How to Integrate sin(1 + cos^2 x)?

[avid7]

First of all, hi I'm new here my name is crisanna. I stumbled upon this site across the web and realized 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 don't even know if this part is correct )

 

[mutton]

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 don't even know if this part is correct )

It is not correct. Use the chain rule carefully.

 

[Gib Z]

Try a substitution rather than integrating by parts.

 

[avid7]

Tried again and i got du/dx = -2sinxcosx using the chain rule. Was that correct??

then I was about to use the substitution method..

 

[Gib Z]

Hopefully that next substitution was u= cos x. =]

 

[Nick89]

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!

 

[FedEx]

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