Integrate sin (1+ cos ^2 x) ? Please help!

  • Thread starter avid7
  • Start date
  • #1
2
0
First of all, hi I'm new here my name is crisanna. I stumbled upon this site across the web and realised this 's a great site!

Anyway , here 's my question. Does anyone know how to integrate sin (1 +cos ^2 x) ?

I tried the method integrate by parts but I got stucked. Below is my attempt :
u= 1+ cos^2 x

du/dx = 1/2 x + 1/2 sinxcosx + c ( TBH i dont even know if this part is correct :bugeye: )
 
Last edited:

Answers and Replies

  • #2
179
0
Does anyone know how to integrate sin (1 +cos ^2) ?

x's are missing.

u= 1+ cos^2 x

du/dx = 1/2 x + 1/2 sinxcosx + c ( TBH i dont even know if this part is correct :bugeye: )

It is not correct. Use the chain rule carefully.
 
  • #3
Gib Z
Homework Helper
3,346
6
Try a substitution rather than integrating by parts.
 
  • #4
2
0
Tried again and i got du/dx = -2sinxcosx using the chain rule. Was that correct??

then I was about to use the substitution method..
 
  • #5
Gib Z
Homework Helper
3,346
6
Hopefully that next substitution was u= cos x. =]
 
  • #6
555
0
Is it
[tex]\sin \left( 1 + \cos^2 x\right)[/tex]

or
[tex]\sin x \cdot \left(1 + \cos^2 x\right)[/tex]

Quite a difference!
 
  • #7
318
0
Substitute

[tex] 1 + cos^2x = u [/tex]

Now you will get a denominator of -sin(2x).

But we [tex] 1 + cos^2x = u [/tex] hence find x in terms of arccos something. Using the rules of inverse trigo find 2x in terms of arcsin something.

Now apply parts Twice
 

Related Threads on Integrate sin (1+ cos ^2 x) ? Please help!

  • Last Post
Replies
7
Views
2K
Replies
2
Views
660
  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
2
Views
1K
Replies
2
Views
2K
Replies
3
Views
2K
  • Last Post
Replies
8
Views
29K
  • Last Post
Replies
2
Views
932
  • Last Post
Replies
9
Views
143K
Top