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# 8.2.22-eg - 1 dt = e t 2 sin(2 t 1-1 2-e t 2 cos(2 t 1 1 2...

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Example Evaluate Z e t cos(2 t + 1) dt . 1. Integrate by parts. If we let u = e t , dv = cos(2 t + 1) dt, du = e t dt, v = 1 2 sin(2 t + 1) , then the integral is of the form R u dv , so we use the integration by parts formula Z u dv = u v - Z v du. Therefore Z e t cos(2 t + 1) dt = e t 2 sin(2 t + 1) - 1 2 Z e t sin(2 t + 1) dt (1) 2. Integrate by parts a second time. To evaluate R e t sin(2 t + 1) dt , we let u = e t , dv = sin(2 t + 1) dt, du = e t dt, v = - 1 2 cos(2 t + 1) , (note that the opposite assignment would return us to the integral we started with!) Use the integration by parts formula again, Z e t sin(2 t + 1) dt = - e t 2 cos(2 t + 1) - Z - e t 2 cos(2 t + 1) dt = - e t 2 cos(2 t + 1) + 1 2 Z e t cos(2 t + 1) dt. (2) Substituting (2) into (1), we have Z e t

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Unformatted text preview: + 1) dt = e t 2 sin(2 t + 1)-1 2-e t 2 cos(2 t + 1) + 1 2 Z e t cos(2 t + 1) dt ! = e t 2 sin(2 t + 1) + e t 4 cos(2 t + 1)-1 4 Z e t cos(2 t + 1) dt. (3) 3. Solve the for unknown integral. Notice that the unknown integral, Z e t cos(2 t + 1) dt appears on both sides of equation (3). Solving for this integral we have 5 4 Z e t cos(2 t + 1) dt = e t 2 sin(2 t + 1) + e t 4 cos(2 t + 1) , or equivalently, Z e t cos(2 t + 1) dt = 2 e t 5 sin(2 t + 1) + e t 5 cos(2 t + 1) . 2...
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8.2.22-eg - 1 dt = e t 2 sin(2 t 1-1 2-e t 2 cos(2 t 1 1 2...

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