Phoebe

2021-11-09

Use a table of integrals to evaluate the following indefinite integrals.
$\int \mathrm{sin}3x\mathrm{cos}2xdx$

pierretteA

Step 1
Given:
$\int \mathrm{sin}3x\mathrm{cos}2xdx$
Step 2
Now,
By using the following identity, we get,
$\mathrm{sin}\left(a\right)\mathrm{cos}\left(b\right)=\frac{1}{2}\left(\mathrm{sin}\left(a-b\right)+\mathrm{sin}\left(a+b\right)\right)$
$\int \mathrm{sin}3x\mathrm{cos}2xdx=\int \frac{1}{2}\left[\mathrm{sin}\left(3x-2x\right)+\mathrm{sin}\left(3x+2x\right)\right]dx$
$=\frac{1}{2}\int \left[\mathrm{sin}\left(x\right)+\mathrm{sin}\left(5x\right)\right]dx$
$=\frac{1}{2}\left[-\mathrm{cos}x-\frac{\mathrm{cos}5x}{5}\right]+C$
Hence,
$\int \mathrm{sin}3x\mathrm{cos}2xdx=\frac{1}{2}\left[-\mathrm{cos}x-\frac{\mathrm{cos}5x}{5}\right]+C$

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