Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Section 7.1 - Integration by Parts - 7.1 Exercises: 14

Answer

= $\frac{xsinh(ax)}{a}$ + $\frac{cosh(ax)}{a^{2}}$+C

Work Step by Step

Note: cosh(ax) is a hyperbolic function; for this question, we can treat it as if we were dealing with cos(x). Treat a as a constant for this problem. Let u=x u' = 1 $\frac{du}{dx}$ = 1 du = 1dx du = dx dv = cosh(ax)dx $\int dv$ = $\frac{1}{a}$ $\int acosh(ax)dx$ [Chain Rule] v = $\frac{1}{a}$ sinh(ax) uv - $\int vdu$ x[$\frac{1}{a}$sinh(ax)] - $\frac{1}{a}$$\int sinh(ax)$dx Let us solve the integral first before putting the entire answer together. $\frac{1}{a}$$\int sinh(ax)$ Let y = ax y' = a $\frac{dy}{dx}$ = a dy = adx dx = $\frac{1}{a}$dy Plug in all the values: ($\frac{1}{a})^{2}$$\int sinh(y)$dy = -($\frac{1}{a})^{2}$cosh(ax) + C Now combine everything: x[$\frac{1}{a}$sinh(ax)] - $\frac{1}{a}$$\int sinh(ax)$dx = x[$\frac{1}{a}$sinh(ax)] + ($\frac{1}{a})^{2}$cosh(ax) + C
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