int x^{3} sin 3 x , dx = | vedclass

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(D) To evaluate the integral $ I = \int x^{3} \sin 3 x \, dx $,we use the method of integration by parts or the tabular method (Bernoulli's formula).

Vedclass · Accepted Answer

$ -\frac{x^{3} \cos 3 x}{3} + \frac{x^{2} \sin 3 x}{3} + \frac{2 x \cos 3 x}{9} - \frac{2 \sin 3 x}{27} + C $

Vedclass · Answer

(D) To evaluate the integral $ I = \int x^{3} \sin 3 x \, dx $,we use the method of integration by parts or the tabular method (Bernoulli's formula). Using the formula $ \int u v \, dx = u v_1 - u' v_2 + u'' v_3 - u''' v_4 + \dots $,where $ u = x^3 $ and $ v = \sin 3x $: $ u = x^3, u' = 3x^2, u'' = 6x, u''' = 6, u'''' = 0 $ $ v_1 = \int \sin 3x \, dx = -\frac{\cos 3x}{3} $ $ v_2 = \int -\frac{\cos 3x}{3} \, dx = -\frac{\sin 3x}{9} $ $ v_3 = \int -\frac{\sin 3x}{9} \, dx = \frac{\cos 3x}{27} $ $ v_4 = \int \frac{\cos 3x}{27} \, dx = \frac{\sin 3x}{81} $ Substituting these into the formula: $ I = x^3 \left( -\frac{\cos 3x}{3} \right) - (3x^2) \left( -\frac{\sin 3x}{9} \right) + (6x) \left( \frac{\cos 3x}{27} \right) - (6) \left( \frac{\sin 3x}{81} \right) + C $ $ I = -\frac{x^3 \cos 3x}{3} + \frac{x^2 \sin 3x}{3} + \frac{2x \cos 3x}{9} - \frac{2 \sin 3x}{27} + C $

$ \int x^{3} \sin 3 x \, dx = $

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