Question

\int \cos^2(2x) dx = \\ a. \frac{x}{2} - \frac{\sin 2x}{4} + c \\ b. \frac{x}{2} + \frac{\sin 4x}{8} + c \\ c. \frac{x}{2} - \frac{\cos 4x}{8} + c \\ d. \frac{x}{2} - \frac{\sin 4x}{8} + c

          \int \cos^2(2x) dx = \\
a. \frac{x}{2} - \frac{\sin 2x}{4} + c \\
b. \frac{x}{2} + \frac{\sin 4x}{8} + c \\
c. \frac{x}{2} - \frac{\cos 4x}{8} + c \\
d. \frac{x}{2} - \frac{\sin 4x}{8} + c

∫cos^2(2x) dx =

a. (x)/(2) - (sin2x)/(4) + c

b. (x)/(2) + (sin4x)/(8) + c

c. (x)/(2) - (cos4x)/(8) + c

d. (x)/(2) - (sin4x)/(8) + c

Added by Jose Carlos W.

Calculus: Early Transcendentals

James Stewart 8th Edition

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Solved by Expert Jerelyn Nevil

Step 1

The given integral is: ∫ fcos(2x) dx sin(zx) sin(ax) + C Show more…

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Question

\int \cos^2(2x) dx = \\ a. \frac{x}{2} - \frac{\sin 2x}{4} + c \\ b. \frac{x}{2} + \frac{\sin 4x}{8} + c \\ c. \frac{x}{2} - \frac{\cos 4x}{8} + c \\ d. \frac{x}{2} - \frac{\sin 4x}{8} + c

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