Explanation Given information: ∫ 0 π / 2 sin n x d x Formula: ∫ u ( x ) v ′ ( x ) d x = u ( x ) v ( x ) − ∫ v ( x ) u ′ ( x ) d x (1) d f ( u ) d x = d f d u ⋅ d u d x ∫ c d x a d x = [ x a + 1 a + 1 ] c d Calculation: Express the integral as below. ∫ 0 π / 2 sin n x d x = ∫ 0 π / 2 cos n x d x Here, the values of function cos x over the interval [ 0 , π 2 ] is the reflection of function sin x over the same interval to the both sides of the line x = π 4 . Hence, the area under both of them will be same. Consider the cosine term from the above equation. ∫ 0 π / 2 cos n x d x = ∫ 0 π / 2 ( cos n − 1 x cos x ) d x (2) Take the term ∫ 0 π / 2 ( cos n − 1 x cos x ) d x from the equation (2). Let us take the term cos n − 1 x as u ( x ) . u ( x ) = cos n − 1 x Differentiate the term u with respect to x . u ′ ( x ) = d ( cos n − 1 x ) d x Apply the chain rule. u ′ ( x ) = d f d u ⋅ d u d x Take the function as f = u n − 1 and term as u = cos ( x ) . Substitute u n − 1 for f and cos ( x ) for u in chain rule. u ′ ( x ) = d d u u n − 1 d d x cos x = ( n − 1 ) u n − 2 ( − sin ( x ) ) = − sin x ( n − 1 ) u n − 2 Substitute cos ( x ) for u. u ′ ( x ) = − sin x ( n − 1 ) cos n − 2 x Let us take the term cos x as v ′ ( x ) . Integrate the term with respect to x . ∫ v ′ ( x ) = ∫ cos x d x v ( x ) = sin x Substitute cos n − 1 x for u ( x ) , − sin x ( n − 1 ) cos n − 2 x for u ′ ( x ) , sin x for v ( x ) and cos x for v ′ ( x ) with the limits in equation (1). ∫ 0 π / 2 cos n − 1 x cos x d x = [ cos n − 1 x ( sin x ) ] 0 π / 2 − ∫ 0 π / 2 ( sin x ) ( − sin x ( n − 1 ) cos n − 2 x ) d x = [ sin x cos n − 1 x ] 0 π / 2 + ( n − 1 ) ∫ 0 π / 2 sin 2 x cos n − 2 x d x Simplify it further as below

University Calculus: Early Transcendentals (4th Edition)

4th Edition

ISBN: 9780134995540

Author: Joel R. Hass, Christopher E. Heil, Przemyslaw Bogacki, Maurice D. Weir, George B. Thomas Jr.

Publisher: PEARSON

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Chapter 8.1, Problem 74E

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To establish the reduction formula using integration by parts.

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Chapter 8.1, Problem 73E

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2.3 Vibration amplitude of a machine plotted against time shown in Fig. T2.3. is described by sin(0.8+) In(t + 2) x(t) = t + 0.5 Here, t is in seconds and x is in millimeters 0.6 0.4 0.2 ° -0.2 mmm -0.4 0 10 20 30 40 50 60 70 Time [s] Fig. T2.3: Machine vibration amplitude variation against time Obtain the vibration velocity as a function of time, x(t), if it is measured by a vibration velocity sensor.

Example(1): (Adiabatic humidification and cooling of air). Air has to be humidified and cooled adiabatically in a honzontal spray chamber with recirculated water. The active part of the chamber is Im #2m #15 m long. Under the operating conditions, the coefficient of heat transfer is expected to be 1300 kcal/(hr)(m2)(°C). 200 m3/min of air at 60 °C and 1 atm pressure with a humidity of 0.018 kg water/kg dry air is to be blown through the spray chamber. Calculate the following (a) the temperature and hunudity of the exit air (b) make-up water to be supplied, windage and blow down are neglected (c) the expected gas-phase mass transfer coefficient, kya (d) the temperature and humidity of the exit air if an identical spray chamber is added in series with the existing one O

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University Calculus: Early Transcendentals (4th Edition)

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