,dx=-\cos[f(x)]+C

Basic integral: ∫sec⁡2x dx=tan⁡x+C\int\sec^2x\,dx=\tan x+C

Composite integral: ∫f′(x)sec⁡2[f(x)] dx=tan⁡[f(x)]+C\int f^{\prime}(x)\sec^2[f(x)]\,dx=\tan[f(x)]+C

Basic integral: ∫csc⁡2x dx=−cot⁡x+C\int\csc^2x\,dx=-\cot x+C

Composite integral: ∫f′(x)csc⁡2[f(x)] dx=−cot⁡[f(x)]+C\int f^{\prime}(x)\csc^2[f(x)]\,dx=-\cot[f(x)]+C

Basic integral: ∫sec⁡xtan⁡x dx=sec⁡x+C\int\sec x\tan x\,dx=\sec x+C

Composite integral: ∫f′(x)sec⁡[f(x)]tan⁡[f(x)] dx=sec⁡[f(x)]+C\int f^{\prime}(x)\sec[f(x)]\tan[f(x)]\,dx=\sec[f(x)]+C

Basic integral: ∫csc⁡xcot⁡x dx=−csc⁡x+C\int\csc x\cot x\,dx=-\csc x+C

Composite integral: ∫f′(x)csc⁡[f(x)]cot⁡[f(x)] dx=−csc⁡[f(x)]+C\int f^{\prime}(x)\csc[f(x)]\cot[f(x)]\,dx=-\csc[f(x)]+C

The source writes cosec for csc. The Teacher labels the basic forms O-level and the composite forms H2-level, emphasising the inner derivative and the minus signs in the sine/cosecant cases; the six composite rules are annotated as not in MF27.

Four logarithmic formulas are marked available in MF27: ∫tan⁡x dx=ln⁡(sec⁡x)+C\int\tan x\,dx=\ln(\sec x)+C and ∫sec⁡x dx=ln⁡(sec⁡x+tan⁡x)+C\int\sec x\,dx=\ln(\sec x+\tan x)+C on ∣x∣<π/2|x|<\pi/2; ∫cot⁡x dx=ln⁡(sin⁡x)+C\int\cot x\,dx=\ln(\sin x)+C and ∫csc⁡x dx=−ln⁡(csc⁡x+cot⁡x)+C\int\csc x\,dx=-\ln(\csc x+\cot x)+C on 0<x<π0<x<\pi. The source remark extends these using absolute values inside the logarithms; work on connected intervals where the original trigonometric functions are defined.

For even powers of sine/cosine, lower the power with sin⁡2A=2sin⁡Acos⁡A\sin2A=2\sin A\cos A and cos⁡2A=2cos⁡2A−1=1−2sin⁡2A\cos2A=2\cos^2A-1=1-2\sin^2A, marked MF27. Rearranging gives cos⁡2A=(1+cos⁡2A)/2\cos^2A=(1+\cos2A)/2 and sin⁡2A=(1−cos⁡2A)/2\sin^2A=(1-\cos2A)/2. Examples 12–13 use these, applying double angle twice for the fourth power. The Teacher crosses out the false power-rule answer cos⁡3x/3+C\cos^3x/3+C to ∫cos⁡2x dx\int\cos^2x\,dx: its derivative is −cos⁡2xsin⁡x-\cos^2x\sin x, not cos⁡2x\cos^2x.

Other printed identities are cos⁡2A+sin⁡2A=1\cos^2A+\sin^2A=1, 1+tan⁡2A=sec⁡2A1+\tan^2A=\sec^2A and 1+cot⁡2A=csc⁡2A1+\cot^2A=\csc^2A. The Teacher adds a worked demonstration: ∫tan⁡2x dx=∫(sin⁡2x/cos⁡2x) dx=∫(sec⁡2x−1) dx=tan⁡x−x+C\int\tan^2x\,dx=\int(\sin^2x/\cos^2x)\,dx=\int(\sec^2x-1)\,dx=\tan x-x+C, with ddxtan⁡x=sec⁡2x\frac{d}{dx}\tan x=\sec^2x. Keep cos⁡x≠0\cos x\ne0 on its interval. This is a Teacher annotation, not a new printed Student exercise. Example 15 retains both supplied product-to-sum identities.

Try these examples

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