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# How do you integrate int sin2x dx?

Dec 17, 2016

$\int \sin \left(2 x\right) \mathrm{dx} = - \frac{1}{2} \cos \left(2 x\right) + C$

#### Explanation:

Using integration by substitution together with the known integral $\int \sin \left(x\right) \mathrm{dx} = - \cos \left(x\right) + C$, we first let $u = 2 x \implies \mathrm{du} = 2 \mathrm{dx}$. Then

$\int \sin \left(2 x\right) \mathrm{dx} = \frac{1}{2} \int \sin \left(2 x\right) 2 \mathrm{dx}$

$= \frac{1}{2} \int \sin \left(u\right) \mathrm{du}$

$= \frac{1}{2} \left(- \cos \left(u\right)\right) + C$

$= - \frac{1}{2} \cos \left(2 x\right) + C$