How can I intergrate: e^(sin theta) cos theta Thanks forany help.
D dangerous_dave New member Joined Mar 13, 2008 Messages 21 May 11, 2008 #1 How can I intergrate: e^(sin theta) cos theta Thanks forany help.
O o_O Full Member Joined Oct 20, 2007 Messages 396 May 11, 2008 #2 ∫esinθ cosθdθ\displaystyle \int e^{\sin \theta} \: \cos \theta d\theta∫esinθcosθdθ u=sinθ\displaystyle u = \sin \thetau=sinθ du=cosθdθ\displaystyle du = \cos \theta d\thetadu=cosθdθ Making the subs: ∫esinθ⏞ucosθdθ⏞du\displaystyle \int e^{\overbrace{\sin \theta}^{u}} \overbrace{\cos \theta d\theta}^{du}∫esinθucosθdθdu
∫esinθ cosθdθ\displaystyle \int e^{\sin \theta} \: \cos \theta d\theta∫esinθcosθdθ u=sinθ\displaystyle u = \sin \thetau=sinθ du=cosθdθ\displaystyle du = \cos \theta d\thetadu=cosθdθ Making the subs: ∫esinθ⏞ucosθdθ⏞du\displaystyle \int e^{\overbrace{\sin \theta}^{u}} \overbrace{\cos \theta d\theta}^{du}∫esinθucosθdθdu
D dangerous_dave New member Joined Mar 13, 2008 Messages 21 May 12, 2008 #3 Are you saying that, when integrated, its the same as before?
skeeter Elite Member Joined Dec 15, 2005 Messages 3,218 May 12, 2008 #4 that's not what o_0 said ... ∫esinθcosθ dθ=esinθ+C\displaystyle \int e^{\sin{\theta}} \cos{\theta} \, d \theta = e^{\sin{\theta}} + C∫esinθcosθdθ=esinθ+C
that's not what o_0 said ... ∫esinθcosθ dθ=esinθ+C\displaystyle \int e^{\sin{\theta}} \cos{\theta} \, d \theta = e^{\sin{\theta}} + C∫esinθcosθdθ=esinθ+C