Question #db777

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Answer 1 · 2017-11-15T20:40:01

$-1/5sin^5(x)+c$

we take

$intsec^5(x)/tan^6(x)dx$

we know

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

$tan^6(x)=sin^6(x)/cos^6(x)$

replacing identities

$int(1/(cos^5(x)))/(sin^6(x)/cos^6(x))dx$

multiplying and simplifying terms

$intcos(x)/sin^6(x)dx$

substitute $u=sin(x) -> du=cos(x)dx -> dx=1/cos(x)du$

$intcos(x)/u^6(1/cos(x)du)$

dividing

$int1/u^6du$

integrating

$int1/u^6du=-1/(5u^5)+c$

but $u=sinx$

$int1/u^6du=-1/(5u^5)+c=-1/(5sin^5(x))+c$

$intsec^5(x)/tan^6(x)dx=-1/(5sin^5(x))+c$
or

$intsec^5(x)/tan^6(x)dx=-1/5 csc^5(x)+c$

Answer 2 · 2017-11-15T21:12:42

$int (secx)^5/(tanx)^6*dx=-(cscx)^5/5+C$

$int (secx)^5/(tanx)^6*dx$

= $int (cotx)^6/(cosx)^5*dx$

= $int cotx*(cscx)^5*dx$

= $int (cscx)^4*(cscx*cotx*dx)$

After using $u=cscx$ and $du=-cscx*cotx*dx$ transforms,

$int -u^4*du$

= $-1/5*u^5+C$

= $-(cscx)^5/5+C$