Ch. 7 – Differential Equations
and Mathematical Modeling
7.3 Antidifferentiation by Parts
• Another strategy you might use to integrate is integration by
parts.


– Use it when substitution won’t work!
u dv  uv  v du
– It’s like Product Rule for integrals!
– When used properly, the 2nd integral is much easier than the 1st
• Ex: Evaluate  x cos x dx .
– Substitution won’t work here because a function and its derivative aren’t
both in this integral.
– Break up the x and cosx into the u and dv parts (make sure the dx goes
with the dv part)…
dv  cos x dx
ux
du  dx
v  sin x
– …then find du and v…
– Lastly, substitute the variable expressions back in and evaluate!
uv   v du  x sin x   sin x dx  x sin x  cos x  C
• Ex: Evaluate  2 xe x dx .
– We can’t use substitution because we won’t get an integral with just u’s.
– Try integration by parts…
dv  e x dx
u  2x
– …then find du and v…
x
du  2dx
ve
– Lastly, substitute the variable expressions back in and evaluate!
uv   v du  2 xe x   2e x dx  2 xe x  2e x  C
– When integrating by parts, your u term should be the one that eventually
differentiates to a constant (like a polynomial function)
• Sometimes you may have to integrate by parts twice in one
problem!
2
x
• Ex: Evaluate sin x dx .
dv  sin x dx
u  x2
– Try integration by parts…
v   cos x
du  2 x dx
– …then find du and v…
– Substitute the variable expressions back in and evaluate!
uv   v du   x 2 cos x   2 x cos x dx
– But now we need to integrate the new integral by parts…
u  2x
du  2 dx
dv  cos x dx
v  sin x
uv   v du  2 x sin x   2sin x dx  2x sin x  2cos x
– Now combine your answers for the final solution…
 x 2 cos x   2 x cos x dx   x 2 cos x  2 x sin x  2 cos x  C