Integrate xcos(5x)

so i set
g'(x) = x
f(x) = cos(5x)

thus, i get
(x^(2)cos(5x)/2) - integral of 5xcos(5x)
so i use substitution
u = 5x
du/5 = dx

therefore,

(x^(2)cos(5x)/2) + (1/5)integral of usin(u)
giving me

(x^(2)cos(5x)/2) + (5x^(2)cos(5x)/10) + c

what am i doing wrong the text book and wolfram alpha claims i have the wrong answer

Match the following terms with their definitions. 1. bisector of a segment ray ba and ray bc are opposite rays if a, b, and c are collinear and b (the endpoint of both rays) is between a and c. 2. opposite rays a ray, , is the set of points beginning at point a and going infinitely in the direction of point b. 3. collinear points a line or segment that intersects the segment at its midpoint. 4. betweenness of points a set of two or more points all on the same line. 5. ray the distance between the endpoints of a segment. 6. space point b is between a and c if a, b, and c are collinear and the equation ab + bc = ac is true, where ab, bc, and ac are the distances between points a and b, b and c, and a and c, respectively. 7. midpoint of a segment a set of two or more points all on the same plane. 8. coplanar points the point on a segment that divides the segment into two equal segments. 9. length of a segment the set of all possible points. 10. line segment the set of two different endpoints and all points between them.

How can you decide witch operations to use to slice a two step equation

Question 7 of 47 1 point what is the equation of a line with a slope of 1/2 that passes through the point, (-2,5)? o a. 1/2x+3 o b. y=1/2x+5 o c. y=x+5 o d. y=1/2x+6