# How do you find the antiderivative of #(cos 2x)e^(cos 2x)#?

I don't think you can (in finite form).

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To find the antiderivative of ( (\cos 2x)e^{\cos 2x} ), you can use integration by substitution. Let ( u = \cos 2x ), then ( du = -2\sin 2x , dx ). This simplifies the integral to ( -\frac{1}{2}\int e^u , du ). Integrating ( e^u ) with respect to ( u ), you get ( -\frac{1}{2}e^u + C ), where ( C ) is the constant of integration. Substituting ( u = \cos 2x ) back in, the antiderivative is ( -\frac{1}{2}e^{\cos 2x} + C ).

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To find the antiderivative of ( (\cos 2x)e^{\cos 2x} ), you can use integration by parts. Let ( u = \cos 2x ) and ( dv = e^{\cos 2x} dx ). Then, ( du = -2\sin 2x dx ) and ( v = \int e^{\cos 2x} dx ).

By integrating ( v ), you can use a substitution or directly apply the antiderivative of ( e^{\cos 2x} ). Once you have ( u ) and ( dv ), apply the integration by parts formula:

[ \int u , dv = uv - \int v , du ]

Substitute ( u ), ( du ), ( v ), and ( dv ) into the formula and evaluate the integral. This will give you the antiderivative of ( (\cos 2x)e^{\cos 2x} ).

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