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So I'm supposed to find the area of this

$$\int_0^{\pi/2} |\sin{x} - \cos{2x}| dx$$

and I try setting $\sin x = \cos 2x$ but have trouble solving that. I tried using trig identities.

Is there a simpler way of evaluating this type of problem or do you have to graph it and split it up and find two different integrals every time?

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I've edited your post to include Latex. Please ensure that it is correct - I'm guessing that you mean either for the integral to run from $0$ to $\pi/2$, or from $-\pi/2$ to $0$. –  user61527 Sep 2 '13 at 22:30
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its supposed to be pi/2 on top and 0 on bottom –  J L Sep 2 '13 at 22:32

2 Answers 2

up vote 4 down vote accepted

Hint: Use the fact that $\cos(2x) = 1 - 2\sin^2x$. Then you will have a quadratic for $\sin x$.

It depends on the situation, but I don't think there is a simpler way of evaluating $\int_a^b|f(x)|dx$ without splitting $[a, b]$ into the sets where $f(x)$ is positive, negative, and zero.

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i tried that and replaced sinx with x to get 2x^2 + x -1 = 0 to get -1 and 1/2 but when i plug those values into the original equation they dont make sense? –  J L Sep 2 '13 at 22:36
    
Don't replace $\sin x$ with $x$ as you are using $x$ to denote two different quantities. Instead, let $a = \sin x$, then you get $a = -1$ and $a = \frac{1}{2}$. This says that when $a = \sin x = -1$ or $a = \sin x = \frac{1}{2}$, $\sin x = \cos(2x)$. For which values of $x \in [0, \frac{\pi}{2}]$ does this occur? These are the points at which the function $\sin x - \cos(2x)$ can change from positive to negative or negative to positive. –  Michael Albanese Sep 2 '13 at 22:38
    
pie/6? ........ –  J L Sep 2 '13 at 22:42
    
It is pi, but yes $x = \frac{\pi}{6}$ is a point such that $\sin x = \frac{1}{2}$ and therefore $\sin x = \cos(2x)$. Now split your interval into two parts: $[0, \frac{\pi}{6}]$ and $[\frac{\pi}{6}, \frac{\pi}{2}]$. You need to determine whether $\sin x - \cos(2x)$ is positive or negative on each of those two intervals. –  Michael Albanese Sep 2 '13 at 22:45
    
from 0 to pie/6 cos2x is higher so from 0 to pi/6 its cos2x-sinx and from pi/6 to 2/pi its sinx-cos2x am I right?? –  J L Sep 2 '13 at 22:48

Note that,

$$ \sin(x)-\cos(2x) \leq 0\quad \forall\, x\in[0,\pi/6] $$

and

$$ \sin(x)-\cos(2x) \geq 0\quad \forall\, x\in[\pi/6,\pi/2]. $$

So, you can split the integral as

$$\int_0^{\pi/2} |\sin{x} - \cos{2x}| dx = \int_0^{\pi/6} -(\sin{x} - \cos{2x}) dx + \int_{\pi/6}^{\pi/2} (\sin{x} - \cos{2x}) dx =\dots\,.$$

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