Evaluate ∫sin 3x dx
Question 1 Report
Evaluate ∫sin 3x dx
Answer Details
To integrate the given function ∫sin 3x dx, we use the trigonometric identity: sin (2a) = 2sin a cos a Using this identity, we can write: sin 3x = sin (2x + x) = sin 2x cos x + cos 2x sin x Therefore, ∫sin 3x dx = ∫(sin 2x cos x + cos 2x sin x) dx Integrating each term separately, we get: = (-cos 2x / 2) cos x + (sin 2x / 2) sin x + c where c is the constant of integration. Simplifying this expression, we get: = -(1/2) cos 2x cos x + (1/2) sin 2x sin x + c = -(1/2) (cos 2x cos x - sin 2x sin x) + c = -(1/2) cos (2x - x) + c = -(1/2) cos x + c Therefore, the answer is: ∫sin 3x dx = -(1/2) cos x + c Hence, the option is: (-1/3) cos 3x + c.
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