Evaluate the integral using the Integration by Parts formula with u = x, v′ = cos(5x).?

The easiest way for me to remember integration by parts is

(Integral) udv = uv - (integral)vdu

u = x therefore du = dx

dv = cos(5x) therefore v = (1/5)sin(5x)

(integral)xcos(5x) = (1/5)xsin(5x) - (integral)(1/5)sin(5x)dx

= (1/5)xsin(5x) +(1/25)cos(5x) + C

ʃx*cos(5x) dx

u = x

du = dx

dv = cos(5x) dx

v = 1/5*sin(5x)

uv - ʃv du

x*sin(5x)/5 + 1/25*cos(5x) + C

in many circumstances to combine with the aid of areas some ?u dv, we do ?u dv = uv - ?v du to %. an appropriate u and dv, we tend to following the handbook of ILATE: Inverse trig Logarithmic Algebraic Trigonometric Exponential. this provides the tough priority of purposes to assign u to be. in many circumstances, you pick for u to be something you are able to definitely differentiate to get du, and additionally you pick for dv to be something you are able to definitely combine to v. So, ?(e^-x)*cos(2x) dx permit u = e^-x and dv = cos(2x), following ILATE. hence, differentiate u and additionally you get du = -e^-x dx and combine dv to get v = ½sin(2x). Now, ?(e^-x)*cos(2x) dx = uv - ?v du ?(e^-x)*cos(2x) dx = ½sin(2x)*e^(-x) - ?½sin(2x)*[-e^(-x)] dx convey the unfavourable and the ½ each and every of ways out of the required to get ?(e^-x)*cos(2x) dx = ½sin(2x)*e^(-x) + ½?sin(2x)*e^(-x) dx Now, combine with the aid of areas back! permit u = e^-x and dv = sin(2x). So du = -e^-x and v = -½cos(2x). So, ?(e^-x)*cos(2x) dx = ½sin(2x)*e^(-x) + ½?[uv - ?v du] ?(e^-x)*cos(2x) dx = ½sin(2x)*e^(-x) + ½[ -½cos(2x)*e^(-x) - ? -½cos(2x)* -e^-x dx] ?(e^-x)*cos(2x) dx = ½sin(2x)*e^(-x) + ½[ -½cos(2x)*e^(-x) - ½?cos(2x)*e^-x dx] Distribute the ½ and additionally you get ?(e^-x)*cos(2x) dx = ½sin(2x)*e^(-x) - ¼cos(2x)*e^(-x) - ¼?cos(2x)*e^-x dx it would seem as in case you continuously combine, yet all you ought to do is upload ¼?cos(2x)*e^-x dx to the two aspects so as which you get (5/4)?(e^-x)*cos(2x) dx = ½sin(2x)*e^(-x) - ¼cos(2x)*e^(-x) And multiply each and everything with the aid of four/5 to get ?(e^-x)*cos(2x) dx = (4/5)[½sin(2x)*e^(-x) - ¼cos(2x)*e^(-x)] ?(e^-x)*cos(2x) dx = (a million/5)[2sin(2x)*e^(-x) - cos(2x)*e^(-x)] + C wish that enables!