 midtlinjeg

2021-10-06

Evaluate the following integrals.
$\int {\mathrm{cos}}^{4}2\theta d\theta$ Consider the provided integral,
$\int {\mathrm{cos}}^{4}2\theta d\theta$
Evaluate the provided integral.
Apply the substitution method,
Let $u=2\theta ⇒du=2d\theta$
We get,
$\int {\mathrm{cos}}^{4}2\theta d\theta =\frac{1}{2}\int {\mathrm{cos}}^{4}udu$
We can write as,
$\int {\mathrm{cos}}^{4}2\theta d\theta =\frac{1}{2}\int {\mathrm{cos}}^{3}u\mathrm{cos}udu$
$\int {\mathrm{cos}}^{4}2\theta d\theta =\frac{1}{2}\int {\mathrm{cos}}^{3}u\mathrm{cos}udu$
Now, apply the integration by parts.
Let $u={\mathrm{cos}}^{3}u$ and $v=\mathrm{cos}u$
$\int {\mathrm{cos}}^{4}2\theta d\theta =\frac{1}{2}\int {\mathrm{cos}}^{3}u\mathrm{cos}udu$
$=\frac{1}{2}\left({\mathrm{cos}}^{3}u\mathrm{sin}u-\int -3{\mathrm{cos}}^{2}u\mathrm{sin}udu\right)+C$
$=\frac{1}{2}\left({\mathrm{cos}}^{3}u\mathrm{sin}u+3\int {\mathrm{cos}}^{2}u\mathrm{sin}udu\right)+C$
$=\frac{1}{2}\left({\mathrm{cos}}^{3}u\mathrm{sin}u+3\int \frac{1-\mathrm{cos}4u}{8}du\right)+C$
Simplifying further,
$\int {\mathrm{cos}}^{4}2\theta d\theta =\frac{1}{2}\left({\mathrm{cos}}^{3}u\mathrm{sin}u+\frac{3}{8}\int 1-\mathrm{cos}4udu\right)+C$
$=\frac{1}{2}\left({\mathrm{cos}}^{3}u\mathrm{sin}u+\frac{3}{8}\left(u-\frac{1}{4}\mathrm{sin}4u\right)\right)+C$
Substitute back $u=2\theta$ we get,
$\int {\mathrm{cos}}^{42}\theta d\theta =\frac{1}{2}\left({\mathrm{cos}}^{32}\theta \mathrm{sin}2\theta +\frac{3}{8}\left(2\theta -\frac{1}{4}\mathrm{sin}8\theta \right)\right)+C$

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