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I am having a difficult time trying to find the area in between $\displaystyle 4\cos^2(x)$ and $\displaystyle \frac{\sec^2(x)}{4}$ from $\displaystyle -\frac{\pi}{3}$ to $\displaystyle \frac{\pi}{3}$.
Here is my work:

$\displaystyle \int^\frac{\pi}{3}_{-\frac{\pi}{3}}\left[4\cos^2(x) - \frac{\sec^2(x)}{4}\right]dx = 2\int^\frac{\pi}{3}_0\left[2\cos(2x)+2 - \frac{1}{2\cos(2x)+2}\right]dx$

$=\displaystyle\left[\sin(2x)+2x - \frac{1}{\sin(2x)+2x}\right] \Biggr|_0^\frac{\pi}{3}$

At this point I get very hairy fractions and division by zero. Can somebody tell me where I am going wrong?

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  • $\begingroup$ One antiderivative of $\sec^2 x$ is $\tan x$. The double angle trick that works nicely for $\cos^2 x$ is not useful for $\sec^2 x$. The expression you got was not the integral of that part. It is not true in general that $\int \frac{1}{f}=\frac{1}{\int f}$. $\endgroup$ – André Nicolas Apr 28 '13 at 18:05
  • $\begingroup$ Did you type one of the limits of integration incorrectly (lower one should be a 0). $\endgroup$ – Amzoti Apr 28 '13 at 18:05
  • $\begingroup$ $\LaTeX$ tip: use "\cos", "\sin", etc. $\endgroup$ – Stefan Smith Apr 29 '13 at 15:47
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The integration of $\sec^2 x$ was done incorrectly. You may recall from derivative days that the derivative of $\tan x$ is $\sec^2 x$.

Remark: The double angle formula that is helpful when we integrate $\cos^2 x$ is not useful in the integration of $\sec^2 x$.

The attempt assumed that $\displaystyle\int \frac{dx}{f(x)}=\frac{1}{\displaystyle \int f(x)\,dx}$. There are very few functions for which this is true.

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$$\int \frac{1}{2\cos (2x)+1}dx \neq \frac{1}{\sin(2x)+x}$$

Just use the fact that $$\int \sec^2(x)dx = \tan x + c$$ It'll be much simpler.

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The key error is this: it is not true that $g(x)=\int f(x) dx$ implies that $\frac{1}{g(x)}=\int \frac{1}{f(x)}dx$. That is, the antiderivative of $\frac{1}{2\cos(2x) + 1}dx$ is not $\frac{1}{\sin(2x)+x}$.

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I think you have a few mistakes mixed in. If $\cos^2x=\frac{\cos2x+1}2$, then $4\cos^2x=2\cos2x+2$. If you wished to replace what you did, you would end up with

$$\int\limits_{-\frac\pi3}^{\frac\pi3}2\cos2x+2-\frac1{2\cos2x+2}dx$$

At first, I had thought you had simply factored out a 2 before realizing the limits changed, so I assume you used the fact that your integrand was an even function. This appears to be done correctly, minus the above mentioned mistake.

The second half of the integral was done incorrectly. I'm not sure what your reasoning is. Of course, the integral would have been a lot easier if you had simply left $\sec^2x$ as is since it can be integrated easily.

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Let $A$ be the area you want. We have

$$ \begin{align} A &= \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \left[4\cos^2(x)-\frac{\sec^2(x)}{4}\right] \mathrm{d}x \\ &= 4 \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \cos^2(x) \mathrm{d}x - \frac{1}{4}\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \frac{\mathrm{d}x}{\cos^2(x)} \\ &= 4 \left[ \frac{x}{2} + \frac{\sin(2x)}{4}\right]_{-\pi/3}^{\pi/3} - \frac{1}{4} \tan(x)|_{-\pi/3}^{\pi/3} \\ &= 4 \left( \frac{\pi}{3} + \frac{\sin(2\pi/3)}{2}\right) - \frac{1}{2} \tan(\pi/3) \\ &= \frac{4\pi}{3} + \sqrt{3} - \frac{\sqrt{3}}{2} \\ &= \frac{4\pi}{3} + \frac{\sqrt{3}}{2} \end{align} $$

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Wolframalpha can be used to obtain a graphical picture:

For $4\cdot cos^2(x)$ from $-\frac{\pi}{3}$ to $\frac{\pi}{3}$:

http://www.wolframalpha.com/input/?i=definite+integral&a=*C.definite+integral-_*Calculator.dflt-&f2=4%28cosx%29%5E2&x=8&y=8&f=DefiniteIntegralCalculator.integrand_4%28cosx%29%5E2&f3=x&f=DefiniteIntegralCalculator.variable_x&f4=-pi%2F3&f=DefiniteIntegralCalculator.rangestart_-pi%2F3&f5=pi%2F3&f=DefiniteIntegralCalculator.rangeend_pi%2F3

For $\frac{sec^2(x)}{4}$ from $-\frac{\pi}{3}$ to $\frac{\pi}{3}$:

http://www.wolframalpha.com/input/?i=definite+integral&a=*C.definite+integral-_*Calculator.dflt-&f2=0.25%28secx%29%5E2&f=DefiniteIntegralCalculator.integrand_0.25%28secx%29%5E2&f3=x&f=DefiniteIntegralCalculator.variable_x&f4=-pi%2F3&f=DefiniteIntegralCalculator.rangestart_-pi%2F3&f5=pi%2F3&f=DefiniteIntegralCalculator.rangeend_pi%2F3

And André Nicolas is right about the part that the antiderivative of $\sec^2 x$ is $\tan x$.

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