Question

Question #db777

Answer 1

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

Explanation:

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

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

Explanation:

`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`