$\displaystyle \int e^(3cos2x) sen2xdx$
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Hello, Let u=3cos2x. Then, du=-6sin2xdx. $\displaystyle \int e^{3\cos 2x}\sin 2xdx =\int e^u\left(-\frac{1}{6}\right)du=-\frac{1}{6}e^u+C =-\frac{1}{6}e^{3\cos 2x}+C.$ ($\displaystyle C$: integral constant) Bye.
This is a very simply and obvious substitution, Apprientice123. This shouts clearly that you just have no desire to do any work. Come on. Step up your game.
ok. thank you
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