If x100sin(x) is implied, there doesn't seem to be a "trick" to solve this as an indefinite integral, but there is a method to solve it, because by integrating by parts, the integrand can be progressively reduced so that the power of x decreases until eventually ∫sin(x)dx is reached, which is readily integrable. The final integral involves terms (such as 100 factorial) which are not readily computable.
If a definite integral is required (for some interval of x) some shortcuts would provide an approximation. To see what would be required, consider x=π/2, then sin(x)=1 and x100sin(x)=(π/2)100 (also applies when x=-π/2) which is a large number. When 0<x<1, x100 is positive but less than 1, mostly considerably less; when x=1, x100sin(x)=0.84 approx. It would be interesting to know the context of the exam question.
Integration delivers the area under a curve, and in this case, the curve has outstanding characteristics, leaping from close to zero to a very large number in a short interval. While sin(x) varies between -1 and 1, x100 is always positive and is either very small or very large, so the gradient shifts from almost horizontal to almost vertical in the same interval. This behaviour provides a clue to evaluation of the definite integral, including "tricks" to perform this evaluation.