1. ## [SOLVED] trig intergrals

So, this is also probably one of those easy problems, I just don't have a good example to work this out by.
$
\int\cos^2x$

where b= $\pi/2$ a= 0

answer= $\pi/4$

I know I'm suppose to use the half angle identity, i just must be doing it wrong because that's not the answer I'm getting.

Thanks for all your help ^.^

2. Can you show your work to see what it is that you're doing wrong?

$\int_{0}^{\frac{\pi}{2}} \cos^2 x \: dx = \int_{0}^{\frac{\pi}{2}} \left[\frac{1}{2} (1 + \cos 2x)\right]dx = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \left[1 + \cos 2x\right]dx =$ ${\color{white}.} \: \frac{1}{2} \left(x + \frac{1}{2}\sin 2x\right) \bigg|_{0}^{\frac{\pi}{2}} = \hdots$

3. $cos^2x = \frac{1 + cos2x}{2}$

$\frac{1}{2}\int{dx} +\frac{1}{2}\int{cos2x}{dx} = \frac{x}{2} +\frac{sin2x}{4}$

Hope this helps.

Oops was to slow lol, just wondering how do you place upper and lower bounds in latex?

4. Originally Posted by 11rdc11
$cos^2x = \frac{1 + cos2x}{2}$

$\frac{1}{2}\int{dx} +\frac{1}{2}\int{cos2x}{dx} = \frac{x}{2} +\frac{sin2x}{4}$

Hope this helps.

Oops was to slow lol, just wondering how do you place upper and lower bounds in latex?
Do:

Code:
$$\int_a^b f(x)\,dx$$
To get $\int_a^b f(x)\,dx$

note if we have limits like $-\ln2$ to $\ln2$, then you need to put these within {} such as:

Code:
$$\int_{-\ln(2)}^{\ln2} e^x\,dx$$
will output $\int_{-\ln(2)}^{\ln2} e^x\,dx$

--Chris

5. $\int_{0}^{4} x \: dx$

is produced by: $$\int_{0}^{4} x \: dx$$

_{0} : produces the lower bound
^{4} : produces the upper bound.

This can applied to other symbols as well:
$\sum_{k = 0}^{4} k = 10$ is produced by $$\sum_{k = 0}^{4} k = 10$$

$\frac{x}{2}\bigg|_{0}^{4}$

etc etc.

6. Originally Posted by o_O
Can you show your work to see what it is that you're doing wrong?

$\int_{0}^{\frac{\pi}{2}} \cos^2 x \: dx = \int_{0}^{\frac{\pi}{2}} \left[\frac{1}{2} (1 + \cos 2x)\right]dx = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \left[1 + \cos 2x\right]dx =$ ${\color{white}.} \: \frac{1}{2} \left(x + \frac{1}{2}\sin 2x\right) \bigg|_{0}^{\frac{\pi}{2}} = \hdots$
Whoa, I totally was thinking that the anti derivative of 1 was 0. That's where I messed up. Silly mistake. Thanks though, I'll remember to post my work next time.

7. $\int_{0}^{\frac{\pi}{2}} \cos^2 x \: dx = \int_{0}^{\frac{\pi}{2}} \left[\frac{1}{2} (1 + \cos 2x)\right]dx = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \left[1 + \cos 2x\right]dx =$ ${\color{white}.} \: \frac{1}{2} \left(x + \frac{1}{2}\sin 2x\right) \bigg|_{0}^{\frac{\pi}{2}} = \hdots$
There's an easier way to deal with these kind of integrals.

Rule: $\int_a^b f(x)~dx = \int_a^b f(a+b-x)~dx$

I'll solve this step by step for you to understand how it works. It actually takes only a nanosecond to solve this.

$I = \int_0^\frac{\pi}{2}\cos^2x~dx = \int_0^\frac{\pi}{2}\cos^2(\frac{\pi}{2}-x)~dx = \int_0^\frac{\pi}{2}\sin^2x~dx$

Now, $I+I = \int_0^\frac{\pi}{2}\cos^2x~dx + \int_0^\frac{\pi}{2}\sin^2x~dx = \int_0^{\frac{\pi}{2}}1~dx = \frac{\pi}{2}$

$I = \frac{\pi}{4}$