# Math Help - Find the careless mistake please

1. ## Find the careless mistake please

I must have made a careless mistake in here somewhere as the book says the answer should be 0.158. Please put me out of my misery. Also, if you know how to do the latex for the "between" part on line 4 then I'd love to know.

$\\\int_{0}^{\frac{\pi }{3}}sin^{2}xcos^{2}x\: dx\\[15]\frac{1}{4}\int_{0}^{\frac{\pi }{3}}\left ( 1+cos2x \right )\left ( 1-sin2x \right )\: dx\\[15]\frac{1}{4}\int_{0}^{\frac{\pi }{3}}1-sin2x+cos2x-cos2xsin2x\: dx\\[15]\left [ \frac{1}{4}\left ( x+\frac{1}{2}cos2x+\frac{1}{2}sin2x-\frac{1}{4}sin^{2}2x \right ) \right ]between\: 0,\, \frac{\pi }{3}\\[15]\frac{1}{4}\left ( \frac{\pi }{3}-\frac{1}{4}+\frac{\sqrt{3}}{4}-\frac{3}{16}-\frac{1}{2} \right )\\[15]= 0.136$

2. ## Re: Find the careless mistake please

Originally Posted by grillage
I must have made a careless mistake in here somewhere as the book says the answer should be 0.158. Please put me out of my misery. Also, if you know how to do the latex for the "between" part on line 4 then I'd love to know.

$\\\int_{0}^{\frac{\pi }{3}}sin^{2}xcos^{2}x\: dx\\[15]\frac{1}{4}\int_{0}^{\frac{\pi }{3}}\left ( 1+cos2x \right )\left ( 1-sin2x \right )\: dx\\[15]\frac{1}{4}\int_{0}^{\frac{\pi }{3}}1-sin2x+cos2x-cos2xsin2x\: dx\\[15]\left [ \frac{1}{4}\left ( x+\frac{1}{2}cos2x+\frac{1}{2}sin2x-\frac{1}{4}sin^{2}2x \right ) \right ]between\: 0,\, \frac{\pi }{3}\\[15]\frac{1}{4}\left ( \frac{\pi }{3}-\frac{1}{4}+\frac{\sqrt{3}}{4}-\frac{3}{16}-\frac{1}{2} \right )\\[15]= 0.136$
\displaystyle \begin{align*} \sin^2{x} \equiv \frac{1}{2}\left( 1 - \cos{2x} \right) \end{align*}, not \displaystyle \begin{align*} \frac{1}{2}\left( 1 - \sin{2x} \right) \end{align*}.

3. ## Re: Find the careless mistake please

Of course. How is it that no matter how long I looked at this I could not see that? Any ideas on the latex question?

4. ## Re: Find the careless mistake please

Yeah, use _0 for the lower limit, then ^{\frac{\pi}{3}} for the upper limit.