[SOLVED] Trigonometric Integral

**Mike9182** · January 26th 2010, 02:30 PM

How is this integrated?

$\int{\cos^2{\Theta}}d\Theta$

**General** · January 26th 2010, 02:41 PM

Originally Posted by Mike9182

How is this integrated?

$\int{\cos^2{\Theta}}d\Theta$

Racall the identity: $cos^2(\theta)=\frac{1}{2}(cos(2\theta)+1)$ .
Hence:
$\int cos^2(\theta) d\theta = \frac{1}{2} \int cos(2\theta) d\theta + \frac{1}{2} \int d\theta$ .

It should be easy now.

**Mike9182** · January 26th 2010, 03:17 PM

How do you integrate $\frac{1}{2} \int(\cos{2\Theta})d\Theta$ ?

**pickslides** · January 26th 2010, 03:27 PM

Originally Posted by Mike9182

How do you integrate $\frac{1}{2} \int(\cos{2\Theta})d\Theta$ ?

$<br /> \int(\cos{k\Theta})d\Theta, k\in \mathbb{R} = \frac{1}{k} \sin{k\Theta}+C<br />$

**General** · January 26th 2010, 03:36 PM

Originally Posted by Mike9182

How do you integrate $\frac{1}{2} \int(\cos{2\Theta})d\Theta$ ?

there is two ways:
1- Substitute $u=2\theta$ .
2- Apply this formula:
$b\int cos(a\theta) d\theta = \frac{b}{a} sin(a\theta) + C$ for $a \neq 0$
in your question: $a=2$ and $b=\frac{1}{2}$ .

Originally Posted by pickslides

$<br /> \int(\cos{k\Theta})d\Theta, k\in \mathbb{R} = \frac{1}{k} \sin{k\Theta}+C<br />$

$k \neq 0$

**Mike9182** · January 26th 2010, 03:49 PM

Thanks, I understand now.

**Archie Meade** · January 26th 2010, 04:24 PM

Originally Posted by Mike9182

How is this integrated?

$\int{\cos^2{\Theta}}d\Theta$

Another way...

$\int{Cos\theta\ Cos\theta}d\theta=\int{u}dv=uv-\int{v}du$

$u=Cos\theta,\ \frac{du}{d\theta}=-Sin\theta,\ du=-Sin\theta\ d\theta$

$dv=Cos\theta\ d\theta,\ v=\int{Cos\theta}d\theta=Sin\theta$

$uv-\int{v}du=Cos\theta Sin\theta+\int{Sin^2\theta}d\theta$

$\Rightarrow\ I=Cos\theta Sin\theta+\int{(1-Cos^2\theta)}d\theta=Cos\theta Sin\theta+\int{(1)}d\theta-I$

$2I=Cos\theta Sin\theta+\theta\ \Rightarrow\ I=\frac{1}{2}Cos\theta Sin\theta+\frac{\theta}{2}+C=\frac{1}{4}Sin2\theta +\frac{\theta}{2}+C$

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