# How do I evaluate the indefinite integral intcot^5(x)*sin^4(x)dx ?

Apr 23, 2018

${\cos}^{2} x + \frac{1}{4} {\sin}^{4} x + \ln \left\mid \sin \right\mid x + \text{c}$

#### Explanation:

${\cot}^{5} x {\sin}^{4} x = {\cos}^{5} \frac{x}{\sin} ^ 5 x {\sin}^{4} x = {\cos}^{5} \frac{x}{\sin} x = {\cos}^{4} x \cot x = {\left(1 - {\sin}^{2} x\right)}^{2} \cot x = \left(1 - 2 {\sin}^{2} x + {\sin}^{4} x\right) \cot = \cot x - 2 \sin x \cos x + {\sin}^{3} x \cos x$

So

$\int {\cot}^{5} x {\sin}^{4} x \mathrm{dx} = \int \cot x \mathrm{dx} - \int 2 \sin x \cos x \mathrm{dx} + \int {\sin}^{3} x \cos x \mathrm{dx}$

Now let $u = \sin x$ and $\mathrm{du} = \cos x \mathrm{dx}$ and $v = \cos x$ and $\mathrm{dv} = - \sin x \mathrm{dx}$

$\int \cot x \mathrm{dx} + \int - 2 \sin x \cos x \mathrm{dx} + \int {\sin}^{3} x \cos x \mathrm{dx} = \ln \left\mid \sin \right\mid x \int 2 v \mathrm{dv} + \int {u}^{3} \mathrm{du} = \ln \left\mid \sin \right\mid x + {v}^{2} + \frac{1}{4} {u}^{4} + \text{c"=cos^2x+1/4sin^4x+lnabssinx+"c}$