Lennie Carroll

2020-10-25

Evaluate the following.
$\int \frac{{\mathrm{cos}}^{5}\left(3z\right)dz}{{\mathrm{sin}}^{2}\left(3z\right)}$

Mitchel Aguirre

We have to find the integral of: $\int \frac{{\mathrm{cos}}^{5}\left(3x\right)}{{\mathrm{sin}}^{2}\left(3x\right)}dx$
Substitute $u=3x⇒\frac{du}{dx}=3⇒\frac{du}{3}$, we obtain $\int \frac{{\mathrm{cos}}^{5}\left(3x\right)}{{\mathrm{sin}}^{2}\left(3x\right)}dx=\frac{1}{3}\int \frac{{\mathrm{cos}}^{5}\left(3u\right)}{{\mathrm{sin}}^{2}\left(3u\right)}du$
Now solving: $\int \frac{{\mathrm{cos}}^{5}\left(3u\right)}{{\mathrm{sin}}^{2}\left(3u\right)}du=\int \frac{{\mathrm{cos}}^{5}\left(U\right)}{{\mathrm{sin}}^{2}\left(U\right)}dU$
Preparing for substitution, we use: ${\mathrm{cos}}^{2}U=1-{\mathrm{sin}}^{2}U$, we obtain
$\int \frac{{\mathrm{cos}}^{5}\left(U\right)}{{\mathrm{sin}}^{2}\left(U\right)}dU$
$=\int \mathrm{cos}\left(U\right)\frac{{\left({\mathrm{sin}}^{2}\left(U\right)-1\right)}^{2}}{{\mathrm{sin}}^{2}\left(U\right)}dU$
Substitute $v=\mathrm{sin}\left(U\right)⇒\frac{dv}{dU}=\mathrm{cos}\left(u\right)⇒du=\frac{1}{\mathrm{cos}\left(u\right)}dv$, we obtain:
$=\int \frac{{\left({v}^{2}-1\right)}^{2}}{{v}^{2}}dv$
$=\int \left({v}^{2}+\frac{1}{{v}^{2}}-2\right)dv$
$=\int \left({v}^{2}\right)dv+\int \left(\frac{1}{{v}^{2}}\right)dv-2\int dv$
$=\frac{{v}^{3}}{3}-\frac{1}{v}+v$
Now, we can undo the substitution $v=\mathrm{sin}\left(u\right)$, we obtain:
$=\frac{{v}^{3}}{3}-\frac{1}{v}-2v$
$=\frac{{\mathrm{sin}}^{3}\left(u\right)}{3}-\frac{1}{\mathrm{sin}\left(u\right)}-2\mathrm{sin}\left(u\right)$
Plug in solved integrals,

Do you have a similar question?