Answered: "Evaluating966∫cos4(16x)dx"

rhedynogh0rp

Answered question

2022-04-05

Evaluating

\frac{96}{6} \int \cos^{4} (16 x) d x

alwadau8ndv

Beginner2022-04-06Added 9 answers

Since

\cos 2 u = 2 \cos^{2} u - 1

we have that

\cos^{4} u = \frac{\frac{\cos 4 u + 1}{2} + 1}{2} = \frac{\cos 4 u}{4} + \frac{3}{4}

This should get you started.

\cos 2 v = 2 \cos^{2} v - 1

therefore

\cos^{2} v = \frac{1 + \cos 2 v}{2}

Now squaring on both sides

\cos^{4} v = \frac{1 + 2 \cos 2 v + \cos^{2} 2 v}{4}

Now use

\cos^{2} 2 v = \frac{\cos 4 v + 1}{2}

\cos^{4} v = \frac{1 + 2 \cos 2 v + \frac{1 + \cos 4 v}{2}}{4}

So the correct formula is:

\cos^{4} v = \frac{3}{8} + \frac{1}{2} \cos (2 u) + \frac{1}{8} \cos (4 u)

Now all terms can be easily integrated.

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