Definite integral of e^(cos x) sin x dx from 0 to pi. u = cos x e^(u) ] -1 to 1
W warwick Full Member Joined Jan 27, 2006 Messages 311 Dec 7, 2006 #1 Definite integral of e^(cos x) sin x dx from 0 to pi. u = cos x e^(u) ] -1 to 1
D daon Senior Member Joined Jan 27, 2006 Messages 1,284 Dec 7, 2006 #2 Your integral: \(\displaystyle \L \int _{0}^\pi e^{cosx}sinx dx\) u=cosxdu=−sinxdx ⇒ −du=sinxdx\displaystyle u = cosx \\ du = -sinxdx \,\, \,\, \Rightarrow \,\,\,\, -du =sinxdxu=cosxdu=−sinxdx⇒−du=sinxdx So, \(\displaystyle \L\int _{0}^\pi e^{cosx}sinx dx \,\, \,\, = \,\, \,\, \int _{1}^{-1} e^{u}(-du)\) Which is the same as: \(\displaystyle \L -\int _{1}^{-1} e^{u}(du) \,\, \,\, = \,\, \,\, \int _{-1}^{1} e^{u}(du)\) Hope that helps, -daon
Your integral: \(\displaystyle \L \int _{0}^\pi e^{cosx}sinx dx\) u=cosxdu=−sinxdx ⇒ −du=sinxdx\displaystyle u = cosx \\ du = -sinxdx \,\, \,\, \Rightarrow \,\,\,\, -du =sinxdxu=cosxdu=−sinxdx⇒−du=sinxdx So, \(\displaystyle \L\int _{0}^\pi e^{cosx}sinx dx \,\, \,\, = \,\, \,\, \int _{1}^{-1} e^{u}(-du)\) Which is the same as: \(\displaystyle \L -\int _{1}^{-1} e^{u}(du) \,\, \,\, = \,\, \,\, \int _{-1}^{1} e^{u}(du)\) Hope that helps, -daon
