Integrating Using Integration by Parts

AP Calculus BC · Integration and Accumulation of Change

Integrating Using Integration by Parts

The product rule backwards. It trades one integral for another, so choosing $u$ is the whole skill.

key idea$\int u\,dv=uv-\int v\,du$

The procedure

Integration by parts is the product rule run backwards. It trades one integral for another, so it only helps when the new integral is easier — which makes the choice of $u$ the whole game.

Split the integrand into $u$ and $dv$

Everything must be assigned: $u$ is one factor, $dv$ is the rest including $dx$. Choose $u$ to be the factor that gets simpler when differentiated.

in the integrand out $u$ and $dv$

Differentiate $u$, integrate $dv$

$du=u'\,dx$ and $v=\displaystyle\int dv$. No $+C$ is needed for $v$ — any antiderivative works.

out $du$ and $v$

Assemble $uv-\int v\,du$

Write the formula with your four pieces in place. The minus sign applies to the whole remaining integral.

out a new integral

Evaluate the new integral

If it is easier, finish. If it is harder, your choice of $u$ was wrong — go back and swap. If it is the same difficulty, parts may need applying again.

out the answer, or a second pass

Priority	Type	Example	Why it makes a good $u$
L	Logarithmic	$\ln x$	differentiates to the much simpler $\tfrac1x$
I	Inverse trig	$\arctan x$	differentiates to an algebraic function
A	Algebraic	$x^{2}$	each differentiation lowers the degree
T	Trigonometric	$\sin x$	neither simplifies nor worsens
E	Exponential	$e^{x}$	never simplifies — usually best as $dv$

LIATE: whichever type appears higher in this list becomes $u$. It is a heuristic, not a theorem, but it succeeds on nearly every AP integral.

$\ln$ PRESENT

the integrand contains $\ln x$let $u=\ln x$ — it has no elementary antiderivative, so it cannot be $dv$

INVERSE TRIG

$\arcsin x$, $\arctan x$ appearslet $u$ be that factor, for the same reason

POLYNOMIAL $\times$ OTHER

$x^{n}e^{x}$, $x^{n}\sin x$let $u=x^{n}$ — differentiating lowers the degree, so repeating parts terminates

EXPONENTIAL $\times$ TRIG

$e^{x}\sin x$, $e^{x}\cos x$either choice works; parts twice returns the original integral, so solve for it algebraically

choosing u

Choose $u$ so that differentiating it simplifies things.
Choose $dv$ so that integrating it is easy.
A polynomial of degree $n$ against $\sin$, $\cos$ or $e^{x}$ needs $n$ passes.
If the original integral reappears, solve for it algebraically.

Worked examples

EX 1

Algebraic times exponential

$\int xe^{x}\,dx$

Evaluate $\displaystyle\int xe^{x}\,dx$.

Apply LIATE

Algebraic ($x$) outranks Exponential ($e^{x}$), so $u=x$ and $dv=e^{x}dx$.

Compute $du$ and $v$

$du=dx$ and $v=e^{x}$.

Assemble

$xe^{x}-\displaystyle\int e^{x}\,dx$.

Finish

$xe^{x}-e^{x}+C=e^{x}(x-1)+C$.

Check the choice

The new integral $\int e^{x}dx$ is trivial. Had we chosen $u=e^{x}$ we would have got $\int\tfrac{x^{2}}{2}e^{x}dx$ — worse.

EX 2

When $\ln$ is present

$\int\ln x\,dx$

Evaluate $\displaystyle\int\ln x\,dx$ — a single factor.

Manufacture a second factor

Write the integrand as $\ln x\cdot1$. LIATE puts Logarithmic first, so $u=\ln x$ and $dv=1\,dx$.

Compute $du$ and $v$

$du=\dfrac{1}{x}dx$ and $v=x$.

Assemble

$x\ln x-\displaystyle\int x\cdot\dfrac{1}{x}\,dx$.

Simplify the new integral

$\displaystyle\int 1\,dx=x$.

Finish

$x\ln x-x+C$. The trick is recognising that $dv=dx$ is a legitimate choice.

EX 3

Applying parts twice

$\int x^{2}\sin x\,dx$

Evaluate $\displaystyle\int x^{2}\sin x\,dx$.

First pass

$u=x^{2}$, $dv=\sin x\,dx$, so $du=2x\,dx$ and $v=-\cos x$.

Assemble

$-x^{2}\cos x+\displaystyle\int 2x\cos x\,dx$ — the degree dropped from $2$ to $1$.

Second pass on the new integral

$u=2x$, $dv=\cos x\,dx$, giving $du=2\,dx$ and $v=\sin x$.

Assemble again

$2x\sin x-\displaystyle\int 2\sin x\,dx=2x\sin x+2\cos x$.

Combine

$-x^{2}\cos x+2x\sin x+2\cos x+C$. Each pass lowers the polynomial degree by one, so $x^{n}$ needs $n$ passes.

EX 4

The circular case

$\int e^{x}\sin x\,dx$

Evaluate $\displaystyle\int e^{x}\sin x\,dx$.

First pass

$u=\sin x$, $dv=e^{x}dx$: $e^{x}\sin x-\displaystyle\int e^{x}\cos x\,dx$.

Second pass, keeping the same choice type

$u=\cos x$, $dv=e^{x}dx$: $\displaystyle\int e^{x}\cos x\,dx=e^{x}\cos x+\int e^{x}\sin x\,dx$.

Substitute back

Let $I=\displaystyle\int e^{x}\sin x\,dx$. Then $I=e^{x}\sin x-e^{x}\cos x-I$.

Solve algebraically

$2I=e^{x}\left(\sin x-\cos x\right)$, so $I=\dfrac{e^{x}\left(\sin x-\cos x\right)}{2}+C$.

The key move

The original integral reappeared. Treat it as an unknown and solve — but you must keep the same $u$-type on both passes, or everything cancels to $0=0$.

Where it goes wrong

Swapping $u$ and $dv$ makes it worse

If the new integral is harder than the original, the choice was backwards. Go back and swap — do not push on.

$dv$ includes the $dx$

Every part of the integrand must be assigned to $u$ or $dv$. Leaving the $dx$ out gives a $v$ with the wrong dimensions.

The minus applies to the whole integral

$uv-\int v\,du$, not $uv-\int v\,du$ with the sign attached to only the first term of the new integrand.

Keep the choice consistent when cycling

In Example 4, switching which factor is $u$ on the second pass returns $I=I$ and proves nothing.

Course map. Topic 6.11 · LO FUN-6.E — calculate integrals using integration by parts (EK FUN-6.E.1). BC-only.