1. ## [SOLVED] Trigonometric Integral

How is this integrated?

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

2. Originally Posted by Mike9182
How is this integrated?

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

It should be easy now.

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

4. Originally Posted by Mike9182
How do you integrate $\displaystyle \frac{1}{2} \int(\cos{2\Theta})d\Theta$?

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

5. Originally Posted by Mike9182
How do you integrate $\displaystyle \frac{1}{2} \int(\cos{2\Theta})d\Theta$?
there is two ways:
1- Substitute $\displaystyle u=2\theta$.
2- Apply this formula:
$\displaystyle b\int cos(a\theta) d\theta = \frac{b}{a} sin(a\theta) + C$ for $\displaystyle a \neq 0$
in your question: $\displaystyle a=2$ and $\displaystyle b=\frac{1}{2}$.

Originally Posted by pickslides
$\displaystyle \int(\cos{k\Theta})d\Theta, k\in \mathbb{R} = \frac{1}{k} \sin{k\Theta}+C$
$\displaystyle k \neq 0$

6. Thanks, I understand now.

7. Originally Posted by Mike9182
How is this integrated?

$\displaystyle \int{\cos^2{\Theta}}d\Theta$
Another way...

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

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

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

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

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

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