Section 7.1: Integration By Parts

Integration techniques

Antidifferentiation is the reverse of differentiation. Anything we can learn about antidifferentation must come from differentiation.

Power rule: (d/(d x))((x^n))=n (x^(n-1)) ⟶ ∫(x^k)d x=(1/(k+1))(x^(k+1))+C if k≠-1.

Chain rule: (d/(d x))(F(u(x)))=F'(u(x))u'(x) ⟶ ∫f(u(x))u'(x)d x=∫f(u)d u

Questions

What is ∫(1/(1+(x^2)))d x?

Integrate[1/(1+x^2),x]

What is ∫sin(2x)(e^(cos(2x)))d x?

Integrate[Sin[2x]E^Cos[2x],x]

Questions

What does the product rule say about ((u(x)v(x)))'.

Antidifferentiate what you obtained.

Integration by parts

If you can identify u(x) and d v(x) in an integral, try rewriting the integral as
	∫u d v=u v-∫v d u

Example

To evaluate ∫x (e^x)d x let’s try integration by parts. Identify u and d v.
	u=x
d v=(e^x)d x
This leads to
	d u=d x
v=∫d v=∫(e^x)d x=((e )^x)
Only one antiderivative is needed; any value of C at this point in “+C” works. Applying the integration by parts formula ∫u d v=u v-∫v d u produces
	∫x (e^x)d x=x (e^x)-∫(e^x)d x
The new integral is one of our basic integrals. The final answer is
	∫x (e^x)d x=x (e^x)-(e^x)+C

Questions

Consider the integral ∫x cos(2x)d x.

What is ∫x cos(2x)d x if you use integration by parts with u=x and d v=cos(2x)d x?

What would happen if you try to use integration by parts with u=cos(2x) and d v=x d x?

Question

How can you use integration by parts to evaluate ∫(x^2)sin(3x)d x?

Integrate[x^2Sin[3x],x]

Question

How can you use integration by parts to evaluate ∫(tan^(-1))(x)d x?

Integrate[ArcTan[x],x]