Question

1 ? Mark for Review \[ \int x \cos (2 x) d x= \] (A) \( \frac{1}{2} x^{2} \sin (2 x)+C \) (B) \( \frac{1}{2} x^{2} \cos (2 x)+\frac{1}{2} \sin (2 x)+C \) (C) \( \frac{1}{2} x \sin (2 x)-\frac{1}{4} \cos (2 x)+C \) (D) \( \frac{1}{2} x \sin (2 x)+\frac{1}{4} \cos (2 x)+C \)

          1 ? Mark for Review
\[
\int x \cos (2 x) d x=
\]
(A) \( \frac{1}{2} x^{2} \sin (2 x)+C \)
(B) \( \frac{1}{2} x^{2} \cos (2 x)+\frac{1}{2} \sin (2 x)+C \)
(C) \( \frac{1}{2} x \sin (2 x)-\frac{1}{4} \cos (2 x)+C \)
(D) \( \frac{1}{2} x \sin (2 x)+\frac{1}{4} \cos (2 x)+C \)

1 ? Mark for Review

∫ x cos (2 x) d x=

(A) (1)/(2) x^2sin (2 x)+C
(B) (1)/(2) x^2cos (2 x)+(1)/(2)sin (2 x)+C
(C) (1)/(2) x sin (2 x)-(1)/(4)cos (2 x)+C
(D) (1)/(2) x sin (2 x)+(1)/(4)cos (2 x)+C

Added by Elisa J.

IIT JEE Super Course in Mathematics: Calculus

Trishna Knowledge Systems 1st Edition

Chapter 3

Instant Answer

Solved by Expert Eduard Sanchez

05/07/2024

Step 1

Step 1: Identify the integral to solve: \[ \int x \cos(2x) \, dx \] Show more…

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1 П Mark for Review \[ \int x \cos (2 x) d x= \] (A) \( \frac{1}{2} x^{2} \sin (2 x)+C \) (B) \( \frac{1}{2} x^{2} \cos (2 x)+\frac{1}{2} \sin (2 x)+C \) (C) \( \frac{1}{2} x \sin (2 x)-\frac{1}{4} \cos (2 x)+C \) (D) \( \frac{1}{2} x \sin (2 x)+\frac{1}{4} \cos (2 x)+C \)