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∫cos⁡(12r)cos⁡(15r) dr\int \cos(12r) \cos(15r) \, dr

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Full solution

Q. ∫cos⁡(12r)cos⁡(15r) dr\int \cos(12r) \cos(15r) \, dr
  1. Apply Product-to-Sum Identity: To solve the integral of the product of two cosines, we can use the product-to-sum identities. The identity states that cos⁡(A)cos⁡(B)=12[cos⁡(A−B)+cos⁡(A+B)]\cos(A)\cos(B) = \frac{1}{2}[\cos(A - B) + \cos(A + B)]. Let's apply this identity to our integral.
  2. Rewrite Integral: Using the identity, we rewrite the integral as follows:\newline∫cos⁡(12r)cos⁡(15r) dr=12∫[cos⁡(12r−15r)+cos⁡(12r+15r)] dr\int \cos(12r)\cos(15r) \, dr = \frac{1}{2} \int [\cos(12r - 15r) + \cos(12r + 15r)] \, dr\newline= 12∫[cos⁡(−3r)+cos⁡(27r)] dr\frac{1}{2} \int [\cos(-3r) + \cos(27r)] \, dr\newlineSince cosine is an even function, cos⁡(−3r)=cos⁡(3r)\cos(-3r) = \cos(3r). So, we can simplify further:\newline= 12∫[cos⁡(3r)+cos⁡(27r)] dr\frac{1}{2} \int [\cos(3r) + \cos(27r)] \, dr
  3. Integrate Each Term: Now we integrate each term separately: 12∫cos⁡(3r) dr+12∫cos⁡(27r) dr\frac{1}{2} \int \cos(3r) \, dr + \frac{1}{2} \int \cos(27r) \, dr
  4. Apply Integral Formula: The integral of cos⁡(kr)\cos(kr) with respect to rr is 1ksin⁡(kr)\frac{1}{k}\sin(kr). Applying this to both terms, we get: 12×13sin⁡(3r)+12×127sin⁡(27r)+C\frac{1}{2} \times \frac{1}{3}\sin(3r) + \frac{1}{2} \times \frac{1}{27}\sin(27r) + C where CC is the constant of integration.
  5. Simplify Coefficients: Simplify the coefficients: (16)sin⁡(3r)+(154)sin⁡(27r)+C(\frac{1}{6})\sin(3r) + (\frac{1}{54})\sin(27r) + C
  6. Combine Terms: Combine the terms to get the final answer: ∫cos⁡(12r)cos⁡(15r) dr=16sin⁡(3r)+154sin⁡(27r)+C\int \cos(12r)\cos(15r) \, dr = \frac{1}{6}\sin(3r) + \frac{1}{54}\sin(27r) + C

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