Answer
$-\dfrac{1}{2} \cos 2x +\cot x+C$
Work Step by Step
Solve. $\int (\sin 2x ) dx - \int (\csc^2 x) dx$
Thus, we have $\int (\sin 2x ) dx - \int (\csc^2 x) dx=-\dfrac{1}{2} \cos 2x -(-\cot x)+C$
or, $=-\dfrac{1}{2} \cos 2x +\cot x+C$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0-32187-896-5
ISBN 13:
978-0-32187-896-0
Chapter 4: Applications of Derivatives - Section 4.7 - Antiderivatives - Exercises 4.7 - Page 239: 45
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
