Question

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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