int_{0}^{frac{pi}{8}} sin 2 x d x

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Step:1 ：The problem is to evaluate the integral of (sin 2x) from 0 to (frac{pi}{8}).Step:2 ：The integral of (sin 2x) is (-frac{1}{2}cos 2x).Step:3 ：We need to evaluate this from 0 to (frac{pi}{8}) and subtract the two results.Step:4 ：The result of the integral is (frac{1}{2} - frac{sqrt{2}}{4}).Step:5 ：Final Answer: (boxed{frac{1}{2} - frac{sqrt{2}}{4}})

Answer

Step: 1 ：The problem is to evaluate the integral of (sin 2x) from 0 to (frac{pi}{8}).Step: 2 ：The integral of (sin 2x) is (-frac{1}{2}cos 2x).Step: 3 ：We need to evaluate this from 0 to (frac{pi}{8}) and subtract the two results.Step: 4 ：The result of the integral is (frac{1}{2} - frac{sqrt{2}}{4}).Step: 5 ：Final Answer: (boxed{frac{1}{2} - frac{sqrt{2}}{4}})

$\int_{0}^{\frac{\pi}{8}} \sin 2 x d x$

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