Use a table of integrals to evaluate the following indefinite integrals. ∫sin⁡3xcos⁡2xdx

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Use a table of integrals to evaluate the following indefinite integrals.  ∫sin⁡3xcos⁡2xdx

pierretteA · Accepted Answer

Step 1  Given:  ∫sin⁡3xcos⁡2xdx  Step 2  Now,  By using the following identity, we get,  sin⁡(a)cos⁡(b)=12(sin⁡(a−b)+sin⁡(a+b))  ∫sin⁡3xcos⁡2xdx=∫12[sin⁡(3x−2x)+sin⁡(3x+2x)]dx  =12∫[sin⁡(x)+sin⁡(5x)]dx  =12[−cos⁡x−cos⁡5x5]+C  Hence,  ∫sin⁡3xcos⁡2xdx=12[−cos⁡x−cos⁡5x5]+C

Use a table of integrals to evaluate the following indefinite integrals. ∫sin⁡3xcos⁡2xdx

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