Let I = ∫x sin3x dxTaking x as the first function and sin3x as the second function, and using integration by parts we getI = x ∫sin3x dx - ∫ddxx ∫sin3x dx dx + C 1Let I1 = ∫sin3x dx = 14∫3 sin x - sin 3x dx as, sin 3x = 3 sin x - 4 sin3x I1 = 14-3 cos x + cos 3x3 from 1, we getI = x 14-3 cos x + cos 3x3 - 14∫-3 cos x + cos 3x3 dx + C=x4-3 cos x + cos 3x3 - 14- 3 sin x + sin 3x9 + C=-3x cos x4 + x cos 3x12 + 3 sin x4 - sin 3x36 + C

integral x sin^3x

Let I = \int x \sin^{3} x d x Taking x as the first function and \sin^{3} x as the second function, and using integration by parts we get I = x \int \sin^{3} x d x - \int [\frac{d}{d x} (x) . \int \sin^{3} x d x] d x + C . . . . . . . . . (1) Let I_{1} = \int \sin^{3} x d x = \frac{1}{4} \int [3 \sin x - \sin 3 x] d x (a s, \sin 3 x = 3 \sin x - 4 \sin^{3} x) I_{1} = \frac{1}{4} [- 3 \cos x + \frac{\cos 3 x}{3}] from (1), we get I = x . \frac{1}{4} [- 3 \cos x + \frac{\cos 3 x}{3}] - \frac{1}{4} \int [- 3 \cos x + \frac{\cos 3 x}{3}] d x + C = \frac{x}{4} [- 3 \cos x + \frac{\cos 3 x}{3}] - \frac{1}{4} [- 3 \sin x + \frac{\sin 3 x}{9}] + C = \frac{- 3 x \cos x}{4} + \frac{x \cos 3 x}{12} + \frac{3 \sin x}{4} - \frac{\sin 3 x}{36} + C

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