- 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 $$\int e^{2x}\sin ⁡x\:dx$$.
$$\displaystyle {5\over4}\int e^{2x}\sin ⁡x\:dx={1\over2}e^{2x}\sin ⁡x-{1\over4}e^{2x}\cos ⁡x$$
$$\displaystyle \int e^{2x}\sin ⁡x\:dx={2\over5} e^{2x}\sin ⁡x-{1\over5} e^{2x}\cos ⁡x+c$$