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# Evaluate the following. intcosbeta(1-cos2beta)^3dbeta

Trigonometry
ANSWERED
asked 2020-12-15
Evaluate the following.
$$\displaystyle\int{\cos{\beta}}{\left({1}-{\cos{{2}}}\beta\right)}^{{3}}{d}\beta$$

## Answers (1)

2020-12-16
Given that :
The integral is $$\displaystyle\int{\cos{\beta}}{\left({1}-{\cos{{2}}}\beta\right)}^{{3}}{d}\beta$$
By using a formula,
$$\displaystyle{{\sin}^{{2}}{x}}={\frac{{{1}-{\cos{{2}}}{x}}}{{{2}}}}$$
To evaluate the integral :
By using a above formula,
$$\displaystyle\int{\cos{\beta}}{\left({1}-{\cos{{2}}}\beta\right)}^{{3}}{d}\beta=\int{\cos{\beta}}\times{\frac{{{8}}}{{{8}}}}{\left({1}-{\cos{{2}}}\beta\right)}^{{3}}{d}\beta$$
$$\displaystyle=\int{\cos{\beta}}\times{8}\times{\frac{{{\left({1}-{\cos{{2}}}\beta\right)}^{{3}}}}{{{8}}}}{d}\beta$$
$$\displaystyle=\int{\cos{\beta}}\times{8}\times{\frac{{{\left({1}-{\cos{{2}}}\beta\right)}^{{3}}}}{{{2}^{{3}}}}}{d}\beta$$
$$\displaystyle=\int{\cos{\beta}}\times{8}\times{\left({\frac{{{\left({1}-{\cos{{2}}}\beta\right)}}}{{{2}}}}\right)}^{{3}}{d}\beta$$
$$\displaystyle={8}\int{\cos{\beta}}\times{\left({\frac{{{1}-{\cos{{2}}}\beta}}{{{2}}}}\right)}^{{3}}{d}\beta$$
$$\displaystyle={8}\int{\cos{\beta}}\times{\left({{\sin}^{{2}}\beta}\right)}^{{3}}{d}\beta$$
$$\displaystyle={8}\int{\cos{\beta}}{{\sin}^{{6}}\beta}{d}\beta$$
Substitute $$\displaystyle{u}={\sin{\beta}}$$, then $$\displaystyle{d}{u}={\cos{\beta}}{d}\beta$$.
Then,
$$\displaystyle={8}\int{\cos{\beta}}{{\sin}^{{6}}\beta}{d}\beta$$
$$\displaystyle={8}\int{u}^{{6}}{d}{u}$$
$$\displaystyle={8}{\left[{\frac{{{u}^{{{6}+{1}}}}}{{{6}+{1}}}}\right]}$$
$$\displaystyle={8}{\left[{\frac{{{u}^{{7}}}}{{{7}}}}\right]}$$
$$\displaystyle={\frac{{{8}{u}^{{7}}}}{{{7}}}}$$
Substitute back $$\displaystyle{u}={\sin{\beta}}$$
$$\displaystyle{\frac{{{8}{u}^{{7}}}}{{{7}}}}={\frac{{{8}{\left({\sin{\beta}}\right)}^{{7}}}}{{{7}}}}$$
$$\displaystyle\Rightarrow\int{\cos{\beta}}{\left({1}-{\cos{{2}}}\beta\right)}^{{3}}{d}\beta={\frac{{{8}{{\sin}^{{7}}\beta}}}{{{7}}}}$$
$$\displaystyle\Rightarrow\int{\cos{\beta}}{\left({1}-{\cos{{2}}}\beta\right)}^{{3}}{d}\beta={\frac{{{8}{{\sin}^{{7}}\beta}}}{{{7}}}}+{C}$$, where C is the integration constant.
Therefore,
$$\displaystyle\Rightarrow\int{\cos{\beta}}{\left({1}-{\cos{{2}}}\beta\right)}^{{3}}{d}\beta={\frac{{{8}{{\sin}^{{7}}\beta}}}{{{7}}}}+{C}$$, where C is the integration constant.

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