Hardish integration (1 Viewer)

[lilkatie]

Ok how do i integrate sin (to the power of 6) from pi/2 to 0
I have no idea i can integrate sin squared, but not to the power of 6
Thanks

 

[googleplex]

id say just keep reusing the double angle theorem like u did with sin^2 (ie a cubic expansion and then reapplying the double angle theorem and then dealing with the cubic one with a substitution)

other than that I can't see anything else except like using one of the even function four unit transformer thingees

yeh... maybe theres a better way *shrug*

 

[lilkatie]

Thanx guys

 

[nike55]

S Sin^6x dx
= S Sin^6xcosx / cosx dx ( * by cosx/cosx)

let u = sinx, du = cosx dx

= S u^6 / cosx du but u = sinx, therefoce sqr(1-u^2) = cosx
= S u^6 / sqr(1-u^2) du - which is simple to integrate

 

[lilkatie]

sweet thanx for the help the answer came out to be 5pi/32...I hope