Integral of Trig function

**superman69** · October 21st 2009, 07:51 PM

2∫cos²x dx

I have trouble solving I tried using u substitution many times and trying to use the identities but I can not some how figure it out. Can someone show me?

**redsoxfan325** · October 21st 2009, 08:21 PM

Originally Posted by superman69

2∫cos²x dx

I have trouble solving I tried using u substitution many times and trying to use the identities but I can not some how figure it out. Can someone show me?

You don't want u-substitution. We have the trig identity: $\cos(2x)=2\cos^2(x)-1$ . Therefore,

$\int2\cos^2(x)\,dx=\int(\cos(2x)+1)\,dx=\frac{1}{2 }\sin(2x)+x+C$

**superman69** · October 21st 2009, 08:28 PM

I didn't get where the 2 went from 2∫cos²x

**redsoxfan325** · October 21st 2009, 08:34 PM

Originally Posted by superman69

I didn't get where the 2 went from 2∫cos²x

$2\int\cos^2x\,dx=\int2\cos^2x\,dx$

It's one of the rules of integrals.

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