Evaluate the following integrals. \int sin^{2}0 \cos^{5} 0d0

Question
Applications of integrals
Evaluate the following integrals.
$$\displaystyle\int{{\sin}^{{{2}}}{0}}{{\cos}^{{{5}}}{0}}{d}{0}$$

2021-02-17
Step 1
Consider the provided integral,
$$\displaystyle\int{{\sin}^{{{2}}}{0}}{{\cos}^{{{5}}}{0}}{d}{0}$$
Evaluate the following integrals.
Using the trigonometric identity,
$$\displaystyle{{\sin}^{{{2}}}{\left({0}\right)}}{{\cos}^{{{5}}}{\left({0}\right)}}{d}{0}=\int{\left({1}-{{\sin}^{{{2}}}{\left({0}\right)}}\right)}^{{{2}}}{\cos{{\left({0}\right)}}}{{\sin}^{{{2}}}{\left({0}\right)}}{d}{0}$$
Apply u-Substitution method,
$$\displaystyle{u}={\sin{{0}}}\Rightarrow{d}{u}={\cos{{0}}}{d}{0}$$
Step 2
Therefore,
$$\displaystyle\int{{\sin}^{{{2}}}{\left({0}\right)}}{{\cos}^{{{5}}}{\left({0}\right)}}{d}{0}=\int{u}^{{{2}}}{\left({1}-{u}^{{{2}}}\right)}^{{{2}}}{d}{u}$$
$$\displaystyle=\int{u}^{{{2}}}-{2}{u}^{{{4}}}+{u}^{{{6}}}{d}{u}$$
$$\displaystyle=\int{u}^{{{2}}}{d}{u}-\int{2}{u}^{{{4}}}{d}{u}+\int{u}^{{{6}}}{d}{u}$$
$$\displaystyle={\frac{{{u}^{{{3}}}}}{{{3}}}}-{\frac{{{2}{u}^{{{5}}}}}{{{5}}}}+{\frac{{{u}^{{{7}}}}}{{{7}}}}+{C}$$
Substitute back,
$$\displaystyle\int{{\sin}^{{{2}}}{\left({0}\right)}}{{\cos}^{{{5}}}{\left({0}\right)}}{d}{0}={\frac{{{{\sin}^{{{3}}}{0}}}}{{{3}}}}-{\frac{{{2}{{\sin}^{{{5}}}{0}}}}{{{5}}}}+{\frac{{{{\sin}^{{{7}}}{0}}}}{{{7}}}}+{C}$$
Hence.

Relevant Questions

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$$\displaystyle\int{{\sin}^{{2}}{0}}{{\cos}^{{5}}{0}}{d}{0}$$
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