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Math III

Trigonometric integrals

The strategy for trig integrals: lower the degree and turn products into sums.

Basic, Standard: Grade 12 · Term 2 / Advanced: Grade 12 · Exam prep

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Basic formulas

∫sin⁡ax dx=−1acos⁡ax+C,∫cos⁡ax dx=1asin⁡ax+C\int \sin ax\,dx = -\frac{1}{a}\cos ax + C,\qquad \int \cos ax\,dx = \frac{1}{a}\sin ax + C
∫dxcos⁡2ax=1atan⁡ax+C\int \frac{dx}{\cos^2 ax} = \frac{1}{a}\tan ax + C

Note the minus sign: the integral of sin⁡\sin is −cos⁡-\cos. It is easy to mix up with the derivative (cos⁡x)′=−sin⁡x(\cos x)' = -\sin x.

Lower the degree of squares with half-angle identities

Half-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}
∫sin⁡2x dx=∫1−cos⁡2x2 dx=x2−14sin⁡2x+C\int \sin^2 x\,dx = \int \frac{1 - \cos 2x}{2}\,dx = \frac{x}{2} - \frac{1}{4}\sin 2x + C

Combine sin⁡xcos⁡x\sin x\cos x with the double-angle identity sin⁡xcos⁡x=12sin⁡2x\sin x\cos x = \frac{1}{2}\sin 2x. For sin⁡4x\sin^4 x or cos⁡4x\cos^4 x, use the half-angle identities twice.

Turn products into sums

Product-to-sum formulas
sin⁡αcos⁡β=12{sin⁡(α+β)+sin⁡(α−β)}\sin\alpha\cos\beta = \frac{1}{2}\{\sin(\alpha+\beta) + \sin(\alpha-\beta)\}
cos⁡αcos⁡β=12{cos⁡(α+β)+cos⁡(α−β)}\cos\alpha\cos\beta = \frac{1}{2}\{\cos(\alpha+\beta) + \cos(\alpha-\beta)\}
sin⁡αsin⁡β=−12{cos⁡(α+β)−cos⁡(α−β)}\sin\alpha\sin\beta = -\frac{1}{2}\{\cos(\alpha+\beta) - \cos(\alpha-\beta)\}

Example: ∫sin⁡3xcos⁡x dx=12∫(sin⁡4x+sin⁡2x) dx=−18cos⁡4x−14cos⁡2x+C\displaystyle\int \sin 3x\cos x\,dx = \frac{1}{2}\int (\sin 4x + \sin 2x)\,dx = -\frac{1}{8}\cos 4x - \frac{1}{4}\cos 2x + C

Odd powers: substitute

For an odd power like sin⁡3x\sin^3 x, keep one factor, rewrite sin⁡2x=1−cos⁡2x\sin^2 x = 1 - \cos^2 x and let t=cos⁡xt = \cos x.

∫sin⁡3x dx=∫(1−cos⁡2x)sin⁡x dx=−cos⁡x+13cos⁡3x+C\int \sin^3 x\,dx = \int (1 - \cos^2 x)\sin x\,dx = -\cos x + \frac{1}{3}\cos^3 x + C

For 1cos⁡x\dfrac{1}{\cos x}, multiply top and bottom by cos⁡x\cos x to get cos⁡x1−sin⁡2x\dfrac{\cos x}{1 - \sin^2 x}, let t=sin⁡xt = \sin x and use partial fractions.

Orthogonality of sine and cosine

For positive integers m,nm, n, the following holds. It is the foundation of Fourier series.

∫02πsin⁡mxsin⁡nx dx=∫02πcos⁡mxcos⁡nx dx={0(m≠n)π(m=n)\int_0^{2\pi} \sin mx\sin nx\,dx = \int_0^{2\pi} \cos mx\cos nx\,dx = \begin{cases} 0 & (m \ne n) \\ \pi & (m = n) \end{cases}
∫02πsin⁡mxcos⁡nx dx=0\int_0^{2\pi} \sin mx\cos nx\,dx = 0