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.
$\displaystyle \int\cos^2x$
where b= $\displaystyle \pi/2$ a= 0

answer= $\displaystyle \pi/4$

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?

$\displaystyle \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 = $$\displaystyle {\color{white}.} \: \frac{1}{2} \left(x + \frac{1}{2}\sin 2x\right) \bigg|_{0}^{\frac{\pi}{2}} = \hdots 3. \displaystyle cos^2x = \frac{1 + cos2x}{2} \displaystyle \frac{1}{2}\int{dx} +\frac{1}{2}\int{cos2x}{dx} = \frac{x}{2} +\frac{sin2x}{4} Hope this helps. Oops was to slow lol, just wondering how do you place upper and lower bounds in latex? 4. Originally Posted by 11rdc11 \displaystyle cos^2x = \frac{1 + cos2x}{2} \displaystyle \frac{1}{2}\int{dx} +\frac{1}{2}\int{cos2x}{dx} = \frac{x}{2} +\frac{sin2x}{4} Hope this helps. Oops was to slow lol, just wondering how do you place upper and lower bounds in latex? Do: Code: $$\int_a^b f(x)\,dx$$ To get \displaystyle \int_a^b f(x)\,dx note if we have limits like \displaystyle -\ln2 to \displaystyle \ln2, then you need to put these within {} such as: Code: $$\int_{-\ln(2)}^{\ln2} e^x\,dx$$ will output \displaystyle \int_{-\ln(2)}^{\ln2} e^x\,dx --Chris 5. \displaystyle \int_{0}^{4} x \: dx is produced by: $$\int_{0}^{4} x \: dx$$ _{0} : produces the lower bound ^{4} : produces the upper bound. This can applied to other symbols as well: \displaystyle \sum_{k = 0}^{4} k = 10 is produced by $$\sum_{k = 0}^{4} k = 10$$ \displaystyle \frac{x}{2}\bigg|_{0}^{4} etc etc. 6. Originally Posted by o_O Can you show your work to see what it is that you're doing wrong? \displaystyle \int_{0}^{\frac{\pi}{2}} \cos^2 x \: dx = \int_{0}^{\frac{\pi}{2}} \left[\frac{1}{2} (1 + \cos 2x)\right]dx = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \left[1 + \cos 2x\right]dx =$$\displaystyle {\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. $\displaystyle \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 =$$\displaystyle {\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: $\displaystyle \int_a^b f(x)~dx = \int_a^b f(a+b-x)~dx$

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

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

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

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