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 )

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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 )

It is not correct. Use the chain rule carefully.

Gib Z
Homework Helper
Try a substitution rather than integrating by parts.

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

Is it
$$\sin \left( 1 + \cos^2 x\right)$$

or
$$\sin x \cdot \left(1 + \cos^2 x\right)$$

Quite a difference!

Substitute

$$1 + cos^2x = u$$

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

But we $$1 + cos^2x = u$$ 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