Integration by substitution ((sin(x))/(1+cos^2(x)))dx

  • Thread starter sapiental
  • Start date
  • #1
118
0
evaluat the indefinite integral ((sin(x))/(1+cos^2(x)))dx

I let

u = 1 + cos^2(x)

then du = -sin^2(x)dx

I rewrite the integral to

- integral sqrt(du)/u

can I set it up like this? should I change u to something else?

I also tried it like this by rewriting the original equation to:

indefinite integral ((sin(x))/(1+cos(x)cos(x)))dx

u = cos(x)

du = -sin(x)dx

then

- integral (du)/(1+(u^2))

Also, can somebody give me directions on how to format equations in this message board to make my questions somewhat clearer.


Thanks alot!
 

Answers and Replies

  • #2
2,063
2
sapiental said:
evaluat the indefinite integral ((sin(x))/(1+cos^2(x)))dx

I let

u = 1 + cos^2(x)

then du = -sin^2(x)dx

I rewrite the integral to

- integral sqrt(du)/u
That's not right. The derivative of [tex]\cos^2{x}[/tex] is NOT [tex]-\sin^2{x}[/tex].

can I set it up like this?

I also tried it like this by rewriting the original equation to:

indefinite integral ((sin(x))/(1+cos(x)cos(x)))dx

u = cos(x)

du = -sin(x)dx

then

- integral (du)/(1+(u^2))
Yes you can. The final integral is pretty straightforward.

Also, can somebody give me directions on how to format equations in this message board to make my questions somewhat clearer.
Download the pdf docs here
https://www.physicsforums.com/showthread.php?t=8997
 
  • #3
10
0
neutrino said:
That's not right. The derivative of [tex]\cos^2{x}[/tex] is NOT [tex]-\sin^2{x}[/tex].



Yes you can. The final integral is pretty straightforward.



Download the pdf docs here
https://www.physicsforums.com/showthread.php?t=8997
neutrino's right - you can't differentiate [tex]\cos^2{x}[/tex] as [tex]-\sin^2{x}[/tex]. If it helps, think of [tex]\cos^2{x}[/tex] as [tex]cos{x} * cos{x}[/tex]. You can then use the chain rule.
 

Related Threads on Integration by substitution ((sin(x))/(1+cos^2(x)))dx

Replies
3
Views
2K
Replies
7
Views
3K
  • Last Post
Replies
10
Views
4K
Replies
6
Views
10K
  • Last Post
Replies
17
Views
42K
  • Last Post
Replies
5
Views
794
  • Last Post
Replies
1
Views
1K
Replies
7
Views
84K
Replies
3
Views
17K
Replies
1
Views
2K
Top