# Question db777

Nov 15, 2017

$- \frac{1}{5} {\sin}^{5} \left(x\right) + c$

#### Explanation:

we take

$\int {\sec}^{5} \frac{x}{\tan} ^ 6 \left(x\right) \mathrm{dx}$

we know

sec^5(x)=1/(cos^5(x)#

${\tan}^{6} \left(x\right) = {\sin}^{6} \frac{x}{\cos} ^ 6 \left(x\right)$

replacing identities

$\int \frac{\frac{1}{{\cos}^{5} \left(x\right)}}{{\sin}^{6} \frac{x}{\cos} ^ 6 \left(x\right)} \mathrm{dx}$

multiplying and simplifying terms

$\int \cos \frac{x}{\sin} ^ 6 \left(x\right) \mathrm{dx}$

substitute $u = \sin \left(x\right) \to \mathrm{du} = \cos \left(x\right) \mathrm{dx} \to \mathrm{dx} = \frac{1}{\cos} \left(x\right) \mathrm{du}$

$\int \cos \frac{x}{u} ^ 6 \left(\frac{1}{\cos} \left(x\right) \mathrm{du}\right)$

dividing

$\int \frac{1}{u} ^ 6 \mathrm{du}$

integrating

$\int \frac{1}{u} ^ 6 \mathrm{du} = - \frac{1}{5 {u}^{5}} + c$

but $u = \sin x$

$\int \frac{1}{u} ^ 6 \mathrm{du} = - \frac{1}{5 {u}^{5}} + c = - \frac{1}{5 {\sin}^{5} \left(x\right)} + c$

$\int {\sec}^{5} \frac{x}{\tan} ^ 6 \left(x\right) \mathrm{dx} = - \frac{1}{5 {\sin}^{5} \left(x\right)} + c$
or

$\int {\sec}^{5} \frac{x}{\tan} ^ 6 \left(x\right) \mathrm{dx} = - \frac{1}{5} {\csc}^{5} \left(x\right) + c$

Nov 15, 2017

$\int {\left(\sec x\right)}^{5} / {\left(\tan x\right)}^{6} \cdot \mathrm{dx} = - {\left(\csc x\right)}^{5} / 5 + C$

#### Explanation:

$\int {\left(\sec x\right)}^{5} / {\left(\tan x\right)}^{6} \cdot \mathrm{dx}$

=$\int {\left(\cot x\right)}^{6} / {\left(\cos x\right)}^{5} \cdot \mathrm{dx}$

=$\int \cot x \cdot {\left(\csc x\right)}^{5} \cdot \mathrm{dx}$

=$\int {\left(\csc x\right)}^{4} \cdot \left(\csc x \cdot \cot x \cdot \mathrm{dx}\right)$

After using $u = \csc x$ and $\mathrm{du} = - \csc x \cdot \cot x \cdot \mathrm{dx}$ transforms,

$\int - {u}^{4} \cdot \mathrm{du}$

=$- \frac{1}{5} \cdot {u}^{5} + C$

=$- {\left(\csc x\right)}^{5} / 5 + C$