Q.205. Indefinite and definite integration.

Question 205.
1) Evaluate ∫ x ln5x dx.
2) Evaluate ∫ x^2 sin(5x+2) dx  from x=0 to x=pi.

Answer 205.
1)
To find the integration of x ln5x, integration of parts is to be used.
According to the formula of integration by parts,
∫ uv dx = u ∫ vdx - ∫ [du/dx ∫ vdx] dx, where u and v are functions of x

Taking ln5x = u and x = v,
∫ x ln5x dx
= ln5x ∫ x dx - ∫ [d/dx(ln5x) ∫ xdx] dx
= (x^2/2) ln5x - ∫ [1/5x * 5 * x^2/2] dx
= (x^2/2) ln5x - (1/2) ∫ x dx
= (x^2/2) ln5x - x^2/4 + c.

2)
Let I = ∫x^2 sin(5x+2) dx
Using integration by parts,
I = x^2 ∫ sin(5x+2) dx - ∫ [d/dx(x^2) ∫ sin(5x+2)dx] dx
= - (1/5) x^2 cos(5x+2) + (2/5) ∫ xcos(5x+2) dx

Using integration by parts again for the second term,
I = - (1/5) x^2 cos(5x+2) + (2/5) [x∫cos(5x+2)dx - ∫d/dx(x)∫cos(5x+2)dx] dx
= - (1/5) x^2 cos(5x+2) + (2/25) xsin(5x+2) - (2/25) ∫sin(5x+2) dx
= - (1/5) x^2 cos(5x+2) + (2/25) xsin(5x+2) + (2/125)cos(5x+2) + c

Plugging the limits x = 0 to x = π
Value of indefinite integral I
= - (1/5) π^2 cos(5π+2) + (2/25)πsin(5π+2) + (2/125)cos(5π+2) - (2/125)cos(2)
= (1/5) π^2 cos(2) - (2/25)πsin(2) - (2/125)cos(2) - (2/125)cos(2)
= (1/5) π^2 cos(2) - (2/25)πsin(2) - (4/125)cos(2).

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