# How do you integrate int sin^5t/sqrtcost?

Jan 14, 2017

$\int {\sin}^{5} \frac{t}{\sqrt{\cos}} t \mathrm{dt} = - 2 {\left(\cos t\right)}^{\frac{1}{2}} + \frac{4}{5} {\left(\cos t\right)}^{\frac{5}{2}} - \frac{2}{9} {\left(\cos t\right)}^{\frac{9}{2}} + C$

#### Explanation:

Sine and cosine functions can frequently be "turned into" one another, especially with the identity ${\sin}^{2} t + {\cos}^{2} t = 1$. In this case, we will be using the form ${\sin}^{2} t = 1 - {\cos}^{2} t$ to take sine terms and turn them into predominantly cosine forms.

Here, we see that ${\sin}^{5} t = {\sin}^{4} t \sin t = {\left({\sin}^{2} t\right)}^{2} \sin t$. Using the identity, we see that ${\sin}^{5} t = {\left(1 - {\cos}^{2} t\right)}^{2} \sin t$.

Then,

$\int {\sin}^{5} \frac{t}{\sqrt{\cos}} t \mathrm{dt} = \int \frac{{\left(1 - {\cos}^{2} t\right)}^{2} \sin t}{\sqrt{\cos}} t \mathrm{dt}$

The whole point of doing this has been to rewrite the integrand into one trig function completely, except while leaving a single of the other function hanging around. This primes us for the substitution $u = \cos t$. This implies that $\mathrm{du} = - \sin t \mathrm{dt}$.

$= - \int \frac{{\left(1 - {\cos}^{2} t\right)}^{2} \left(- \sin t\right)}{\sqrt{\cos}} t \mathrm{dt} = - \int {\left(1 - {u}^{2}\right)}^{2} / \sqrt{u} \mathrm{du}$

At this point, expand the squared binomial then divide by $\sqrt{u}$.

$= - \int \frac{1 - 2 {u}^{2} + {u}^{4}}{u} ^ \left(\frac{1}{2}\right) \mathrm{du} = - \int \left({u}^{- \frac{1}{2}} - 2 {u}^{\frac{3}{2}} + {u}^{\frac{7}{2}}\right) \mathrm{du}$

Integrate term-by-term using $\int {u}^{n} \mathrm{du} = {u}^{n + 1} / \left(n + 1\right)$:

$= - \left({u}^{\frac{1}{2}} / \left(\frac{1}{2}\right) - 2 \left({u}^{\frac{5}{2}} / \left(\frac{5}{2}\right)\right) + {u}^{\frac{9}{2}} / \left(\frac{9}{2}\right)\right)$

$= - \left(2 {u}^{\frac{1}{2}} - \frac{4}{5} {u}^{\frac{5}{2}} + \frac{2}{9} {u}^{\frac{9}{2}}\right)$

From $u = \cos t$:

$= - 2 {\left(\cos t\right)}^{\frac{1}{2}} + \frac{4}{5} {\left(\cos t\right)}^{\frac{5}{2}} - \frac{2}{9} {\left(\cos t\right)}^{\frac{9}{2}} + C$