Integration by parts formula, choosing u and dv, repeated application.
Answer at least 3 of 3 correctly to complete this section.
Substitution works when you have a composite function (a function inside a function). But what about a product of two unrelated functions, like xsinx or x2ex? These do not fit the substitution pattern.
Integration by parts is the reverse of the product rule. It is the go-to method when you need to integrate a product of two functions where one becomes simpler when differentiated.
This technique is essential at A-Level — you will meet it in pure maths, differential equations, and even some mechanics problems. It is also the only way to integrate lnx.
The product rule for differentiation says:
dxd(uv)=udxdv+vdxdu
Rearranging:
udxdv=dxd(uv)−vdxdu
Integrating both sides with respect to x:
∫udxdvdx=uv−∫vdxdudx
This is the integration by parts formula. It replaces one integral with another — the aim is to make the new integral simpler than the original.
The key decision is: which part of the product do you call u, and which do you call dxdv?
Choose u as the function that appears earlier in this list:
| Priority | Type | Example |
|---|---|---|
| 1st | Logarithms | lnx, logx |
| 2nd | Inverse trig | arctanx, arcsinx |
| 3rd | Algebraic (polynomials) | x, x2, 3x+1 |
| 4th | Trigonometric | sinx, cosx |
| 5th | Exponential | ex, 2x |
The idea: u should be the part that simplifies when differentiated. Polynomials eventually differentiate to zero; lnx differentiates to x1 (simpler). Exponentials and trig functions do not simplify on differentiation, so they are better as dxdv.
Find ∫xcosxdx.
Using LIATE: x is algebraic (A), cosx is trigonometric (T). Since A comes before T, let u=x.
| u | dxdv | |
|---|---|---|
| Choose: | x | cosx |
| Then: | dxdu=1 | v=sinx |
Apply the formula:
∫xcosxdx=xsinx−∫sinx⋅1dx=xsinx−(−cosx)+c
=xsinx+cosx+c
Check by differentiating: dxd(xsinx+cosx)=sinx+xcosx−sinx=xcosx ✓
Find ∫lnxdx.
This looks like it has only one function, but write it as a product: ∫lnx⋅1dx.
Using LIATE: lnx is a logarithm (L), so let u=lnx and dxdv=1.
| u | dxdv | |
|---|---|---|
| Choose: | lnx | 1 |
| Then: | dxdu=x1 | v=x |
∫lnxdx=xlnx−∫x⋅x1dx=xlnx−∫1dx=xlnx−x+c
Key result: ∫lnxdx=xlnx−x+c. This is worth memorising — it appears frequently.
Evaluate ∫0πxsinxdx.
Let u=x, dxdv=sinx. Then dxdu=1, v=−cosx.
∫0πxsinxdx=[−xcosx]0π−∫0π(−cosx)dx
=[−xcosx]0π+∫0πcosxdx
=(−πcosπ−0)+[sinx]0π
=(−π(−1))+(sinπ−sin0)=π+0=π
Answer at least 3 of 3 correctly to complete this section.
If the integrand contains xn multiplied by ex or a trig function, you may need to apply integration by parts n times — once for each power of x.
Find ∫x2exdx.
First application: Let u=x2, dxdv=ex.
∫x2exdx=x2ex−∫2xexdx
We still have xex to integrate — apply parts again.
Second application: Let u=2x, dxdv=ex.
∫2xexdx=2xex−∫2exdx=2xex−2ex+c
Putting it all together:
∫x2exdx=x2ex−(2xex−2ex)+c=x2ex−2xex+2ex+c
=ex(x2−2x+2)+c
Watch the signs carefully. When you substitute the second integral back into the first, you are subtracting the entire expression. Use brackets to avoid sign errors.
For ∫xneaxdx or ∫xnsin(ax)dx, you can set up a table:
For ∫x2exdx:
| Differentiate u | Integrate dxdv | Sign |
|---|---|---|
| x2 | ex | + |
| 2x | ex | − |
| 2 | ex | + |
| 0 | ex | — |
Read off diagonally with alternating signs:
=(+)x2ex+(−)2xex+(+)2ex+c=ex(x2−2x+2)+c
This gives the same answer much faster once you are comfortable with it.
Find ∫x2sinxdx.
First round: u=x2, dxdv=sinx, so v=−cosx.
∫x2sinxdx=−x2cosx−∫(−cosx)(2x)dx=−x2cosx+2∫xcosxdx
Second round: u=x, dxdv=cosx, so v=sinx.
∫xcosxdx=xsinx−∫sinxdx=xsinx+cosx+c
Substituting back:
∫x2sinxdx=−x2cosx+2(xsinx+cosx)+c
=−x2cosx+2xsinx+2cosx+c
Answer at least 3 of 3 correctly to complete this section.
Some integrals, like ∫exsinxdx, do not simplify no matter which you choose as u. Instead, after two applications of integration by parts, the original integral reappears. The trick is to call the original integral I, and solve the resulting equation.
Find ∫exsinxdx.
Let I=∫exsinxdx.
First application: Let u=ex, dxdv=sinx.
I=−excosx−∫(−cosx)(ex)dx=−excosx+∫excosxdx
Second application: On ∫excosxdx, let u=ex, dxdv=cosx.
∫excosxdx=exsinx−∫exsinxdx=exsinx−I
Substituting back:
I=−excosx+exsinx−I
Now solve for I:
2I=ex(sinx−cosx)
I=2ex(sinx−cosx)+c
Critical point: You must apply parts the same way both times. If you use u=ex the first time, use u=ex the second time too. If you switch (using u=sinx the second time), you will undo the first step and get 0=0.
Find ∫excosxdx.
Let I=∫excosxdx.
First: u=ex, dxdv=cosx, so v=sinx.
I=exsinx−∫exsinxdx
Second: On ∫exsinxdx, let u=ex, dxdv=sinx, so v=−cosx.
∫exsinxdx=−excosx+∫excosxdx=−excosx+I
Substituting back:
I=exsinx−(−excosx+I)=exsinx+excosx−I
2I=ex(sinx+cosx)
I=2ex(sinx+cosx)+c
Notice the same cycling pattern: the original integral I reappears after two rounds, and you solve the resulting equation.
This uses the same cycling technique but with messier coefficients.
Let I=∫e2xcos3xdx.
First: u=e2x, dxdv=cos3x, so v=3sin3x.
I=3e2xsin3x−32∫e2xsin3xdx
Second: u=e2x, dxdv=sin3x, so v=−3cos3x.
∫e2xsin3xdx=−3e2xcos3x+32I
Substituting:
I=3e2xsin3x−32(−3e2xcos3x+32I)
913I=93e2xsin3x+2e2xcos3x
I=13e2x(3sin3x+2cos3x)+c
Answer at least 3 of 3 correctly to complete this section.
Find ∫x2lnxdx.
By LIATE: lnx is L, x2 is A. So u=lnx, dxdv=x2.
| u | dxdv | |
|---|---|---|
| Choose: | lnx | x2 |
| Then: | dxdu=x1 | v=3x3 |
∫x2lnxdx=3x3lnx−∫3x3⋅x1dx=3x3lnx−31∫x2dx
=3x3lnx−9x3+c=9x3(3lnx−1)+c
Evaluate ∫1elnxdx.
Using the result ∫lnxdx=xlnx−x:
∫1elnxdx=[xlnx−x]1e=(elne−e)−(1⋅ln1−1)
=(e−e)−(0−1)=0+1=1
| Integral | Method |
|---|---|
| ∫xneaxdx | Parts n times (u=xn) |
| ∫xnsin(ax)dx | Parts n times (u=xn) |
| ∫xnlnxdx | Parts once (u=lnx) |
| ∫eaxsin(bx)dx | Parts twice, then solve for I |
| ∫lnxdx | Parts with dxdv=1 |
Exam tip: When the question says “use integration by parts”, follow it. But when it does not specify a method, look for the patterns above. If the integrand is a product where one factor simplifies on differentiation, try parts.
Answer at least 3 of 4 correctly to complete this section.
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