Answered: "Evaluate the following. ∫cos⁡β(1−cos⁡2β)3dβ"

Daniaal Sanchez

Answered question

2020-12-15

Evaluate the following.

\int \cos β {(1 - \cos 2 β)}^{3} d β

nitruraviX

Skilled2020-12-16Added 101 answers

Given that :
The integral is

\int \cos β {(1 - \cos 2 β)}^{3} d β

By using a formula,

\sin^{2} x = \frac{1 - \cos 2 x}{2}

To evaluate the integral :
By using a above formula,

\int \cos β {(1 - \cos 2 β)}^{3} d β = \int \cos β \times \frac{8}{8} {(1 - \cos 2 β)}^{3} d β

= \int \cos β \times 8 \times \frac{{(1 - \cos 2 β)}^{3}}{8} d β

= \int \cos β \times 8 \times \frac{{(1 - \cos 2 β)}^{3}}{2^{3}} d β

= \int \cos β \times 8 \times {(\frac{(1 - \cos 2 β)}{2})}^{3} d β

= 8 \int \cos β \times {(\frac{1 - \cos 2 β}{2})}^{3} d β

= 8 \int \cos β \times {(\sin^{2} β)}^{3} d β

= 8 \int \cos β \sin^{6} β d β

Substitute

u = \sin β

, then

d u = \cos β d β

.
Then,

= 8 \int \cos β \sin^{6} β d β

= 8 \int u^{6} d u

= 8 [\frac{u^{6 + 1}}{6 + 1}]

= 8 [\frac{u^{7}}{7}]

= \frac{8 u^{7}}{7}

Substitute back

u = \sin β

\frac{8 u^{7}}{7} = \frac{8 {(\sin β)}^{7}}{7}

\Rightarrow \int \cos β {(1 - \cos 2 β)}^{3} d β = \frac{8 \sin^{7} β}{7}

\Rightarrow \int \cos β {(1 - \cos 2 β)}^{3} d β = \frac{8 \sin^{7} β}{7} + C

, where C is the integration constant.
Therefore,

\Rightarrow \int \cos β {(1 - \cos 2 β)}^{3} d β = \frac{8 \sin^{7} β}{7} + C

, where C is the integration constant.

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