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Mathematics · Ch 7 — Integrals

Integration Using Trigonometric Identities

7.3.2

Integration Using Trigonometric Identities

7.3.2 Integration Using Trigonometric Identities

When the integrand is a power of a trigonometric function, or a product of sines and cosines, the standard integration formulas usually cannot be applied directly. The strategy is to first rewrite the integrand — using a suitable trigonometric identity — as a sum of terms that each match a known integral (typically sin⁡kx\sin kx or cos⁡kx\cos kx), and then integrate term by term.

Reducing squares (double-angle identities)

sin⁡2x=1−cos⁡2x2,cos⁡2x=1+cos⁡2x2\sin^2 x = \frac{1 - \cos 2x}{2}, \qquad \cos^2 x = \frac{1 + \cos 2x}{2}

For example, ∫sin⁡2x dx=∫1−cos⁡2x2 dx=x2−sin⁡2x4+C.\displaystyle\int \sin^2 x\,dx = \int \frac{1-\cos 2x}{2}\,dx = \frac{x}{2} - \frac{\sin 2x}{4} + C.

Reducing cubes (triple-angle identities)

sin⁡3x=3sin⁡x−sin⁡3x4,cos⁡3x=3cos⁡x+cos⁡3x4\sin^3 x = \frac{3\sin x - \sin 3x}{4}, \qquad \cos^3 x = \frac{3\cos x + \cos 3x}{4}

Converting products into sums (product-to-sum identities)

2sin⁡Acos⁡B=sin⁡(A+B)+sin⁡(A−B)2\sin A\cos B = \sin(A+B) + \sin(A-B)

2cos⁡Acos⁡B=cos⁡(A+B)+cos⁡(A−B)2\cos A\cos B = \cos(A+B) + \cos(A-B)

2sin⁡Asin⁡B=cos⁡(A−B)−cos⁡(A+B)2\sin A\sin B = \cos(A-B) - \cos(A+B)

These let an integral such as ∫sin⁡3xcos⁡2x dx\int \sin 3x\cos 2x\,dx be split into 12∫[sin⁡5x+sin⁡x] dx\tfrac{1}{2}\int[\sin 5x + \sin x]\,dx, which integrates immediately.

Other useful reductions …

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← Previous · 7.3.1Q39 — ∫ (dx)/(sin² x cos² x) equals (A) tan x + cot x + C (B) tan x - cot x …Next · 7.3.2 →Example 7 — Find (i) ∫ cos² x dx (ii) ∫ sin 2x cos 3x dx (iii) ∫ sin³ x dx