Question #db777

2 Answers
Nov 15, 2017

-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

Nov 15, 2017

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