Integrating sin(2x)dx: Solving for the Integral of a Trigonometric Function

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Homework Statement


\int sin(2x)dx

Homework Equations


I know the integral of sin(x)dx = -cos(x) + C

The Attempt at a Solution


What I did was to say that the integral is -cos(2x) +C, which isn't the correct answer...I should have gotten -1/2(cos(2x)) +C. I can see that this is the correct answer when I differentiate it via chain rule and get sin(2x), however I can't seem to integrate the problem to get the right answer. Can someone walk me through it please...
 
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Use substitution, with u = 2x, du = 2dx. Integration by substitution is the reverse of the chain rule in differentiation.
 
ah so that's where I went wrong...I tried u substitution but I used u=sin2x instead of u=2x. thanks mark44.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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