Phoebe

2021-11-09

Use a table of integrals to evaluate the following indefinite integrals.

$\int \mathrm{sin}3x\mathrm{cos}2xdx$

pierretteA

Skilled2021-11-10Added 102 answers

Step 1

Given:

$\int \mathrm{sin}3x\mathrm{cos}2xdx$

Step 2

Now,

By using the following identity, we get,

$\mathrm{sin}\left(a\right)\mathrm{cos}\left(b\right)=\frac{1}{2}(\mathrm{sin}(a-b)+\mathrm{sin}(a+b))$

$\int \mathrm{sin}3x\mathrm{cos}2xdx=\int \frac{1}{2}[\mathrm{sin}(3x-2x)+\mathrm{sin}(3x+2x)]dx$

$=\frac{1}{2}\int [\mathrm{sin}\left(x\right)+\mathrm{sin}\left(5x\right)]dx$

$=\frac{1}{2}[-\mathrm{cos}x-\frac{\mathrm{cos}5x}{5}]+C$

Hence,

$\int \mathrm{sin}3x\mathrm{cos}2xdx=\frac{1}{2}[-\mathrm{cos}x-\frac{\mathrm{cos}5x}{5}]+C$

Given:

Step 2

Now,

By using the following identity, we get,

Hence,

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