Porisma

Worked solution

Product-to-sum before integration

Given 2cos⁡Asin⁡B=sin⁡(A+B)−sin⁡(A−B)2\cos A\sin B=\sin(A+B)-\sin(A-B), find ∫cos⁡4θsin⁡2θ dθ\displaystyle\int\cos4\theta\sin2\theta\,d\theta.

CJC · Discussion · Question 12

Answer

−112cos⁡6θ+14cos⁡2θ+C-\dfrac1{12}\cos6\theta+\dfrac14\cos2\theta+C.

Apply the given identity

Set A=4θ and B=2θ. The product is 12(sin⁡6θ−sin⁡2θ)\dfrac12(\sin6\theta-\sin2\theta). Thus the integral is 12[−cos⁡6θ/6+cos⁡2θ/2]+C=−cos⁡6θ/12+cos⁡2θ/4+C\dfrac12[-\cos6\theta/6+\cos2\theta/2]+C=-\cos6\theta/12+\cos2\theta/4+C.

How did this part go?

Common mistakes

Keep the minus sign between the two sine terms and divide by each inner frequency.

Read the hint