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Evaluate ` int x sin 3x dx`. |
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Answer» Here, both the functions, viz., `x` and `sin 3x` are easily integrable and the derivative of x is one, a less complicated function. Therefore, we take x as the first function and `sin 3x` as the second function. Thus, `int underset (I)(x)underset(II)(sin)3x dx` `=x{intsin 3xdx}-int{(d)/(dx)(x)intsin 3x dx}dx` `= -x (cos 3x)/(3)-int1 {-(cos 3x)/(3)}dx` `= -(1)/(3)x cos 3x +(1)/(3) int cos3x dx` `= -(1)/(3)x cos 3x +(1)/(9) sin 3x +C` |
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