1. ## Textbook Error?

The textbook is Chandler/Bostock, Core Course for A Level. The question asks the reader to integrate sin^2 x from 0 to pi/6. The answer given is (2pi - 3 sqrt(3))/24. I get an answer of (pi - 3 sqrt(3))/12

I know it is more likely that I have the mistake but this is driving me nuts.

Andrew

2. ## Re: Textbook Error?

Originally Posted by grillage
The textbook is Chandler/Bostock, Core Course for A Level. The question asks the reader to integrate sin^2 x from 0 to pi/6. The answer given is (2pi - 3 sqrt(3))/24. I get an answer of (pi - 3 sqrt(3))/12

I know it is more likely that I have the mistake but this is driving me nuts.

Andrew
\displaystyle \displaystyle \begin{align*} \int_0^{\frac{\pi}{6}}{\sin^2{x}\,dx} &= \int_0^{\frac{\pi}{6}}{\frac{1}{2} - \frac{1}{2}\cos{2x}\,dx} \\ &= \left[\frac{1}{2}x - \frac{1}{4}\sin{2x}\right]_0^{\frac{\pi}{6}} \\ &= \left[\frac{1}{2}\left(\frac{\pi}{6}\right) - \frac{1}{4}\sin{\left(\frac{2\pi}{6}\right)}\right] - \left[\frac{1}{2}(0) - \frac{1}{4}\sin{(2\cdot 0)}\right] \\ &= \left(\frac{\pi}{12} - \frac{1}{4}\sin{\frac{\pi}{3}}\right) - \left( 0 - 0 \right) \\ &= \frac{\pi}{12} - \frac{1}{4}\left(\frac{\sqrt{3}}{2}\right) \\ &= \frac{\pi}{12} - \frac{\sqrt{3}}{8} \\ &= \frac{2\pi - 3\sqrt{3}}{24} \end{align*}

3. ## Re: Textbook Error?

Thanks very much. I see the silly mistake I made now.

4. ## Re: Textbook Error?

You can very often use wolframalpha to check your work.

integrate sin(x)^2 x=0,pi/6 - Wolfram|Alpha

5. ## Re: Textbook Error?

That looks like a very interesting resource, thanks for sharing. Although that would have confirmed that the textbook was indeed correct, I needed the response from Prove it to easily spot my error.

Andrew