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Evaluate the integral. $ \displaystyle \int e^…

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Problem 17 Medium Difficulty

Evaluate the integral.

$ \displaystyle \int e^{2 \theta} \sin 3 \theta d \theta $

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Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 7

Techniques of Integration

Section 1

Integration by Parts

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Video Transcript

The problem is you evaluated the integral into 2 theta times sine 3 theta d theta for this problem we will use the method of integration by parts, so the formula is integral: u times, prime dx is equal to? U times, v minus the integral of? U. Prime times v d x for our problem, we can let? U is equal to 22 theta and the way prime is equal to sine 3 theta. Then prime is equal to 2 times e to 2. Theta and v is equal to 1. Third cos negative 1. Third. Cosine 3. The now this integral is equal to u times v, but this is negative. 1. Third to 2 theta sine 3 theta minus the integral of prim times v. So this is negative. 2. Third, this plus 2 third times to 2 theta cosine 3 theta d. Theta point now for this integral we can use also use the method of integration by parts that the? U is equal to e to 2 theta nopolis equal to cosine, 3 theta? Then? U prais equal to 2 times e to 2 theta and v is equal to 1 third times sine 3 theta. Now this is equal to negative 1 over 3 into 2 theta cosine 3 theta plus 2 over 3 times. U times so. This is 1. Third to 2 theta sine 3 theta minus integral f you prime times, so this is 2 third to 2 theta times sine 3 thetasothis is equal to negative 1 third into 2 theta cosine 3 theta plus 2 over 9 into 2 theta am sine 3 theta minus 4, over 9 integrite 2 theta in sin, 3, theta d, theta now locatintegral. This is the same as this 1, so we move this term to left hand side and we have 1 plus 4. Over 9 is integral into 2 theta times sine 3 theta d. This is equal to negative 1. Third, third into 2 third 2 times theta cosine 3 times theta plus 2 over 9 into 2 times theta times sine 3 times thetaso this integral this constant is 13 over 9 point. So this integral this integral to 2 theta sine theta d theta is equal to negative 1 third times on, so we multiply 9 over 13 for both sides, and this is negative. 3. Over 13 e to 2 theta cosine, 3 theta plus 2 over 13 into 2 theta sine 3 theta and plus to const.

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Calculus: Early Transcendentals

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