1. Dec 6, 2008

### avid7

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 )

Last edited: Dec 7, 2008
2. Dec 6, 2008

### mutton

x's are missing.

It is not correct. Use the chain rule carefully.

3. Dec 7, 2008

### Gib Z

Try a substitution rather than integrating by parts.

4. Dec 7, 2008

### 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..

5. Dec 24, 2008

### Gib Z

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

6. Dec 24, 2008

### Nick89

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

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

Quite a difference!

7. Dec 24, 2008

### FedEx

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