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Worked solution

Use the supplied product-to-sum identities

Given 2cos⁡Acos⁡B=cos⁡(A+B)+cos⁡(A−B)2\cos A\cos B=\cos(A+B)+\cos(A-B) and 2cos⁡Asin⁡B=sin⁡(A+B)−sin⁡(A−B)2\cos A\sin B=\sin(A+B)-\sin(A-B), find (a) ∫cos⁡3xcos⁡x dx\int\cos3x\cos x\,dx; (b) ∫cos⁡4xsin⁡x dx\int\cos4x\sin x\,dx.

CJC · Lecture · Example 15

(a)

cos⁡3xcos⁡x=(cos⁡4x+cos⁡2x)/2\cos3x\cos x=(\cos4x+\cos2x)/2, so the integral is 12(sin⁡4x/4+sin⁡2x/2)+C=18sin⁡4x+14sin⁡2x+C\frac12(\sin4x/4+\sin2x/2)+C=\frac18\sin4x+\frac14\sin2x+C.

How did this part go?

(b)

cos⁡4xsin⁡x=(sin⁡5x−sin⁡3x)/2\cos4x\sin x=(\sin5x-\sin3x)/2, so the integral is 12[−cos⁡5x/5−(−cos⁡3x/3)]+C=−110cos⁡5x+16cos⁡3x+C\frac12[-\cos5x/5-(-\cos3x/3)]+C=-\frac1{10}\cos5x+\frac16\cos3x+C.

How did this part go?

Common mistakes

In (b), subtracting the negative cosine antiderivative makes the cos 3x term positive.

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