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1-2-x-2-sin3x-dx-




Question Number 945 by tera last updated on 04/May/15
∫_1 ^2 x^2 sin3x dx =.....
$$\int_{\mathrm{1}} ^{\mathrm{2}} {x}^{\mathrm{2}} {sin}\mathrm{3}{x}\:{dx}\:=….. \\ $$
Answered by prakash jain last updated on 04/May/15
Integrate by parts  x^3 (−((cos 3x)/3))+∫x^2 cos 3x  =−((x^3 cos 3x)/3)+((2xsin 3x)/3)−∫((2xsin 3x)/3)dx  =−((x^3 cos 3x)/3)+((2xsin 3x)/3)+((2xcos 3x)/9)−∫((2cos 3x)/9)  =−((x^3 cos 3x)/3)+((2xsin 3x)/3)+((2xcos 3x)/9)−((2sin 3x)/(27))
$$\mathrm{Integrate}\:\mathrm{by}\:\mathrm{parts} \\ $$$${x}^{\mathrm{3}} \left(−\frac{\mathrm{cos}\:\mathrm{3}{x}}{\mathrm{3}}\right)+\int{x}^{\mathrm{2}} \mathrm{cos}\:\mathrm{3}{x} \\ $$$$=−\frac{{x}^{\mathrm{3}} \mathrm{cos}\:\mathrm{3}{x}}{\mathrm{3}}+\frac{\mathrm{2}{x}\mathrm{sin}\:\mathrm{3}{x}}{\mathrm{3}}−\int\frac{\mathrm{2}{x}\mathrm{sin}\:\mathrm{3}{x}}{\mathrm{3}}{dx} \\ $$$$=−\frac{{x}^{\mathrm{3}} \mathrm{cos}\:\mathrm{3}{x}}{\mathrm{3}}+\frac{\mathrm{2}{x}\mathrm{sin}\:\mathrm{3}{x}}{\mathrm{3}}+\frac{\mathrm{2}{x}\mathrm{cos}\:\mathrm{3}{x}}{\mathrm{9}}−\int\frac{\mathrm{2cos}\:\mathrm{3}{x}}{\mathrm{9}} \\ $$$$=−\frac{{x}^{\mathrm{3}} \mathrm{cos}\:\mathrm{3}{x}}{\mathrm{3}}+\frac{\mathrm{2}{x}\mathrm{sin}\:\mathrm{3}{x}}{\mathrm{3}}+\frac{\mathrm{2}{x}\mathrm{cos}\:\mathrm{3}{x}}{\mathrm{9}}−\frac{\mathrm{2sin}\:\mathrm{3}{x}}{\mathrm{27}} \\ $$$$ \\ $$

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