intergrate: e^(sin theta) cos theta

esinθcosθdθ\displaystyle \int e^{\sin \theta} \: \cos \theta d\theta

u=sinθ\displaystyle u = \sin \theta
du=cosθdθ\displaystyle du = \cos \theta d\theta

Making the subs:
esinθucosθdθdu\displaystyle \int e^{\overbrace{\sin \theta}^{u}} \overbrace{\cos \theta d\theta}^{du}
 
that's not what o_0 said ...

esinθcosθdθ=esinθ+C\displaystyle \int e^{\sin{\theta}} \cos{\theta} \, d \theta = e^{\sin{\theta}} + C
 
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