Khaleesi Herbert

2020-12-28

Use the table of integrals at the back of the text to evaluate the integrals $\int 8\mathrm{sin}\left(4t\right)\mathrm{sin}\left(\frac{t}{2}\right)dt$

yunitsiL

Step 1
Let the given integral is,
$\int 8\mathrm{sin}\left(4t\right)\mathrm{sin}\left(\frac{t}{2}\right)dt$
By using the formula,
$\mathrm{sin}\left(a\right)\mathrm{sin}\left(b\right)=\frac{-\mathrm{cos}\left(a+b\right)+\mathrm{cos}\left(a-b\right)}{2}$
$\int 8\left(\frac{\mathrm{cos}\left(4t-\frac{t}{2}\right)-\mathrm{cos}\left(4t+\frac{t}{2}\right)}{2}\right)dt$
$⇒8\int \left(\frac{\mathrm{cos}\left(\frac{7t}{2}\right)-\mathrm{cos}\left(\frac{9t}{2}\right)}{2}\right)dt$
Step 2
By separating the integrals,
$⇒\frac{8}{2}\int \left(\mathrm{cos}\left(\frac{7t}{2}\right)\right)dt-\int \left(\mathrm{cos}\left(\frac{9t}{2}\right)\right)dt$
Simplifying this,
$⇒\int 8\mathrm{sin}\left(4t\right)\mathrm{sin}\left(\frac{t}{2}\right)dt=4\left[\frac{2}{7}\mathrm{sin}\left(\frac{7t}{2}\right)-\frac{2}{9}\mathrm{sin}\left(\frac{9t}{2}\right)\right]+C$

Jeffrey Jordon

Answer is given below (on video)

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