Trig Integration help! (1 Viewer)

[Finx]

1. Show that secxcosecx = (sec²x)/(tanx)

2. Find ∫sinx°dx

3. Use the substitution u = cosx to find ∫sin³x dx


These should be simple, but I'm stuck!
Thanks in advance, your help is very much appreciated =]

 

[lychnobity]

1. Show that secxcosecx = (sec²x)/(tanx)

2. Find ∫sinx°dx

3. Use the substitution u = cosx to find ∫sin³x dx


These should be simple, but I'm stuck!
Thanks in advance, your help is very much appreciated =]

1) SecxCosecx = secx/sinx
= secx/tanxcosx
=(1/cosx)/(tanxcosx)
=1/tanxcos²x
=sec²x/tanx

2) ∫sinx°dx = ∫sin(pi/180)x dx
= -(180/pi)cos(pi/180)x + c

3) u = cos x, du = -sinx dx

∫sin³x dx = -∫(1- u²)du
As sin³x = sinxsin²x = sinx(1 - cos²x) = -du(1 - u²)
= -u + u³/3 + C
= -cos x + cos³x/3 + C

 

[Finx]

Thanks for your help!

For 2, is that an accepted rule, or is there working to it?

For 3, what did you do here: ∫sin³x dx = -∫(1- u²du ?

 

[Drongoski]

Lychnobity:

I think Q2 needs more work. Argument in degrees not radians ??

 

[lychnobity]

Thanks for your help!

For 2, is that an accepted rule, or is there working to it?

For 3, what did you do here: ∫sin³x dx = -∫(1- u²du ?

Hold a minute, is that a degree sign? I assumed it was a typo, or else it would be:

2) ∫sinx°dx = -180Cosx/pi + c

Convert the x° into x(pi)/180

3) umm, I sort of subbed in the substitutions I got for u & du already. I edited the post

 

[GUSSSSSSSSSSSSS]

yes as drongoski said...q2 is being measured in degrees NOT in radians, so simply saying cosx + C wud not work:

since 1 degree = pi/180 radians
then u can convert it to: integral of [sin(pi/180)x dx]
which is then -(180/pi)cos(pi/180)x + c

 

[lychnobity]

Oh shit, sorry man. Will edit post.

 

[GUSSSSSSSSSSSSS]

LOL sorry i just realised you already corrected it


I TYPE TOO SLOW =[[[