- Integration by Parts

To understand Integration by Parts we need to recap the Product Rule from differentiation.

\(\displaystyle {d\over dx} (uv)=u {dv\over dx}+v {du\over dx}\)
\(\displaystyle u {dv\over dx}={d\over dx} (uv)-v {du\over dx}\:\:\:\:\: (1)\)
This is the basis for Integration by Parts. Integrating both sides of (1) we get:
\(\displaystyle \int u {dv\over dx} dx=uv-\int v {du\over dx} dx\)
\(\int u \:dv=uv-\int v \:du\)
When choosing u we use the following acronym.
L – Logarithms
I – Inverses
A – Algebra
T – Trigonometric Ratios
E – Exponentials

LESSON 1
Determine
\[\int xe^x \:dx\]
SOLUTION
Since Algebra precedes Exponentials in the acronym LIATE we choose \(x\) as \(u\). Therefore, \(dv=e^x\).
\(u=x\), \(dv=e^x\)
Therefore
\(du=1\). \(v=e^x\)
Integrate by parts,
\(\int u \:dv=uv-\int v\:du\)
\(\int xe^x\:dx=xe^x-\int e^x \:dx\)
\(=xe^x-e^x+c\)

LESSON 2
Determine
\[\int \ln ⁡x \:dx\]
SOLUTION
  • Rewrite \(\int \ln ⁡x\:dx\) as \(\int 1(\ln⁡ x)\:dx\) before integrating by parts.
\(\int 1.\ln ⁡x\:dx\)
Logarithms precedes Algebra in LIATE.
\(\displaystyle u=\ln⁡x → du={1\over x}\)
\(dv=1 → v=x\)
  • Integrate by parts
\(\int u \:dv=uv-\int v\:du\)
\(\int \ln ⁡x\:dx=x \ln⁡x-\int 1\:dx\)
\(=x \ln ⁡x-x+c\)

LESSON 3
Evaluate
\[\displaystyle \int {3x+1\over \sqrt{1-2x}}\:dx\]
SOLUTION
  • Rewrite integral
\(\displaystyle \int {3x+1\over \sqrt{1-2x}}\:dx=\int {(3x+1)(1-2x)^{-{1\over2}}}\:dx\)
  • Since both parts are Algebra we choose the easier part to differentiate as \(u\).
\(u=3x+1 → du=3\)
\(\displaystyle {dv\over dx}=(1-2x)^{-{1\over2}} → v=-(1-2x)^{{1\over2}}\)
  • Integrate by parts
\(\int u \:dv=uv-\int v\:du\)
\(\displaystyle \int {3x+1\over \sqrt{1-2x}}\:dx=-(3x+1) (1-2x)^{{1\over2}}-\int -3(1-2x)^{{1\over2}}\:dx\)
\(\displaystyle =-(3x+1) (1-2x)^{{1\over2}}-(1-2x)^{{3\over2}}\)
\(\displaystyle =(1-2x)^{{1\over2}} (-3x-1-1+2x)\)
\(\displaystyle =-(1-2x)^{{1\over2}} (x+2)+c\)

LESSON 4
Determine
\[\int x^2 \sin ⁡x \:dx\]
SOLUTION
\(\int x^2 \sin ⁡x\:dx\)
Let
\(u=x^2 → du=2x\)
\(dv=\sin ⁡x → v=-\cos ⁡x\)
  • Use integration by parts
\(\int u \:dv=uv-\int v \: du\)
\(\int x^2 \sin ⁡x\:dx=-x^2 \cos ⁡x-\int 2x(-\cos ⁡x) \:dx\)
\(=-x^2 \cos ⁡x+2\int x \cos ⁡x\:dx\:\:\:\:\: (1)\)
  • At this stage we realise that we need to use Integration by Parts again to determine \(\int x \cos ⁡x\: dx\).
\(\int x \cos ⁡x\:dx\)
Let
\(u=x → du=1\)
\(dv=\cos⁡x → v=\sin ⁡x\)
  • Use Integration by Parts
\(\int u \:dv=uv-\int v \:du\)
\(\int x \cos ⁡x\:dx=x \sin ⁡x-\int \sin ⁡x\:dx\)
\(\int x \cos ⁡x\:dx=x \sin ⁡x+\cos⁡ x\:\:\:\:\: (2)\)
  • We now substitute (2) into (1)
\(\int x^2 \cos ⁡x\:dx=-x^2 \cos ⁡x+2[x \sin ⁡x+\cos ⁡x]+c\)
\(=-x^2 \cos ⁡x+2x \sin ⁡x+2 \cos ⁡x+c\)

INTEGRATION BY PARTS WITH LIMITS
LESSON 1
Determine
\[\int_1^4 x \ln ⁡x\:dx\]
SOLUTION
  • We begin in the same manner as if we determining an indefinite integral using Integration by Parts.
\(\displaystyle \int_1^4 x \ln ⁡x\:dx\)
Let
\(\displaystyle u=\ln ⁡x → du={1\over x}\)
\(\displaystyle dv=x → v={x^2\over2}\)
  • Use the result
\[\int_a^b u\:dv=[uv]_a^b-\int_a^b v \:du\]
\(\displaystyle ⟹\int_1^4 x \ln ⁡x\:dx=\left[{x^2\over2} \ln ⁡x\right]_1^4-\int_1^4 \left({x^2\over2}\right)\left({1\over x}\right)\:dx\)
\(\displaystyle ⟹\int_1^4 x \ln ⁡x\:dx=\left[{x^2\over2}\ln ⁡x\right]_1^4-\int_1^4 {x\over2}\:dx\)
  • Evaluate both parts of the integral individually.
\(\displaystyle \int_1^4 x\ln ⁡x\:dx=\left[{4^2\over2}\ln ⁡4-{1^2\over2} \ln⁡1\right]-\left[{x^2\over4}\right]_1^4\)
\(\displaystyle \int_1^4 x \ln ⁡x\:dx=\left[{4^2\over2}\ln⁡4-{1^2\over2}\ln⁡1\right]-\left[{4^2\over4}-{1^2\over4}\right]\)
\(\displaystyle \int_1^4 x \ln ⁡x\:dx=8 \ln⁡4-{15\over4}\)
It is best to state the answer in its EXACT form.

LESSON 2
Evaluate
\[\int_0^π (x-π)^2 \sin ⁡x\:dx\]
SOLUTION
\(\displaystyle \int_0^π (x-π)^2 \sin ⁡x\:dx\)
Let
\(u=(x-π)^2 → du=2(x-π)\)
\(\displaystyle {dv\over dx}=\sin ⁡x → v=-\cos ⁡x\)
  • Use
\[\int_a^b u\:dv=[uv]_a^b-\int_a^b v\:du\]
\(\displaystyle \int_0^π (x-π)^2 \sin ⁡x\:dx=[-(x-π)^2 \cos ⁡x]_0^\pi-\int_0^π -2(x-π) \cos⁡ x\:dx\)
  • Evaluating \([-(x-π)^2 \cos ⁡x]_0^\pi\) now gives us
\(\displaystyle \int_0^π (x-π)^2 \sin ⁡x\:dx=π^2+2\int_0^π (x-π) \cos ⁡x\:dx\)
  • Determine \(\int_0^π (x-π) \cos ⁡x\:dx\) using Integration by Parts.
\(u=(x-π) → du=1\)
\(\displaystyle {dv\over dx}=\cos ⁡x → v=\sin ⁡x\)
  • Use
\[\int_a^b u\:dv=[uv]_a^b-\int_a^b v\:du\]
\(\displaystyle \int_0^π (x-π) \cos ⁡x \:dx=[(x-π) \sin ⁡x]_0^\pi-\int_0^π \sin ⁡x\:dx\)
  • Evaluating \([(x-π)\sin ⁡x]_0^\pi\) now gives us
\(\displaystyle \int (x-π) \cos ⁡x\:dx=-\int_0^π \sin ⁡x\:dx\)
\(=[\cos⁡x]_0^\pi\)
\(=-2\)
  • Resubstitute to get
\(\displaystyle \int_0^π (x-π)^2 \sin ⁡x\:dx=π^2+2(-2)\)
\(=π^2-4\)

INTEGRATION BY PARTS USING AN EQUATION
LESSON 1
Evaluate
\[\int e^{2x}\sin ⁡x\:dx\]
SOLUTION
\(\int e^{2x}\sin ⁡x\:dx\)
\(u=\sin ⁡x \to du=\cos x\) dv=e^2x
\(\displaystyle dv=\cos ⁡x \to v={1\over2}e^{2x}\)
\(\displaystyle \int e^{2x}\sin ⁡x\:dx={1\over2}e^{2x}\sin ⁡x-{1\over2}\int e^{2x}\cos ⁡x\:dx\)
  • Integrate \(\int e^{2x}\cos ⁡x\:dx\) by parts
\(u=\cos ⁡x → du=-\sin ⁡x\)
\(\displaystyle dv=e^{2x} → v={1\over2}e^{2x}\)
\(\displaystyle \int e^{2x}\cos ⁡x\:dx={1\over2}e^{2x}+{1\over2}\int e^{2x}\sin ⁡x\:dx\)
  • Resubstitute
\(\displaystyle \int e^{2x}\sin ⁡x\:dx={1\over2}e^{2x}\sin ⁡x-{1\over2} \left[{1\over2} e^{2x}\cos ⁡x+{1\over2}\int e^{2x}\sin ⁡x\:dx\right]\)
\(\displaystyle \int e^{2x}\sin ⁡x\:dx={1\over2}e^{2x}\sin ⁡x-{1\over4} e^{2x}\cos ⁡x-{1\over4} \int e^{2x}\sin ⁡x\:dx\)
We have the same integral on both sides so at this stage we can form an equation which we can solve for \(\int e^{2x}\sin ⁡x\:dx\).
\(\displaystyle {5\over4}\int e^{2x}\sin ⁡x\:dx={1\over2}e^{2x}\sin ⁡x-{1\over4}e^{2x}\cos ⁡x\)
\(\displaystyle \int e^{2x}\sin ⁡x\:dx={2\over5} e^{2x}\sin ⁡x-{1\over5} e^{2x}\cos ⁡x+c\)