Calculus: Early Transcendentals 8th Edition

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

Chapter 13 - Section 13.2 - Derivatives and Integrals of Vector Functions - 13.2 Exercise - Page 861: 39


$\tan ti+\frac{1}{8}(t^2+1)^4j+(\dfrac{1}{3}t^3 \ln t-\dfrac{1}{9}t^3)k+C$

Work Step by Step

Given: $\int (\sec^2 ti+t(t^2+1)^3j+t^2 \ln t k)dt$ In order to evaluate the integral we will have to integrate each component of the function individually. Let $I=\int (\sec^2 ti+t(t^2+1)^3j+t^2 \ln t k)dt$ Thus, $I=\int \sec^2 tdti+\int t(t^2+1)^3dtj+\int t^2 \ln t dtk$ $I=\tan ti+ Aj+Bk$ ... (1) Here, $A=\int t(t^2+1)^3dt$ and $B=\int t^2 \ln t dt$ Consider $A=\int t(t^2+1)^3dt$ suppose $p=t^2+1 \implies dp=2tdt$ Thus, $A=\frac{1}{2}\int p^3dp=\frac{1}{8}p^4=\frac{1}{8}(t^2+1)^4$ Now, consider $B=\int t^2 \ln t dt=\ln t (1/3t^3)-\int (1/3t^3) \dfrac{1}{t} dt=\dfrac{1}{3}t^3 \ln t-\dfrac{1}{9}t^3$ Plug in the values of A and B in equation (1) and we get $I=\tan ti+\frac{1}{8}(t^2+1)^4j+(\dfrac{1}{3}t^3 \ln t-\dfrac{1}{9}t^3)k+C$ Here, C is the constant of integration.
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