Trigonometric functions and integrals

  • Thread starter Any Help
  • Start date
  • #1
79
2

Homework Statement


I'm searching for the integral that gives arcosu

Homework Equations


as we know : ∫u'/[1-u^2]^0.5 dx = arcsinu
derivative of arccosu = -u'/[1-u^2]^0.5 + C
derivative of arcsinu= u'/[1-u^2]^0.5

The Attempt at a Solution


when I type the -u'/[1-u^2]^0.5 on the online integral calculator it always gives me -arcsinu +C
Why it doesn't gives arccosu
is the arccosu=-arcsinu?!
and what is the integral that gives arcosu?
 

Answers and Replies

  • #2
cnh1995
Homework Helper
Gold Member
3,444
1,144

Homework Statement


I'm searching for the integral that gives arcosu

Homework Equations


as we know : ∫u'/[1-u^2]^0.5 dx = arcsinu
derivative of arccosu = -u'/[1-u^2]^0.5 + C
derivative of arcsinu= u'/[1-u^2]^0.5

The Attempt at a Solution


when I type the -u'/[1-u^2]^0.5 on the online integral calculator it always gives me -arcsinu +C
Why it doesn't gives arccosu
is the arccosu=-arcsinu?!
and what is the integral that gives arcosu?
cos-1u= -sin-1u+90°.
Sine and cosine functions have a phase difference of 90°.
 
  • #3
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722
cos-1u= -sin-1u+90°.
Sine and cosine functions have a phase difference of 90°.

No, they differ by ##\pi/2## radians. It is a great mistake to mix trigonometric integrals with degree measurements; the calculus of trig functions is in radian units only!
 
  • #4
35,428
7,287
@Any Help, please post questions about calculus in the Calculus & Beyond section, not the Precalc section. I have moved your thread.
 
  • #5
79
2
No, they differ by π/2π/2\pi/2 radians. It is a great mistake to mix trigonometric integrals with degree measurements; the calculus of trig functions is in radian units only!
then what's the integral that gives arccos u?
 
  • #6
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,722
then what's the integral that gives arccos u?

There are two antiderivatives of ##1/\sqrt{1-x^2}##; they are ##\arcsin(x)+C## and ##-\arccos(x)+C##. Typically, sources like tables of integrals, or computer algebra packages, give only ONE of the two, and those seem to always be the ##\arcsin(x)## version. When you do it manually you are certainly allowed to give ##-\arccos(x) + C## as an answer; it is absolutely correct, because it is the same as ##\arcsin(x) + K## for some constant ##K## that is different from the constant ##C##.
 
  • Like
Likes Any Help and cnh1995

Related Threads on Trigonometric functions and integrals

  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
2
Views
1K
Replies
2
Views
1K
Replies
1
Views
2K
Replies
11
Views
770
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
1
Views
787
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
8
Views
2K
Top