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Answer 1 · 2017-11-15T20:40:01

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

we take

$\int \frac{{sec}^{5} (x)}{{tan}^{6} (x)} d x$ $intsec^5(x)/tan^6(x)dx$

we know

${sec}^{5} (x) = \frac{1}{{cos}^{5} (x)}$ $sec^5(x)=1/(cos^5(x)$

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

replacing identities

$\int \frac{\frac{1}{{cos}^{5} (x)}}{\frac{{sin}^{6} (x)}{{cos}^{6} (x)}} d x$ $int(1/(cos^5(x)))/(sin^6(x)/cos^6(x))dx$

multiplying and simplifying terms

$\int \frac{cos (x)}{{sin}^{6} (x)} d x$ $intcos(x)/sin^6(x)dx$

substitute $u = sin (x) \to d u = cos (x) d x \to d x = \frac{1}{cos (x)} d u$ $u=sin(x) -> du=cos(x)dx -> dx=1/cos(x)du$

$\int \frac{cos (x)}{u^{6}} (\frac{1}{cos (x)} d u)$ $intcos(x)/u^6(1/cos(x)du)$

dividing

$\int \frac{1}{u^{6}} d u$ $int1/u^6du$

integrating

$\int \frac{1}{u^{6}} d u = - \frac{1}{5 u^{5}} + c$ $int1/u^6du=-1/(5u^5)+c$

but $u = sin x$ $u=sinx$

$\int \frac{1}{u^{6}} d u = - \frac{1}{5 u^{5}} + c = - \frac{1}{5 {sin}^{5} (x)} + c$ $int1/u^6du=-1/(5u^5)+c=-1/(5sin^5(x))+c$

$\int \frac{{sec}^{5} (x)}{{tan}^{6} (x)} d x = - \frac{1}{5 {sin}^{5} (x)} + c$ $intsec^5(x)/tan^6(x)dx=-1/(5sin^5(x))+c$
or

$\int \frac{{sec}^{5} (x)}{{tan}^{6} (x)} d x = - \frac{1}{5} {csc}^{5} (x) + c$ $intsec^5(x)/tan^6(x)dx=-1/5 csc^5(x)+c$

Answer 2 · 2017-11-15T21:12:42

$\int \frac{{(sec x)}^{5}}{{(tan x)}^{6}} \cdot d x = - \frac{{(csc x)}^{5}}{5} + C$ $int (secx)^5/(tanx)^6*dx=-(cscx)^5/5+C$

$\int \frac{{(sec x)}^{5}}{{(tan x)}^{6}} \cdot d x$ $int (secx)^5/(tanx)^6*dx$

= $\int \frac{{(cot x)}^{6}}{{(cos x)}^{5}} \cdot d x$ $int (cotx)^6/(cosx)^5*dx$

= $\int cot x \cdot {(csc x)}^{5} \cdot d x$ $int cotx*(cscx)^5*dx$

= $\int {(csc x)}^{4} \cdot (csc x \cdot cot x \cdot d x)$ $int (cscx)^4*(cscx*cotx*dx)$

After using $u = csc x$ $u=cscx$ and $d u = - csc x \cdot cot x \cdot d x$ $du=-cscx*cotx*dx$ transforms,

$\int - u^{4} \cdot d u$ $int -u^4*du$

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

= $- \frac{{(csc x)}^{5}}{5} + C$ $-(cscx)^5/5+C$