Mathematics

Evaluate :
$$\int x \sin 3x \ dx$$


ANSWER

$$-\dfrac{1}{3}x \cos 3x+\dfrac{1}{9} \sin 3x +C$$


SOLUTION
$$I=\int x \sin 3x \ dx$$
$$I=x[\int \sin 3x \ dx]-\int [\dfrac{d}{dx}(x) \times \int \sin 3x \ dx]dx$$

$$I=x \times -\dfrac{1}{3} \cos 3x -\int [-\dfrac{1}{3} \cos 3x] dx$$

$$I=-\dfrac{1}{3}x \cos 3x +\dfrac{1}{3} \int \cos 3x \ dx =-\dfrac{1}{3}x\cos 3x +\dfrac{1}{9} \sin 3x +C$$
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Single Correct Medium Published on 17th 09, 2020
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