# Trigonometric integral Evaluate the following integrals. int sin^2 0cos^5 0d0

Trigonometric integral Evaluate the following integrals.
$\int {\mathrm{sin}}^{2}0{\mathrm{cos}}^{5}0d0$
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Step 1
Consider the provided integral,
$\int {\mathrm{sin}}^{2}0{\mathrm{cos}}^{5}0d0$
Evaluate the following integrals.
Using the trigonometric identity,
$\int {\mathrm{sin}}^{2}\left(0\right){\mathrm{cos}}^{5}\left(0\right)d0=\int {\left(1-{\mathrm{sin}}^{2}\left(0\right)\right)}^{2}\mathrm{cos}\left(0\right)d0$
Apply u-Substitution method,
$u=\mathrm{sin}0⇒du=\mathrm{cos}0d0$
Step 2
Therefore,
$\int {\mathrm{sin}}^{2}\left(0\right){\mathrm{cos}}^{5}\left(0\right)d0=\int {u}^{2}{\left(1-{u}^{2}\right)}^{2}du$
$=\int {u}^{2}-2{u}^{4}+{u}^{6}du$
$=\int {u}^{2}du-\int 2{u}^{4}du+\int {u}^{6}du$
$=\frac{{u}^{3}}{3}-\frac{2{u}^{5}}{5}+\frac{{u}^{7}}{7}+C$
Substitute back,
$\int {\mathrm{sin}}^{2}\left(0\right){\mathrm{cos}}^{5}\left(0\right)d0=\frac{{\mathrm{sin}}^{3}0}{3}-\frac{2{\mathrm{sin}}^{5}0}{5}+\frac{{\mathrm{sin}}^{7}0}{7}+C$
Hence.