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integral sin (2x) dx from 0 to pi/8
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\[\Large\bf\sf \int\limits_0^{\pi/8} \sin(\color{orangered}{2x})\;dx\] Make the substitution:\[\Large\bf\sf \color{orangered}{u=2x}\]
Taking the derivative of our substitution: \[\Large\bf\sf du=2\;dx\]Solving for dx, we'll divide each side by 2,\[\Large\bf\sf \frac{1}{2}du=dx\]
Then we just plug in our pieces.\[\Large\bf\sf \int\limits \sin(\color{orangered}{u})\;\frac{1}{2}du\]
Careful, our integral is now in terms of u, not x. So we either have need to: ~Change the boundaries so they correspond to u or ~After integrating, we can simply undo our substitution and use the original boundaries.
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